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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
Yesterday at 11:16 PM
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

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[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
Yesterday at 11:16 PM
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
Estimate on number of progressions
Assassino9931   0
20 minutes ago
Source: RMM Shortlist 2024 C4
Let $n$ be a positive integer. For a set $S$ of $n$ real numbers, let $f(S)$ denote the number of increasing arithmetic progressions of length at least two all of whose terms are in $S$. Prove that, if $S$ is a set of $n$ real numbers, then
\[ f(S) \leq \frac{n^2}{4} + f(\{1,2,\ldots,n\})\]
0 replies
1 viewing
Assassino9931
20 minutes ago
0 replies
Popular children at camp with algebra and geometry
Assassino9931   0
23 minutes ago
Source: RMM Shortlist 2024 C3
Fix an odd integer $n\geq 3$. At a maths camp, there are $n^2$ children, each of whom selects
either algebra or geometry as their favourite topic. At lunch, they sit at $n$ tables, with $n$ children
on each table, and start talking about mathematics. A child is said to be popular if their favourite
topic has a majority at their table. For dinner, the students again sit at $n$ tables, with $n$ children
on each table, such that no two children share a table at both lunch and dinner. Determine the
minimal number of young mathematicians who are popular at both mealtimes. (The minimum is across all sets of topic preferences and seating arrangements.)
0 replies
1 viewing
Assassino9931
23 minutes ago
0 replies
Triangles in dissections
Assassino9931   0
24 minutes ago
Source: RMM Shortlist 2024 C2
Fix an integer $n\geq 3$ and let $A_1A_2\ldots A_n$ be a convex polygon in the plane. Let $\mathcal{M}$ be the set of all midpoints $M_{i,j}$ of segments $A_iA_j$ where $i\neq j$. Assume that all of these midpoints are distinct, i.e. $\mathcal{M}$ consists of $\frac{n(n-1)}{2}$ elements. Dissect the polygon $M_{1,2}M_{2,3}\ldots M_{n,1}$ into triangles so that the following hold:

(1) The intersection of every two triangles (interior and boundary) is either empty or a common
vertex or a common side.
(2) The vertices of all triangles lie in M (not all points in M are necessarily used).
(3) Each side of every triangle is of the form $M_{i,j}M_{i,k}$ for some pairwise distinct indices $i,j,k$.

Prove that the total number of triangles in such a dissection is $3n-8$.
0 replies
Assassino9931
24 minutes ago
0 replies
IMO Shortlist Problems
ABCD1728   1
N 27 minutes ago by mrtheory
Source: IMO official website
Where can I get the official solution for ISL before 2005? The official website only has solutions after 2006. Thanks :)
1 reply
ABCD1728
Today at 12:44 PM
mrtheory
27 minutes ago
Process on scalar products and permutations
Assassino9931   0
29 minutes ago
Source: RMM Shortlist 2024 C1
Fix an integer $n\geq 2$. Consider $2n$ real numbers $a_1,\ldots,a_n$ and $b_1,\ldots, b_n$. Let $S$ be the set of all pairs $(x, y)$ of real numbers for which $M_i = a_ix + b_iy$, $i=1,2,\ldots,n$ are pairwise distinct. For every such pair sort the corresponding values $M_1, M_2, \ldots, M_n$ increasingly and let $M(i)$ be the $i$-th term in the list thus sorted. This denes an index permutation of $1,2,\ldots,n$. Let $N$ be the number of all such permutations, as the pairs run through all of $S$. In terms of $n$, determine the largest value $N$ may achieve over all possible choices of $a_1,\ldots,a_n,b_1,\ldots,b_n$.
0 replies
1 viewing
Assassino9931
29 minutes ago
0 replies
Tangency geo
Assassino9931   0
32 minutes ago
Source: RMM Shortlist 2024 G1
Let $ABC$ be an acute triangle with $\angle ABC > 45^{\circ}$ and $\angle ACB > 45^{\circ}$. Let $M$ be the midpoint of the side $BC$. The circumcircle of triangle $ABM$ intersects the side $AC$ again at $X\neq A$ and the circumcircle of triangle $ACM$ intersects the side $AB$ again at $Y\neq A$. The point $P$ lies on the perpendicular bisector of the segment $BC$ so that the points $P$ and $A$ lie on the same side of $XY$ and $\angle XPY = 90^{\circ} + \angle BAC$. Prove that the circumcircles of triangles $BPY$ and $CPX$ are tangent.
0 replies
+1 w
Assassino9931
32 minutes ago
0 replies
Inequalities in real math research
Assassino9931   0
36 minutes ago
Source: RMM Shortlist 2024 A3
For a positive integer $n$ denote $F_n(x_1,x_2,\ldots,x_n) = 1 + x_1 + x_1x_2 + \cdots +x_1x_2\ldots x_n$. For any real numbers $x_1\geq x_2 \geq \ldots \geq x_k \geq 0$ prove that
\[ \prod_{i=1}^k F_i(x_{k-i+1},x_{k-i+2},\ldots,x_k) \geq \prod_{i=1}^k F_i(x_i,x_i,\ldots,x_i)\]
0 replies
1 viewing
Assassino9931
36 minutes ago
0 replies
A folklore polynomial game
Assassino9931   0
39 minutes ago
Source: RMM Shortlist 2024 A1, also Bulgaria Regional Round 2016, Grade 12
Fix a positive integer $d$. Yael and Ziad play a game as follows, involving a monic polynomial of degree $2d$. With Yael going first, they take turns to choose a strictly positive real number as the value of one of the coecients of the polynomial. Once a coefficient is assigned a value, it cannot be chosen again later in the game. So the game
lasts for $2d$ rounds, until Ziad assigns the final coefficient. Yael wins if $P(x) = 0$ for some real
number $x$. Otherwise, Ziad wins. Decide who has the winning strategy.
0 replies
Assassino9931
39 minutes ago
0 replies
Two circles, a tangent line and a parallel
Valentin Vornicu   104
N 2 hours ago by cubres
Source: IMO 2000, Problem 1, IMO Shortlist 2000, G2
Two circles $ G_1$ and $ G_2$ intersect at two points $ M$ and $ N$. Let $ AB$ be the line tangent to these circles at $ A$ and $ B$, respectively, so that $ M$ lies closer to $ AB$ than $ N$. Let $ CD$ be the line parallel to $ AB$ and passing through the point $ M$, with $ C$ on $ G_1$ and $ D$ on $ G_2$. Lines $ AC$ and $ BD$ meet at $ E$; lines $ AN$ and $ CD$ meet at $ P$; lines $ BN$ and $ CD$ meet at $ Q$. Show that $ EP = EQ$.
104 replies
Valentin Vornicu
Oct 24, 2005
cubres
2 hours ago
Austrian Regional MO 2025 P4
BR1F1SZ   2
N 2 hours ago by NumberzAndStuff
Source: Austrian Regional MO
Let $z$ be a positive integer that is not divisible by $8$. Furthermore, let $n \geqslant 2$ be a positive integer. Prove that none of the numbers of the form $z^n + z + 1$ is a square number.

(Walther Janous)
2 replies
BR1F1SZ
Apr 18, 2025
NumberzAndStuff
2 hours ago
Austrian Regional MO 2025 P3
BR1F1SZ   1
N 2 hours ago by NumberzAndStuff
Source: Austrian Regional MO
There are $6$ different bus lines in a city, each stopping at exactly $5$ stations and running in both directions. Nevertheless, for every two different stations there is always a bus line connecting these two stations. Determine the maximum number of stations in this city.

(Karl Czakler)
1 reply
BR1F1SZ
Apr 18, 2025
NumberzAndStuff
2 hours ago
Austrian Regional MO 2025 P2
BR1F1SZ   2
N 2 hours ago by NumberzAndStuff
Source: Austrian Regional MO
Let $\triangle{ABC}$ be an isosceles triangle with $AC = BC$ and circumcircle $\omega$. The line through $B$ perpendicular to $BC$ is denoted by $\ell$. Furthermore, let $M$ be any point on $\ell$. The circle $\gamma$ with center $M$ and radius $BM$ intersects $AB$ once more at point $P$ and the circumcircle $\omega$ once more at point $Q$. Prove that the points $P,Q$ and $C$ lie on a straight line.

(Karl Czakler)
2 replies
BR1F1SZ
Apr 18, 2025
NumberzAndStuff
2 hours ago
Austrian Regional MO 2025 P1
BR1F1SZ   2
N 2 hours ago by NumberzAndStuff
Source: Austrian Regional MO
Let $n \geqslant 3$ be a positive integer. Furthermore, let $x_1, x_2,\ldots, x_n \in [0, 2]$ be real numbers subject to $x_1 + x_2 +\cdots + x_n = 5$. Prove the inequality$$x_1^2 + x_2^2 + \cdots + x_n^2 \leqslant 9.$$When does equality hold?

(Walther Janous)
2 replies
BR1F1SZ
Apr 18, 2025
NumberzAndStuff
2 hours ago
positive integers forming a perfect square
cielblue   0
3 hours ago
Find all positive integers $n$ such that $2^n-n^2+1$ is a perfect square.
0 replies
cielblue
3 hours ago
0 replies
Perfect Square Function
Miku3D   16
N Apr 18, 2025 by MathLuis
Source: 2021 APMO P5
Determine all Functions $f:\mathbb{Z} \to \mathbb{Z}$ such that $f(f(a)-b)+bf(2a)$ is a perfect square for all integers $a$ and $b$.
16 replies
Miku3D
Jun 9, 2021
MathLuis
Apr 18, 2025
Perfect Square Function
G H J
G H BBookmark kLocked kLocked NReply
Source: 2021 APMO P5
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Miku3D
57 posts
#1 • 3 Y
Y by centslordm, oVlad, megarnie
Determine all Functions $f:\mathbb{Z} \to \mathbb{Z}$ such that $f(f(a)-b)+bf(2a)$ is a perfect square for all integers $a$ and $b$.
This post has been edited 1 time. Last edited by Miku3D, Jun 9, 2021, 6:34 AM
Reason: Typo
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gghx
1072 posts
#2 • 4 Y
Y by centslordm, tck_darkness, ZHEKSHEN, luutrongphuc
Let $P(a,b): f(f(a)-b)+bf(2a)$ is a perfect square.
Lemma: $f(0)=0$, $f(a)$ is a perfect square

Let $f(0)=c$
$P(0,f(0)):c^2+c=k^2$ for some $k$ so $(2c+1)^2=(2k)^2+1$.
The only time two perfect squares can be consecutive is $0,1$ so $k=0$ and thus $c=0/-1$

If $f(0)=-1$,
$P(0,-1-b): f(b)+b+1$ is a perfect square, $P(a,0): f(f(a))$ is a perfect square.
Thus, $f(f(a))+f(a)+1$ is a perfect square.
If $f(a)$ is $1\pmod{4}$, then $f(f(a))+f(a)+1\equiv 2/3 \pmod{4}$, so it cannot be a perfect square.
Thus $f(a)$ is not $1 \pmod{4}$

Now $P(0,0): f(-1)$ is a perfect square hence it is $0\pmod{4}$.
But $P(-1,f(-1)): f(-1)f(-2)-1$ is a perfect square but it is $3$ mod 4, contradiction.

This proves that $f(0)=0$.
Now $P(0,-a): f(a)$ is a perfect square.
Case 1: $f$ has a positive zero.
Suppose $f(c)=0, c>0$.
Then $P(a,f(a)-c): (f(a)-c)f(2a)=0$.
Taking $a=c$, since $f(c)-c=-c<0$, $f(2c)=0$.
Similarly, $f(4c),f(8c),...$ are all $0$.

Now $P(a,f(a)-2^kc): (f(a)-2^kc)f(2a)=0$. Since $f(a)-2^kc$ is negative for large enough $k$, we must have $f(2a)=0$ for all $a$. This gives the solution $f(2x)=0$, $f(2x+1)=g(x)^2$ for any function $g:\mathbb{Z}\rightarrow \mathbb{Z}$.
Case 2: $f(a)>0$ for any $a>0$
$P(a,f(a)-2a): (f(a)-2a+1)f(2a)$ is a perfect square. Hence for positive $a$, $f(a)-2a+1$ is a perfect square.
Take $k=\frac{p+1}{2}$ where $p$ is prime, since $f(k)$ and $f(k)-p$ are both perfect squares, $f(k)=k^2$. So $f(a)=a^2$ for infinitely many $a$.

Also, this means $f(a)\le a^2$ for positive $a$, as bigger than that, the difference of squares would be more than $2a-1$ so $f(a)-2a+1$ won't be a perfect square.

Take any $x,y$ such that $f(x)=x^2,f(y)=y^2$ and $x,y$ positive.
$P(x,x^2-y): y^2+(x^2-y)f(2x)=y^2-yf(2x)+x^2f(2x)$ is a perfect square. Fix $x$ and increase $y$ towards infinity, we have that $4(y^2-yf(2x)+x^2f(2x))=(2y-f(2x))^2-f(2x)^2+4x^2f(2x)$ which means $f(2x)=4x^2$.

Thus, taking any integer $a$, $P(x,x^2-a): f(a)+(x^2-a)(4x^2)$ is a perfect square, so $4x^4-4ax^2+f(a)=(2x^2-a)^2+f(a)-a^2$ is a perfect square. Taking $x$ to infinity, we have $f(a)=a^2$ for all integers $a$, the second solution.

1/3 done during the test, 1/3 done at home, 1/3 my senior told me
This post has been edited 1 time. Last edited by gghx, Jun 9, 2021, 6:46 AM
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TheUltimate123
1740 posts
#3 • 2 Y
Y by centslordm, HamstPan38825
Say \(f\) is even-vanishing if it sends all evens to zero, and let \(Q\) be the set of all \(q\) such that \(|q|=1\) or \(|q|\) is prime.

The answer is \(f(x)\equiv x^2\) or \(f\) is an even-vanishing function that sends odds to perfect squares. All such functions clearly work, so we show they are the only solutions. Let \(P(a,b)\) denote the assertion.
  • From \(P(0,f(0))\), we know \(f(0)+f(0)^2\) is a perfect square. If \(f(0)\ne0\), then we can bound this between \(f(0)^2\) and \((f(0)+1)^2\), so \(f(0)=0\).
  • Applying \(P(0,-x)\), we find that \(f(x)\) is a perfect square for all \(x\).
  • By \(P(x,f(x)-2x)\),we have \(f(2x)(f(x)-2x+1)\) is always a perfect square. In particular, for all \(x\), either \(f(2x)=0\) or \(f(x)-2x+1\) is a square. Call this property \((\star)\).

Claim 1: For \(q\in Q\), either \(f(q+1)=0\) or \(f(\frac{q+1}2)=(\frac{q+1}2)^2\).

Proof. By taking \(x=\frac{q+1}2\) in \((\star)\), we know that \(f(q+1)=0\) or \(f(\frac{q+1}2)-q\) is a square. In the latter case, take \(f(\frac{q+1}2)=u^2\) and \(f(\frac{q+1}2)-q=v^2\) for \(u,v\ge0\), so \(q=u^2-v^2=(u-v)(u+v)\). It follows that \(u+v=|q|\), and the conclusion readily follows. \(\blacksquare\)

Claim 2: If \(n\equiv1\pmod4\), if \(f(n+1)=0\), then \(f\) is even-vanishing.

Proof. Take \(P(f(a),f(a)-n-1)\) for arbitrary \(a\). Then \(f(2a)(f(a)-n-1)\) is a square, so either \(f(2a)=0\) or \(f(a)-n-1\) is a square. In the latter case, \(n+1\equiv2\pmod4\) is the difference between two squares, absurd. \(\blacksquare\)

Claim 3: If \(f(n)=0\) for any \(n\ne0\), then \(f\) is even-vanishing.

Proof. Take \(P(f(a),f(a)-n)\) for arbitrary \(a\). Then \(f(2a)(f(a)-n)\) is a square, so either \(f(2a)=0\) or \(f(a)-n-1\) is a square.

Assume for contradiction \(f\) is not even-vanishing, so we may choose a large prime \(q\equiv1\pmod4\) with \(f(q+1)\ne0\) by Claim 2, i.e.\ \(f(\frac{q+1}2)=(\frac{q+1}2)^2\) by Claim 1. Select \(a=\frac{q+1}2\), so \((\frac{q+1}2)^2-n\) is a square. For \(q\gg n\) this is absurd. \(\blacksquare\)

Assume \(f\) is not even-vanishing, so the goal is to show \(f(x)\equiv x^2\).

Claim 4: \(f(\frac{q+1}2)=(\frac{q+1}2)^2\) for all \(q\in Q\).

Proof. From Claim 3 above, we know \(f(n)\ne0\) for \(n\ne0\), and thus the desired result follows from Claim 1. \(\blacksquare\)

It is now easy to generalize via induction; first, we verify the following:

Claim 5: If \(f(n)=n^2\) then \(f(2n)=4n^2\).

Proof. There are infinitely many \(k\) with \(f(k)=k^2\) by Claim 4; for each \(k\), consider \(P(n,f(n)-k)\). Then \(k^2-f(2n)k+n^2f(2n)\) is a perfect square. Since this is a perfect square for infinitely many \(k\), it must itself be a perfect square trinomial in \(k\), so \(f(2n)^2=4n^2f(2n)\), i.e.\ \(f(2n)=4n^2\). \(\blacksquare\)

Now we proceed by induction to show \(f(n)=f(-n)=n^2\) for \(n\ge0\); base cases \(n=0,1,2,3\) readily follow from Claim 4. Let \(a=n-1\), and consider \(P(f(a),f(a)-n)\): we know \(f(n)+4a^2(a^2-n)\) is a square.

Recall from \((\star)\) that \(f(n)-2n+1\) is a square; hence \(0\le f(n)\le n^2\). Then it is easy to verify that \[\left(2a^2-n-1\right)^2<f(n)+4a^2(a^2-n)\le\left(2a^2-n\right)^2,\]so equality holds on the right and \(f(n)=n^2\).

Analogously, \(P(f(a),f(a)+n)\) will show \(f(-n)=n^2\), so the induction is complete.
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MarkBcc168
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#4 • 2 Y
Y by centslordm, Mango247
All answers are the benign $f(x) = x^2$ and any $f$ that sends $\text{even}\mapsto 0$ and $\text{odd}\mapsto\text{perfect square}$, all of which clearly works. We divide the proof into multiple claims.
Claim: $f(0)=0$ and $f(x)$ is perfect square for any $x$

Proof: Plugging $a=0$ and $b=f(0)$ gives $f(0)^2+f(0)$ is a perfect square, which is enough to force that $f(0)=0,-1$. To rule out $-1$, we plug in $a=0$ to find out that $f(x)+x+1$ is a perfect square. Moreover, plugging in $b=0$ gives $f(f(x))$ is a perfect square. Thus, we have that
\begin{align*}
f(f(f(2))) + f(f(2)) + 1 &= \text{ perfect square} \\
f(f(2)) + f(2) + 1 &= \text{ perfect square} \\
f(2) + 3 &= \text{ perfect square} \\
\end{align*}Considering the first equation in mod $4$ forces $f(f(f(2))) \equiv f(f(2)) \equiv 0\pmod 4$. The second equation then gives $f(2)\equiv -1,0\pmod 4$, and lastly, the third equation gives the contradiction. Hence, $f(0)=0$.

Finally, plugging in $a=0$ gives the second assertion. $\blacksquare$

Now, we give the main step.

Claim: Either one of the following assertions must be true.
  • There exists an infinite sequence of positive integers $x_1<x_2<\hdots$ such that $f(2x_1) = f(2x_2) = \hdots = 0$.
  • There exists an infinite sequence of positive integers $x_1<x_2<\hdots$ such that $f(x_i) = x_i^2$ and $f(2x_i)\ne 0$ for all $i$.
Proof: The main idea of the proof hinges from the substitution $b=f(a)-2a$. This gives
$$f(2a) + (f(a)-2a)f(2a) = f(2a)(f(a)-2a+1) \text{ is a perfect square}.$$Therefore, for each $a$, either $f(2a)=0$ or $f(a)-2a+1$ is a perfect square. Motivated by this, we commit to select $a = \tfrac{p+1}{2}$ for an odd prime $p$. Thus, either $f(p+1)=0$ or $f\left(\tfrac{p+1}{2}\right) = p$ is a perfect square.

However, if the latter happens, then both $f\left(\tfrac{p+1}{2}\right)$ and $f\left(\tfrac{p+1}{2}\right)-p$ must be a perfect square, which forces (by difference of squares) $f\left(\tfrac{p+1}{2}\right) = \left(\tfrac{p+1}{2}\right)^2$. In conclusion, for any prime $p$, either $f(p+1)=0$ or $f\left(\tfrac{p+1}{2}\right) = \left(\tfrac{p+1}{2}\right)^2$, which is enough to imply the claim. $\blacksquare$
Of course, we do both cases separately. The following claim finishes the first case.

Claim: Suppose the first assertion: there exists an infinite sequence of positive integers $x_1<x_2<\hdots$ such that $f(2x_1) = f(2x_2) = \hdots = 0$. Then, $f(\text{even}) = 0$.

Proof: We plug in $b = f(a)-2x_i$ to get that $(f(a)-2x_i)f(2a)$ is a perfect square. However, if $2x_i > f(a)$, then it forces $f(2a)=0$ for sign reason. $\blacksquare$
Now, assume the second case: there exists an infinite sequence of positive integers $x_1<x_2<\hdots$ such that $f(x_i) = x_i^2$ and $f(2x_i)\ne 0$ for all $i$ . We need the following well-known lemma.

Lemma: Suppose that $a,b$ are integers such that $a^2\ne 4b$. Then, there are finitely many $x$ such that $x^2+ax+b$ is perfect square.

Proof: Just complete the square. I don't bother writing the details. $\blacksquare$

To finish, we prove the following two claims.

Claim: For any $i$, $f(2x_i) = 4x_i^2$.

Proof: Vary $j$ through all positive integers and plug in $a=x_i$ and $b=x_i^2-x_j$ to obtain that
$$f(x_j) + (x_i^2-x_j)f(2x_i) = x_j^2 - x_jf(2x_i) + x_i^2f(2x_i) \text{ is a perfect square},$$so by the lemma, $f(2x_i)^2 = 4x_i^2f(2x_i)\implies f(2x_i) = 4x_i^2$ (reminder: $f(2x_i)\ne 0$). $\blacksquare$

Claim: $f(k)=k^2$ for any $k$

Proof: Vary $i$ through all positive integers and plug in $a = x_i$ and $b = x_i^2-k$ to obtain that
$$f(k) + 4x_i^2(x_i^2-k) = (2x_i^2)^2 + (2x_i^2)\cdot 2k + f(k) \text{ is perfect square},$$so by the lemma once again, we get that $(2k)^2 = 4f(k)\implies f(k)=k^2$. $\blacksquare$
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Mr.C
539 posts
#5 • 2 Y
Y by centslordm, Wizard0001
My solution at the exam :

Let
$$P(a,b)^2=f(b)+f(a)f(2a)-bf(2a)$$from $P(0,0)^2$ we have $f(0) \in \{0,-1\}$.
Assume $f(0)=0$. now if $a>0,f(a)=0$ we have from $P(a,a)^2$ that $-af(2a)$ is a perfect square.
Also from $P(0,2a)^2$ we have $f(2a)$ is a perfect square aswell.
Hence $f(2a)=0$. now from $P(x,2^n a)^2$ we have $f(x)f(2x)-2^naf(2x)$ is a perfect square. hence $f(2x)=0$ for all integers $x$.
now the condition goes : $P(a,b)^2=f(b)$ hence $f(2k+1)=g(k)^2,f(2k)=0$ works for every arbitrary function $g$.
Now if $f(a)$ is non zero for positive $a$ we have :
$$P(a,2a)^2=f(2a)(1+f(a)-2a)$$$$P(0,2a)^2=f(2a)>0$$hence $T(a)^2=1+f(a)-2a$ also let $f(a)=g(a)^2$ clearly $(g(a)-1)^2 \ge f(a)-2a+1$ hence $a \ge g(a)$. which gives $f(1)=1$.
Now from $P(a,1)=1+f(a)f(2a)-f(2a)$ is a perfect square. if $f(2a)=1$ then $f(2a)-4a+1 <-1$ so it is not a perfect square. hence :
$$(g(a)g(2a)-1)^2 \ge f(a)f(2a)-f(2a)+1$$hence :
$$f(2a) \ge 4f(a)$$which gives clearly :
$$f(2^n) \ge 4^n$$but we also had $g(a) \le a$ so $f(2^n)=4^n$. now from $P(2^n,b)^2=f(b)+(2^{2n+1})^2-2b.2^{2n+1}$ so we have :
$$P(2^n,b)^2+b^2=f(b)+(2^{2n+1} -b)^2$$so $f(b)-b^2$ is the difference of infinitely many perfect squares hence it is $0$. so $f(x)=x^2$ is also a solution.
Now if $f(0)=-1$ we have :
$$P(a,0)^2=f(a)f(2a)-1$$hence if $q$ is a positive devisor of $f(a)$ for some $a$, then $q=4k+2$ or $q=4k+1$.
Now from $P(0,a)^2=f(b)+b+1$ now $f(4k)+1+4k$ is a perfect square. so $f(4k) \in \{4a,4a-1\}$ which is imposible if $f(4k)>0$ so $f(4k)$ should be negative.
Now we prove $f(2a) \le 0$ which is trivial since $f(4a)f(2a)-1$ is perfect square hence non negative. assume $f(2a)=0$ then we have $f(b)$ is a perfect square itself so $f(4k)=0$ hence $f(0)=0$ and it is a contradiction. so $f(2a)<0$ now since
$f(2a)f(a)-1$ is a perfect square we get that for all integers $n$ we have $f(n) \le 0$ now we have $f(1)+2$ is a perfect square.
So $f(1) \in \{-1,-2\}$ if $f(1)=-2$ we have $-2f(2)-1$ is a perfect square. also $f(2)+3$ is a perfect square aswell. which is impossible .
Hence $f(1)=-1$ and it also gives $f(2)=-2$ . now from $P(1,1)^2=-1+4$ is a perfect square which is a contradiction.
so in this case we have no solutions.
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Orestis_Lignos
558 posts
#6 • 2 Y
Y by centslordm, SerdarBozdag
Let $P(x,y)$ be the assertion that $f(f(x)-y)+yf(2x)$ is a perfect square for all $x,y \in \mathbb{Z}$.

Then, we proceed with some Claims.

Claim 1: $f(0)=0$.
Proof: Considering $P(0,f(0))$ we obtain that $f^2(0)+f(0)$ is a perfect square, hence $f(0) \in \{-1,0 \}$.

Suppose FTSOC that $f(0)=-1$. Then, $P(0,-x-1)$ implies $f(x)+x+1$ is a perfect square. In addition, by $P(x,f(x)$ we obtain that $f(x)f(2x)-1$ is also a perfect square.

Now, take a $x \equiv 0 \pmod 4$. Then, $f(x)+x+1 \equiv f(x)+1 \pmod 4$, therefore $f(x) \equiv 0 \, \, \rm or \, \,  3 \pmod 4$.

If $f(x) \equiv 0 \pmod 4$, then $f(x)f(2x)-1 \equiv 3 \pmod 4$, which can never be a perfect square.
If $f(x) \equiv 3 \pmod 4$, then if $f(x)f(2x)-1=m^2$, we obtain that $f(x) \mid (m^2+1)$, hence if $p \equiv 3 \pmod 4$ is a prime dividing $f(x)$, we obtain $p \mid (m^2+1)$. By Fermat's Christmas Theorem, this gives $p \mid 1$, a contradiction $\blacksquare$

Claim 2: $f(x)$ is a perfect square for all $x$.
Proof: Consider $P(0,-x)$ $\blacksquare$

Now, we may let $f(x)=g^2(x)$, with $g: \mathbb{Z} \rightarrow \mathbb{N}$ i.e., $g$ takes only nonnegative values.

$P(x,y)$ rewrites as $g^2(g^2(x)-y)+yg^2(2x)$ is a perfect square. Taking $y \rightarrow g^2(x)-y$ and then swapping $x$ and $y$ we obtain that $g^2(x)+(g^2(y)-x)g^2(2x)$ is a perfect square, for all $x,y \in \mathbb{Z}$. Let this be $Q(x,y)$.

We distinguish two cases.

Case 1: There exists a $k \neq 0$ such that $g(k)=0$. Then, we make the following Claim:

Claim 3: $g(2k)=0$.
Proof: Suppose not. Then, $g(2k) \neq 0$. If there exists an $\ell >0$ such that $g(\ell)=0$, then $Q(\ell,k)$ implies $-\ell g^2(2k)$ is a perfect square. Since $g(2k) \neq 0$ and $\ell>0$, this expression is $<0$, a contradiction.

Therefore, $g(x) >0$, for all $x>0$.

Now, take a $x>0$ and consider $Q(k,x)$. This gives that $(g^2(x)-k)g^2(2x)$ is a perfect square. Since $g(2x)>0$ from the above discussion, we obtain that $g^2(x)-k$ is also a perfect square, which implies that $g(x)$ takes finitely many values for all $x>0$ (specifically, $g(x)$ is of the form $\dfrac{d_i+d_j}{2}$ where $d_i,d_j$ are divisors of $k$ such that $d_id_j=k$).

Let the maximum value of $g(x)$ for $x>0$ be $S$. Then, fixing $y$ and taking a pretty large $x$, we obtain that $$g^2(x)+(g^2(y)-x)g^2(2y) \leq S^2+g^2(y)g^2(2y)-xg^2(2y)<0,$$a contradiction.

Hence, the Claim is proved $\blacksquare$

To the problem, the Claim implies that $g(k)=0 \Rightarrow g(2k)=0 \Rightarrow g(4k)=0 \Rightarrow \ldots g(2^tk)=0$, for all $t>0$.

Hence, $Q(2^tk,x)$ implies that $(g^2(x)-2^tk)g(2x)$ is a perfect square.

If, for some $x$, $g(2x) \neq 0$, then $g^2(x)>2^tk$ for all $t$, which is an immediate contradiction.

Therefore, $g(2x)=0$ for all $x \in \mathbb{Z}$. It is trivial to verify now that any function of the form $
g(x) =
\begin{cases}
0 &  \text{if} \,\, x \,\, \text{is even} \\
\text{anything} & \text{if} \,\, x  \,\, \text{is odd}
\end{cases}
$ satisfies.

Equivalently, we obtain that $
f(x) =
\begin{cases}
0 &  \text{if} \,\, x \,\, \text{is even} \\
\text{any perfect square} & \text{if} \,\, x  \,\, \text{is odd}
\end{cases}
$ is a solution to the problem.

Case 2: $g(x)=0$ only if $x=0$. Then $g(x) >0$ when $x \neq 0$. Considering $Q(2x,x)$ when $x \neq 0$, we obtain that $(g^2(x)-2x+1)g^2(2x)$ is a perfect square, hence $g^2(x)-2x+1$ is also a perfect square $(*)$.

Now, we prove a series of Claims.

Claim 4: $g(\dfrac{p+1}{2})=\dfrac{p+1}{2}$ for all odd primes $p$.
Proof: Taking $x=\dfrac{p+1}{2}$ in $(*)$ we obtain that $g^2(\dfrac{p+1}{2})-p=a^2$, hence $$(g(\dfrac{p+1}{2})-a)(g(\dfrac{p+1}{2})+a)=p,$$implying that $$g(\dfrac{p+1}{2})-a=1, \,\, g(\dfrac{p+1}{2})+a=p,$$which readily gives the desired $\blacksquare$

Claim 5: $g(2x)$ is even for all $x$.
Proof: By $(*)$, $g^2(2x)-4x+1$ is a perfect square. If $g(2x)$ is odd, then $g^2(2x)-4x+1 \equiv 2 \pmod 4$, a contradiction $\blacksquare$

Claim 6: $g(2x)=2g(x)$.
Proof: Fix a $x$ and let $g(x)=M$ and $g(2x)=N$. Then, by considering $Q(\dfrac{p+1}{2},x)$, we obtain that $$M^2N^2-\dfrac{p+1}{2}N^2+(\dfrac{p+1}{2})^2$$is a perfect square for all odd primes $p$.

The latter rewrites as $$(\dfrac{N^2}{2}-\dfrac{p+1}{2})^2+M^2N^2-\dfrac{N^4}{4},$$where the expressions in the brackets and the one outside of them are integers due to Claim 5.

If $M^2N^2-\dfrac{N^4}{4} \neq 0$, then by letting $p \rightarrow +\infty$ we obtain a clear contradiction, since then $\dfrac{N^2}{2}-\dfrac{p+1}{2}$ should attain finitely many values, a contradiction.

Therefore, $M^2N^2-\dfrac{N^4}{4}=0$, which easily rearranges to $N=2M$, i.e. $g(2x)=2g(x)$, as desired $\blacksquare$

Claim 7: $g^2(x)=x^2$.
Proof: Using Claim 6, $Q(x,y)$ rewrites as $g^2(x)+4(g^2(y)-x)g^2(y)$ being a perfect square.

Taking in the latter $y=\dfrac{p+1}{2}$ and fixing $x$ and rearranging, we obtain that $$(\dfrac{p+1}{2}-x)^2+g^2(x)-x^2$$is a perfect square, which when $g^2(x)-x^2 \neq 0$ is a contradiction after taking $p \rightarrow +\infty$ $\blacksquare$

Therefore, we conclude that $g^2(x)=x^2$, i.e. $f(x)=x^2$ which is a solution.

To conclude, $f(x)=x^2$ and $
f(x) =
\begin{cases}
0 &  \text{if} \,\, x \,\, \text{is even} \\
\text{any perfect square} & \text{if} \,\, x  \,\, \text{is odd}
\end{cases}
$ are the only solutions to the problem.
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hyay
181 posts
#7 • 2 Y
Y by centslordm, somebodyyouusedtoknow
The only such functions are $f(x) = x^2$ for all $x$ and $f(x) = \begin{cases} 0 & \text{if } 2 \mid x \\ g(x)^2 & \text{otherwise} \end{cases}$ for all $x$, where $g(x)$ is any function from the integers to itself.

Let $P(a, b) = f(f(a) - b) + bf(2a)$.

First of all, $P(0, 0) = f(0)(f(0) + 1)$ is a square. If $f(0) \neq 0, -1$, then both $|f(0)|$ and $|f(0) + 1|$ are squares since they are relatively prime, but this is only possible when $f(0) = 0/-1$, contradiction. If $f(0) = -1$, then $P(0, -x - 1) = f(x) + x + 1$ and $P(x, 0) = f(f(x))$ are both squares for all $x$. Recall that $n^2 \equiv 0/1 \pmod{4}$ for all $n$, so since $f(2) + 3$ is a square, then $f(2) \equiv 1/2 \pmod{4}$. Also, since $f(f(f(2))) + f(f(2)) + 1$ is a square and all of the terms are squares, then we must have $f(f(f(2))) \equiv f(f(2)) \equiv 0 \pmod{4}$. But then $f(f(2)) + f(2) + 1 \equiv 2/3 \pmod{4}$ is a square, contradiction. So, $f(0) = 0$.

Then, $P(0, -x) = f(x)$ is a square for all $x$. Furthermore, $P(x, f(x) - 2x) = f(2x)(f(x) - 2x + 1)$, so either $f(2x) = 0$ or $f(x) - 2x + 1$ is a square.

Case 1. $f(2x) = 0$ for infinitely many positive integers $x$.

We claim that $f(2x) = 0$ for all $x$. Assume that there exists $t$ such that $f(2t) \neq 0$. For all $k$ with $f(2k) = 0$, $P(t, f(t) - 2k) = (f(t) - 2k)f(2t)$ is a square. However by taking $k$ sufficiently large the value becomes negative (recall that $f(2t)$ is a square), contradiction. Hence, $f(x) = 0$ for all even $x$, while $f(x)$ is a square for all $x$. This is exactly the second solution.

Case 2. There exists $N$ such that $f(x) - 2x + 1$ is a square for all $x \geq N$

Suppose $x \geq N$ and $2x - 1$ is a prime, then let $f(x) = a^2$ and $f(x) - 2x + 1 = b^2$ where $a, b \geq 0$. Then $2x - 1 = (a + b)(a - b)$, which implies $a + b = 2x - 1$ and $a - b = 1$, thus $a = x$ and $f(x) = x^2$. This implies there exists infinitely many $x \geq 0$ such that $f(x) = x^2$.

Then, for all $a$ and $x \geq 0$ with $f(x) = x^2$, $P(a, f(a) - x) = x^2 - xf(2a) + f(a)f(2a)$ is a square. If $f(2a) = 2k + 1$ for some $k \in \mathbb{Z}$, then notice that $(x - k - 1)^2 < P(a, f(a) - x) < (x - k)^2$ for all $x$ sufficiently large, contradiction. So, $f(2a) = 2k$ and $(x - k - 1)^2 < P(a, f(a) - x) < (x - k + 1)^2$ for all large $x$, which means $x^2 - xf(2a) + f(a)f(2a) = (x - f(2a)/2)^2$ for all large $x$. Comparing coefficients, we get $f(2a) = 4f(a)$.

Now, for all $a$ with $f(a) = a^2$, $P(a, f(a) - x) = f(x) - 4a^2x + 4a^4$ is a square. Then for all $a$ sufficiently large, we have $(2a^2 - x - 1)^2 < P(a, f(a) - x) < (2a^2 - x + 1)^2$, so $P(a, f(a) - x) = (2a^2 - x)^2$ for all large $a$, and thus $f(x) = x^2$ for all $x$.

Done.
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Supercali
1261 posts
#8 • 3 Y
Y by centslordm, PRMOisTheHardestExam, Mango247
Kinda annoying problem ngl.

By substituting $x=f(a)-b$, rewrite the original condition as $$f(x)-xf(2a)+f(a)f(2a) \text{ is a perfect square}$$Let $P(x,a)$ denote the above. Also we will use "PS" for perfect square.

Claim 1: $f(0)=0$
Proof: $P(0,0)$ $\implies$ $f(0)(f(0)+1)$ is a PS $\implies$ $f(0) =0$ or $-1$. Assume FTSOC that $f(0)=-1$. Then $P(-4,0)$ $\implies$ $f(-4)-3$ is a PS; in particular, $f(-4)>0$. Taking this modulo $4$, we get $f(-4) \equiv 3 \text{ or } 0 \pmod 4$. Now $P(0,-4)$ gives $f(-4)f(-8) =k^2+1$ for some $k \in \mathbb{Z}$. It is well known that all positive factors of a number of the form $k^2+1$ are congruent to $1$ or $2$ modulo $4$, which contradicts the previous congruence. So $f(0)=0$. $\blacksquare$

Claim 2: $f(x)$ is a PS for all integers $x$.
Proof: $P(x,0)$. $\blacksquare$

Let $y=\frac{p+1}{2}$ for any odd prime $p$. Then $P(2y,y)$ gives $f(2y)(f(y)-p)$ is a PS. This means either $f(2y)=0$, or $f(y)-p$ is a PS. But it is easy to see that $p$ can be written in a unique way as a difference of two squares - $\left ( \frac{p+1}{2} \right)^2-\left ( \frac{p-1}{2} \right)^2$, so we must have $$f(y)=\left ( \frac{p+1}{2} \right)^2=y^2$$in this case. Using the deep, highly non-trivial, and recently discovered result known as "the infinitude of primes", we can say that either $f(2y)=0$, or $f(2y) \neq 0$ and $f(y)=y^2$ for infinitely many positive integers $y$.


Case I: $f(2y)=0$ for infinitely many $y>0$

Fix any arbitrary integer $a$, and choose a $y>f(a)+200000$ satisfying $f(2y)=0$. Then $P(2y,a)$ gives $f(2a)(f(a)-2y)$ is a PS. Since $f(a)-2y<0$, this can only be true if $f(2a)=0$. Thus $f$ is $0$ on even numbers, and using Claim 2, we get the following solution: $$\boxed{f(x) = \left( \frac{1-(-1)^x}{2} \right ) g(x)^2 \ \ \forall x \in \mathbb{Z} }$$for any arbitrary $g: \mathbb{Z} \rightarrow \mathbb{Z}$.


Case II $f(2y) \neq 0$ and $f(y)=y^2$ for infinitely many $y>0$

Call all such $y$ good.

Claim 3: $f(2y)=4y^2$ for all good $y$.
Proof: Fix any arbitrary good $y$, and let $q(t)$ be the polynomial $t^2-tf(2y)+y^2f(2y)$. There exists a positive integer $M$ satisfying $(t-M)^2<q(t)<(t+M)^2$ for all $t>0$ (for instance, you can take $M=(y^2f(2y))^{1000}+1000+y^{999}+99f(2y)^9$). If $t$ is good, then by $P(t,y)$, $q(t)$ is a PS $\implies$ $q(t)=(t+n)^2$ for some integer $n \in [-M,M]$. But since there are infinitely many good $t$, by another deep result known as "pigeonhole principle", there exists an integer $k \in [-M,M]$ such that $q(t)=(t+k)^2$ for infinitely many good $t$. Since this is a polynomial equation with infinitely many roots, we can compare coefficients to get $f(2y)^2=4y^2f(2y)$ $\implies$ $f(2y)=4y^2$ since $f(2y) \neq 0$. $\blacksquare$

Now, let $x \in \mathbb{Z}$ be arbitrary. Then for any good $y$, $P(x,y)$ gives $$f(x)-4xy^2+4y^4=f(x)-x^2+(2y^2-x)^2 \text{  is a PS}$$Assume $f(x)-x^2 = m \neq 0$ for some $x$. Take a large good $y$, then there must be some PS at a distance of $m$ from $(2y^2-x)^2$. Since the nearest perfect square to $(2y^2-x)^2$ is $2(2y^2-x)-1$ away, we must have $m \geq 2(2y^2-x)-1$, which clearly cannot hold for sufficiently large $y$, contradiction! Therefore we get the solution $$\boxed{f(x)=x^2 \ \ \forall x \in \mathbb{Z}}$$$\blacksquare$
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MathForesterCycle1
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dame dame
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jasperE3
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Let $\mathbb S=\{x^2\mid x\in\mathbb Z\}$, and let $P(a,b)$ be the assertion $f(f(a)-b)+bf(2a)\in\mathbb S$.
$P(0,f(0))\Rightarrow f(0)+f(0)^2=k_1^2$
If $f(0)\ge1$ then $f(0)^2<k_1^2<(f(0)+1)^2$, contradiction. Similarly, if $f(0)\le-2$ then $(f(0)+1)^2<k_1^2<f(0)^2$, another contradiction, so this leaves $f(0)\in\{-1,0\}$.

$\textbf{Case 1: }f(0)=-1$
$P(0,-1-f(x))\Rightarrow f(f(x))+f(x)+1\in\mathbb S$, so $f(x)+1$ is a difference of squares since $P(x,0)\Rightarrow f(f(x))\in\mathbb S$. Since squares are always either $0$ or $1\pmod4$, we have that $f(x)+1\not\equiv2\pmod4$, so $f(x)\not\equiv1\pmod4$. Note that if $f(x)\in\mathbb S$ now, we must have $f(x)\equiv0\pmod4$. An example is $P(0,0)\Rightarrow f(-1)\equiv0\pmod4$. However, then we have $P(-1,f(-1))\Rightarrow f(-1)f(-2)-1=k_2^2$, so taking this equation$\pmod4$ gives us $k_2^2\equiv3$, absurd.

$\textbf{Case 2: }f(0)=0$
$P(0,-x)\Rightarrow f(x)\in\mathbb S\Rightarrow f(x)\ge0$

$\textbf{Case 2.1: }\exists k>0:f(k)=0$
$P(k,-2k)\Rightarrow(1-2k)f(2k)\ge0$, so $f(2k)\le0$, hence $f(2k)=0$ with $2k>0$. Simple induction gives $f(2^nk)=0$ for all $n\in\mathbb N$.
$P(x,f(x)-2^nk)\Rightarrow f(2x)(f(x)-2^nk)\ge0$, but since $f(2x)\ge0$ and $f(x)-2^nk$ is negative for sufficiently large $n$, we must have $f(2x)=0$. Combined with the fact that $f$ is always square, we have the solution $\boxed{f(x)=\left(\frac{1-(-1)^x}2\right)h(x)}$, where $h$ is any function from $2\mathbb Z+1\to\mathbb S$, which works since $P(a,b)$ reduces to $f(f(a)-b)\in\mathbb S$, which is true since $f$ is always a square.

$\textbf{Case 2.2: }f(x)>0\forall x>0$
$P(x,f(x)-2x)\Rightarrow f(2x)(f(x)-2x+1)\in\mathbb S$
Choose $x=\frac{p+1}2$ for an odd prime $p$, then $f(p+1)(f(x)-p)\in\mathbb S$, and since $f(p+1)$ is square, either $f(p+1)=0$ or $f(x)-p$ is a square. The latter must be true in this case. Since $f(x)$ and $f(x)-p$ are both square, we get $p=m^2-n^2$ for $m,n$ such that $f(x)=m^2$ and $f(x)-p=n^2$. Then, since $p$ is prime, we require $m=n+1$, so $f(x)=n^2+2n+1=n^2+p$, so $p=2n+1=2x-1$, hence $x=n+1=m$. Then $f(x)=m^2=x^2$.
To summarize, $\exists g:\mathbb N\to\mathbb N$ strictly increasing such that $f(g(n))=g(n)^2$ for all $n\in\mathbb N$ (where $g(n)=\frac{p+1}2$ and $p$ is the $n$th odd prime).

Now $P(x,f(x)-2x)\Rightarrow f(x)-2x+1\in\mathbb S\Rightarrow f(x)\le x^2$
$P(g(x),g(x)^2-g(y))\Rightarrow g(y)^2-g(y)f(2g(x))+g(x)^2f(2g(x))\in\mathbb S$, setting $y$ such that $g(y)$ is sufficiently large gives that $f(2x)=4x^2$.
Then $P(x,y)$ becomes $Q(x,y):f(f(x)-y)+4x^2y$.
$Q(x,f(x)-y)\Rightarrow f(y)+4x^2f(x)-4x^2y\in\mathbb S$, setting $x=g(n)$ large enough gives us that $(2g(n)^2-y)^2+f(y)-y^2\in\mathbb S$, so $\boxed{f(x)=x^2}$, which also works since $f(f(a)-b)+bf(2a)=(a^2+b)^2\in\mathbb S$.
This post has been edited 4 times. Last edited by jasperE3, Apr 18, 2025, 6:59 AM
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ZETA_in_olympiad
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#12 • 2 Y
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Let $P(a,b)$ denote the given assertion and let $\mathcal{T}=\{t^2: t\in \mathbb{Z}\},$ i.e \ the set of squares.

$P(0,f(0))$ gives $f(0)+f(0)^2\in \mathcal{T}.$ So $f(0)=0$ or $f(0)=-1.$ Assume the latter, $P(0,a)$ gives $f(a)+a+1 \in \mathcal{T}.$ And $P(a,0)$ gives $f(f(x))\in \mathcal{T}.$ But comparing $P(0,2)$ then $P(0,f(f(2)))$ then $P(0,f(2))$ gives $f(2)\equiv 1\pmod{4}$ and $f(2)\equiv -1$ or $0 \pmod{4},$ absurd. So $f(0)=0.$ And $P(0,-a)$ gives $f(a)\in \mathcal{T}.$

Incomplete...
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ZETA_in_olympiad
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Completing...

A. $f(x)=0$ for some $x>0:$ Fix some arbitrary $u$ and let $f(v)=0.$ $P(u,f(u)-v)$ gives $f(2u)(f(u)-v)\in \mathcal{T}.$ Taking large $v$ forces $f(2u)=0.$ So $f(x_{2\mid x})=0$ and $f(x_{2\nmid x})\in \mathcal{T}$ which works.

B. $f(x)> 0~~\forall x>0:$ Then let $p$ be a prime. $P(\frac{p+1}{2}, f(\frac{p+1}{2})-p+1)$ yields $(f(\frac{p+1}{2})-p)(f(p+1))\in \mathcal{T}.$ Taking $p\to \infty$ gives $f(p+1)\in \mathcal{T}.$ It follows that $f(\frac{p+1}{2})-p \in \mathcal{T}$ and $f(\frac{p+1}{2})\in \mathcal{T}.$ Let $f(\frac{p+1}{2})=a^2$ and $f(\frac{p+1}{2})-p=b^2.$ Then solving for $p$ gives $a=\frac{p+1}{2}.$ So $f(\frac{p+1}{2})=(\frac{p+1}{2})^2.$

Take fixed $x$ and an arbitrary $y$ such that $f(x)=x^2$ and $f(y)=y^2.$ $P(x,f(x)-y)$ gives $y^2+x^2f(2x)-yf(2x).$ Note that taking $y\to \infty$ (it is not hard) forces $f(2x)=4x^2.$ Now $P(y,f(y)-x)$ gives $f(x)-x^2+(2y^2-x)^2\in \mathcal{T}.$ Again $y\to \infty$ gives $f(x)=x^2,$ which works.
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Inconsistent
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Solution is $f(x) = 0$ for all even $x$ and equals any perfect square for all odd $x$.

First, notice $f(0)^2 + f(0)$ is a perfect square. So if $x = f(0), y^2 = f(0)^2+f(0)$, then $(2x+1)^2-(2y)^2 = 1$ so $y = 0$ so $x = 0, -1$.

If $x = -1$, then plugging $b = f(a)$ gives $f(a)f(2a)-1$ is a perfect square. Plugging $a = 0$ gives $f(x)+x+1$ is a perfect square. Thus $f(4k) \equiv 0, 3 \pmod 4$, but if $4 \mid f(4k)$ then $f(4k)f(8k) - 1 \equiv 3 \pmod 4$ contradiction, so $f(4k) \equiv 3 \pmod 4$ for all $k$. However, $f(4k+2) \equiv 1, 2 \pmod 4$ for all $k$ but if $f(4k + 2) \equiv 1 \pmod 4$ then $f(4k+2)f(8k+4) - 1 \equiv 2 \pmod 4$ contradiction so $f(4k+2) \equiv 2 \pmod 4$ for all $k$. However since with $b = 0$ we have $f(f(a))$ is a perfect square for all $a$, it follows that $f(f(4k+2))$ is a perfect square but is $2 \pmod 4$, which is a contradiction.

Thus $x = 0$. Assume there exists $c \neq 0$ such that $f(c) = 0$. With $a = 0$ we get $f(x)$ is a perfect square for all $x$. Now notice plugging $b = f(a) - 2a$ gives $f(2a)(f(a)-2a+1)$ is a perfect square for all $a$. So either $f(2a) = 0$ or $f(a) \geq 2a - 1$. Consider the set $S$ of $a$ such that $f(2a) \neq 0$. If $S$ is infinite then it is unbounded. However plugging in, for $c \neq 0$ such that $f(c) = 0$, $a = f(a) - c$ gives $(f(a)-x)f(2a)$ is a perfect square, so $f(a) - x$ is a perfect square for all $a$ in $S$. However since $f(a)$ and $f(a) - x$ are both perfect squares for fixed $x$, $f(a)$ is bounded. However since $f(a) \geq 2a - 1$ for all $a \in S$, it follows that $S$ is bounded, contradiction. Thus $S$ is finite, so for all $a > N$ for some fixed $N$, we have $f(2a) = 0$.

Now suppose $f(2a) \neq 0$ for some $a$. Then plugging in $c = -b$ gives $f(f(a)+c) - cf(2a) \geq 0$. However for large positive $c$ of certain parity, $f(f(a)+c) = 0$, so it follows that $f(2a) = 0$. Contradiction. Thus $f(2a)$ is $0$ for all $a$, giving the desired solution.

Finally consider the case where $f(x) \neq 0$ for all $x \neq 0$. Then as before, it follows that $f(a)-2a+1$ is a perfect square for all $a$. So $f(1) = 1, f(2) = 4$ and $f(a) \leq a^2$. Thus $f(x) - 4x+4$ is a perfect square with $a = 1$. It follows for $n = 3$ that $f(3) = 9$, $f(4) = 16$ by size. Thus again we have $f(x) - 16x + 64$ is a perfect square for all $x$. Again by size and $f(a) - 2a + 1$, $f(x) = x^2$ for all $x \leq 8$. It is easy to verify from small cases that $f(9) = 81$ and $f(10) = 100$. Now we strong induct: supposing it is true up to $2a$, we have

$f(c) - 4a^2c+4a^4$ is a perfect square, but LHS differs at most $c^2$ from $(2a^2 - c)^2$. Thus for $c = 2a+1$, we have that $f(c) = c^2$, $4a^2+4a+1 \geq 2(2a^2 - 2a - 1) - 1$ (possibly true), or $4a^2+4a+1 \geq 4(2a^2-2a-1)-4 = 8a^2 - 8a-8 \Longleftrightarrow 4a^2-12a-9 \leq 0$ which fails for $a \geq 5$. But in the middle case, it follows that $f(2a+1)^2 = 8a+4$, but then $f(2a+1)-16(2a+1)+64 < 0$ contradiction. Thus $f(2a+1) = (2a+1)^2$. Now for $c = 2a+2$, we have that $f(c) = c^2$, $4a^2+8a+4 \geq 2(2a^2-2a-1)-1$ (possibly true), or $4a^2+8a+4 \geq 4(2a^2-2a-1)-4 = 8a^2-8a-8 \Longleftrightarrow 4a^2-16a-12 \leq 0$ which fails for $a \geq 5$. But in the middle case, it follows that $f(2a+2)^2 =  12a + 7$. This is impossible mod $4$, so it must be that $f(2a+2) = (2a+2)^2$. Now the inductive step is complete, so $f(x) = x^2$ for all positive $x$.

Plugging $b = f(a)+c$ for positive $a$ gives $f(-c) + (f(a)+c)f(2a) = f(-c)+ 4a^2c+4a^2$ is a perfect square, so through the analogous argument as above (same rule used), the same conclusion follows through induction.
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CircleInvert
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There are two types of solutions: one is $f(x)=x^2$ and the other is $f(x)=0$ for all even $x$ and for all odd $x$ we choose $f(x)$ to be any perfect square (choice may be dependent on $x$). Clearly both types of solutions work. It remains to prove they are the only ones.

Plug in $a=0$ and $b=f(0)$. This gives $f(0)+(f(0))^2=f(0)(f(0)+1)$ is a perfect square. Now, observe that the factors are relatively prime; thus either one is $0$ or both are positive perfect squares, but no two positive perfect squares differ by exactly $1$, so either $f(0)=0$ or $f(0)=-1$.

Let's deal with $f(0)=-1$: plugging in $b=f(a)$, we get $f(a)f(2a)-1$ is always a perfect square, say $f(a)f(2a)=k^2+1$. Then $f(a)$ must be either $1$ or $2$ mod $4$ for all $a$, but now if we plug in $a=0$ and $b=3$, we get $f(-4)-3$ is a perfect square, so it is $0$ or $1$ mod $4$, but this gives that $f(-4)$ is $0$ or $3$ mod $4$, contradiction.

Hence $f(0)=0$.

Plugging in $a=0$ we get that the range of $f$ is a subset of the set of perfect squares.

Assume that there exists $2c$ such that $f(2c)\ne 0$ (otherwise we are in a solution set we have already identified). Then as $f(c)$ is a perfect square, $f(c)>0$.

Plugging in $a=c$, we get $f(f(c)-b)+bf(2c)$ is always a perfect square, and thus $f(f(c)-b)+bf(2c)\ge 0$. Then $f(f(c)-b)\ge -bf(2c)$. Hence, if we send $b$ to $-\infty$, this shows that as $x\to\infty$ we have $f(x)\to\infty$.

Suppose for the sake of contradiction that $k\ne 0$ and $f(k)=0$. Then then, letting $b=f(a)-k$ we get that for every $a$ we have $(f(a)-k)f(2a)$ is a perfect square. For sufficiently large (positive) $a$, we have $f(2a)>0$, and of course $f(2a)$ is a perfect square, so this implies that $f(a)-k$ is a perfect square. However, as $a\to\infty$, $f(a)\to\infty$, and $f(a)$ is also a perfect square, so this would imply arbitrarily large perfect squares that are a constant $k$ apart, which is absurd, hence we get our contradiction, and conclude that $f$ is positive besides at $0$.

Now, letting $b=f(a)-2a$, we get that $f(2a)(1+f(a)-2a)$ is a perfect square. This means that for $a\ne 0$, $1+f(a)-2a$ is a perfect square (and actually $a=0$ can be checked seperately so this statement just holds for all $a$). Now, suppose $2a-1$. Then let $1+f(a)-2a=A^2$ and $f(a)=B^2$ where $A,B\ge 0$. We then have $2a-1=B^2-A^2=(B-A)(B+A)$, so $B-A=1$ and $B+A=2a-1$, hence $B=a$ so $f(a)=a^2$. Hence, for infinitely many positive $a$, we have $f(a)=a^2$.

Now, changing $b$ to $f(a)-b$ in the original equation gives $f(b)-bf(2a)+f(a)f(2a)$ is always a perfect square. Now, letting $a$ range over the infinitely many positive integers for which we have shown $f(a)=a^2$, this becomes $f(b)-4a^2b+4a^4$ is always a perfect square. Now, let $r=f(b)-b^2$, and observe that $b^2-4a^2b+4a^4=(b-2a^2)$ is always a perfect square, and is always $r$ away from $f(b)-4a^2b+4a^4$. As $f(b)-4a^2b+4a^4$ can get arbitrarily large (positive), the only square within $r$ away will be $f(b)-4a^2b+4a^4$ itself, so in fact $r=0$ and hence $f(b)=b^2$ for all $b$. This concludes the proof.
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megarnie
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The only solutions are $\boxed{f(x) = x^2}$ and $\boxed{ \begin{cases} f(x)  = 0  & \text{ if } x \text{ is even } \\
f(x) \text{ is a perfect square } & \text{ if } x \text{ is odd} \\ \end{cases}}$, where the perfect squares can be arbitrarily chosen. It's easy to check that these work. Now we prove they are the only solutions.


Let $P(a,b)$ denote the given assertion.

$P(a,0): f(f(a))$ is a perfect square.

Claim: $f(0) = 0$.
Proof: Suppose otherwise.
$P(0,f(0)): f(0) + f(0)^2$ is a perfect square. If $f(0)^2 + f(0) = m^2$, then $(2f(0) + 1)^2 = (2m)^2 + 1$, meaning that $(2f(0) + 1 - 2m)(2f(0) + 1 + 2m) = 1$. Since any two integers multiplying to one are equal we have that $2f(0) + 1 - 2m = 2f(0) + 1 + 2m$, so $m = 0$. Since $f(0) \ne 0$, $f(0) = -1$ must hold.

$P(0,a): f(-a - 1) - a$ is a perfect square, or in other words $f(a) +  a + 1$ is a perfect square.
Then $f(f(a)) + f(a) + 1$ and $f(f(a))$ are perfect squares. Since the difference between two squares cannot be $2\pmod 4$, $f(a)$ cannot be $1\pmod 4$ for any $a$.

$P(a, f(a)): f(a) f(2a) - 1$ is a perfect square. This implies that if $f(a) \equiv 0\pmod 4$, then there is a perfect square that is $3\pmod 4$, absurd. Hence for each $a$, $f(a) \equiv \{2,3 \}\pmod 4$.

Consider $f(3)$. Since $f(3) + 3 + 1$ is a perfect square, $f(3)$ is a quadratic residue mod $4$, so it must be $0$ or $1$ modulo $4$, contradiction. Hence $f(0) = -1$ is impossible, so $f(0) = 0$ must hold. $\square$

$P(0,-a)$ implies that $f(a)$ is a perfect square for each $a$, so also $f(a) \ge 0$ for each $a$.

$P(a, f(a) - 2a): f(2a)(f(a) - 2a + 1)$ is a perfect square, so either $f(2a) = 0$ or $f(a) - 2a + 1$ is a perfect square.

Case 1: There exists a positive root of $f$.
Claim: For any positive integer $k$ with $f(k) = 0$, $f(2k) = 0$ must hold.
Proof: We have $f(2k)(f(k) - 2k + 1)$ is a perfect square, meaning that $f(2k)(1 - 2k)$ is a perfect square. If $f(2k) \ne 0$, then $f(2k)(1 - 2k) < 0$ (because $f(2k) \ge 0$), contradiction. $\square$

$P(a,  f(a) - k): (f(a) - k) f(2a)$ is a perfect square for any integer $a$. If some integer $a$ satisfied $f(2a) \ne 0$, then $f(a) - k > 0$ must hold. However, by our previous claim, we can replace $k$ with $k \cdot 2^n$ for any positive integer $n$. Choosing $n$ such that $k \cdot 2^n > f(a)$ gives a contradiction. Thus, $f(2a) = 0$ for all integers $a$, so $f(x) = 0$ for all even $x$. Since we had already proven $f(a)$ is a perfect square, this case implies the second solution described in the beginning.

Case 2: There does not exist a positive root of $f$.
Then for any positive integer $a$, $f(a) - 2a + 1$ is a perfect square.

Call positive integers $x$ with $f(x) = x^2$ special.

Claim: There are infinitely many special positive integers $x$.
Proof: Consider any positive integer $x$ where $2x - 1 = p$ is a prime (clearly infinitely many $x$ exist).

Then $f(x) - 2x + 1$ and $f(x)$ are perfect squares with a difference of $2x - 1$. Notice if $m^2 - n^2 = p$ for some prime $p$, then $(m-n)(m+n) = p$ hence $(m-n) + (m+n) = (-1) + (-p)$ or $(m-n) + (m+n) = (1 + p)$, meaning that $m =\pm \frac{p+1}{2}$.

Since $f(x) - (f(x) - 2x + 1) = 2x -1$ is a prime, we have $f(x) =  \left( \frac{2x - 1 + 1}{2} \right) = x^2$. $\square$

Claim: If $x$ is special, then $2x$ is special.
Proof: If $x$ is special, then $f(x^2  - a) + af(2x)$ is always a perfect square and $x^2 + (f(a) - x) f(2a)$ is also a perfect square.

If $x^2 -a $ is special then $(x^2 - a) ^2 + a f(2x)$ is a square. Let $f(2x) = 4x^2 + c$. $(2x-1)^2 + c$ is a square

We have $x^4 - 2ax^2 + a^2 + 4x^2 a + a\cdot c$ is a square.

$x^4  + 2ax^2 + a^2 + a\cdot c = (x^2 + a)^2 + a\cdot c$ is a square for infinitely many (negative) $a$. Hence $P(a) = (a-x^2)^2  - a\cdot c$ is a perfect square for infinitely many positive integers $a> x^2$.

We have $P(a) = a^2 - (2x^2 + c) a + x^4$ is a perfect square.

If $f(2x)$ was odd, then $f(2x) - 4x + 1\equiv 2\pmod 4$, so it isn't a perfect square, absurd. Hence $f(2x)$ is even, implying that $c$ is also even. Let $c = 2m$.

Hence we have $P(a) = a^2 - 2(x^2 + m) a + x^4 = (a - (x^2 + m))^2  - ((x^2 + m)^2 - x^4))$.

Therefore, $(a - (x^2 + m))^2 - P(a) = (x^2 + m)^2 - x^4$, so infinitely many distinct squares have a difference of $(x^2 + m)^2 - x^4$, meaning that $(x^2 + m)^2 = x^4$, so $m \in \{-2x^2, 0\}$.

If $m = -2x^2$, then $f(2x) = 4x^2 - 2m = 0$, so contradiction since $x$ is a positive integer. Thus, $m = 0$ and $2x$ is also special. $\square$

Now we prove that $f(x) = x^2$ for all integers $x$. Suppose otherwise. Let $k$ be any integer with $f(k) \ne k^2$ and let $f(k) = k^2 + c$.

For any special $x$, $P(x, x^2 - k): k^2 + c + 4x^2 (x^2 - k) = (2x^2 - k)^2  + c$ is a perfect square.

This implies that infinitely many pairs of two squares have a difference of $c$, so $c=0$, absurd since $f(k) \ne k^2 $.

Therefore, all integers $x$ satisfy $f(x) = x^2$ and we are done.
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bin_sherlo
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#21 • 1 Y
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Answer are $f(a)=a^2$ for all positive integers and $f(2a)=0, \ f(2a+1)\in \mathbb{Z}^2$. We have $f(b)+(f(a)-b)f(2a)\in \mathbb{Z}^2$.
Lemma: If $k^2x^2+px+q$ is a perfect square for infinitely many positive integer $x$, then there exists an integer $r$ such that $2r=p, \ r^2=q$.

Claim: $f(0)=0$.
Proof: $P(0,0)$ gives $f(0)^2+f(0)$ is a perfect square hence $f(0)\in \{0,-1\}$. Suppose that $f(0)=-1$. Plugging $P(0,b)$ yields $f(b)+b+1\in \mathbb{Z}^2$. Also $P(a,0)$ implies $f(a)f(2a)-1\in \mathbb{Z}^2$ thus, $4\not | f(a)$. Hence $f(4a)\equiv -1(mod \ 4)$. Since $f(2)+3\in \mathbb{Z}^2$ and $f(2)f(4)-1\in \mathbb{Z}^2$ we have $f(2)\equiv 2(mod \ 4)$. However, $P(2,2)$ gives $f(2)+f(2)f(4)-2f(4)\equiv 2(mod \ 4)$ cannot be a perfect square. Thus, $f(0)=0$.
Claim: $f(a)$ is perfect square.
Proof: $P(0,b)$ gives the result. Let $f(a)=g(a)^2$ where $g\geq 0$.
\[g(2a)^2g(a)^2-bg(2a)^2+g(b)^2\in \mathbb{Z}^2\]Claim: If $g(m)=0$ for $m\neq 0$, then $g(2a)=0$ holds.
Proof: Suppose that $g(2l)\neq 0$. $P(l,m)$ gives $g(2l)^2(g(l)^2-m)\in \mathbb{Z}^2$ or $g(l)^2-m\in \mathbb{Z}^2$. We observe that $g(m)=0$ implies $m<g(a)^2$ hence $g(x)>0$ for $x>g(a)^2$. Pick sufficiently large $x$. Since $g(2x)\neq 0$, we get $g(x)^2-m\in \mathbb{Z}^2$ and this implies $g(x)<m+10^{15!}$. By $P(x,2x)$ we also see that $g(x)^2-2x+1\in \mathbb{Z}^2$ however $g(x)-2x+1<0$.
Since all even integers map to zero, $f(b)\in \mathbb{Z}^2$ sufficies for odds.

Now suppose that $g$ is injective at $0$.
Claim: $g(\frac{p-1}{2})=\frac{p-1}{2}$.
Proof: Plugging $P(\frac{p-1}{2},p-1)$ gives $g(\frac{p-1}{2})^2-p\in \mathbb{Z}^2$ and since $x^2-y^2=p$ or $(x-y)(x+y)=p$ implies $x=\frac{p+1}{2}, \ y=\frac{p-1}{2}$ over positive integers, we get $g(\frac{p-1}{2})=\frac{p-1}{2}$.
Claim: $g(2a)=2g(a)$.
Proof: $P(a,\frac{p-1}{2})$ implies $(\frac{p-1}{2})^2-g(2a)^2.(\frac{p-1}{2})+g(a)^2g(2a)^2\in \mathbb{Z}^2$. Hence $x^2-g(2a)^2x+g(a)^2g(2a)^2\in \mathbb{Z}^2$ has infinitely many positive integer solutions. By the lemma, $2|g(a)g(2a)|=g(2a)^2$ or $2g(a)=g(2a)$.
Claim: $g(a)=|a|\iff f(a)=a^2$.
Proof: We obtain $4g(a)^4-4g(a)^2b+g(b)^2\in \mathbb{Z}^2$. Now pick $a=\frac{p-1}{2}$ to see that $4x^2-4bx+g(b)^2\in \mathbb{Z}^2$ has infinitely many solutions over positive integers. Thus, $g(b)=|b|$ or $f(a)=a^2$ as desired. $\blacksquare$
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MathLuis
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Oh boi, time to have some real fun and avenge child Luis.
Denote $P(a,b)$ the assertion of the given F.E.
$P(0,f(0))$ gives $f(0)^2+f(0)=k^2$ but then $(2f(0)+1)^2=(2k)^2+1$ which means that since only consecutive squares are $0,1$ we must have that $k=0$ and thus $f(0)=0$ or $f(0)=-1$.
If $f(0)=-1$ then from $P(a,0)$ we always get (regardless of the case btw) that $f(f(a))$ is a perfect square but also $P(0,-b-1)$ gives that $f(b)+b+1$ is always a perfect square and thus $f(f(a))+f(a)+1$ is always a perfect square so if $f(a)$ was ever $1 \pmod 4$ then we would have $c^2 \equiv f(f(a))+2 \equiv 2,3 \pmod 4$ and of course this is a contradiction, therefore $f(a)$ is not $1 \pmod 4$ at any time which means that $4 \mid f(f(a))$ always holds and thus $4 \mid f(-1)$ and also $f(a)$ will always be $-1,0 \pmod 4$ and thus by $P(a,4k)$ we have that $f(f(a)-4k) \equiv 0 \pmod 4$ and thus $4 \mid f(4\ell)$ for any integer $\ell$ (follows by fixing $a$) however now $P(2\ell, b)$ gives $f(f(2\ell)-b)+bf(4\ell)$ is a square congrutn to $f(f(2\ell)-b)) \pmod 4$ and this can be $0,1$ but all $f$'s are $-1,0$ so since $b$ is any integer we must have $4 \mid f(z)$ for all integers $z$ and of course $4 \not \; \mid f(0)=-1$ thus a contradiction!.
Therefore $f(0)=0$ and from $P(0,a)$ we get that $f(a)$ is always a perfect square.
Now consider $P(a,f(a)-2a)$ which gives that $f(2a)=0$ or $f(a)-2a+1$ is a perfect square, now let $\mathcal D$ the set of numbers with at most two positive divisors then for any $d \in \mathcal D$ that is not $\pm 2$ it happens that $f(d+1)=0$ or $f \left(\frac{d+1}{2} \right)-d$ is a perfect square and in the latter case it means that there exists $k, \ell$ such that $d=k^2-\ell^2$ which leads to $k+\ell=\pm d$ and $k-\ell=\pm 1$ and in either case we would end up having that either $f(d+1)=0$ or $f \left(\frac{d+1}{2} \right)= \left(\frac{d+1}{2} \right)^2$.
Now the next thing to see is that a difference of squares is always $0 \pmod 4$ or odd this leaves us to consider that if there was a zero $c$ then we would have from $P(a, f(a)-c)$ that $(f(a)-c)f(2a)$ is a perfect square and if $f(a)-c$ was a perfect square then $-c$ is a difference of squares so $c \equiv 2 \pmod 4$ is never going to happen, this also proves that $f(2) \ne 0$ and thus $f(1)=1$ but also for every prime $p \equiv 1 \pmod 4$ we have that $f \left(\frac{p+1}{2} \right)= \left(\frac{p+1}{2} \right)^2$ or else we would have $f(p+1)=0$ in which case it is a contradiction unless we always had $f(2k)=0$ and $f(2k+1)$ a perfect square for all $k \in \mathbb Z$ which indeed works so we will stop considering this solution. And now we also have for all primes $q \equiv 3 \pmod 4$ that $f \left(\frac{1-q}{2} \right)=\left(\frac{1-q}{2} \right)^2$, and using this and the thing above we will check that if such $c \ne 0$ existed then the sol we said we won't mention anymore will happen again.
Indeed if it weren't the case then we would have that $\left(\frac{p+1}{2} \right)^2-c$ is always a perfect square but setting $p$ large enough we can get its bounded between two squares so we get a contradiction!.
Therefore this means for every $d \in \mathcal D$ we have that $f \left(\frac{d+1}{2} \right)=\left(\frac{d+1}{2} \right)^2$ in this case, the most useful part from this is getting infinitely many $\ell \in \mathbb Z$ such that $f(\ell)=\ell^2$ and plug $P(a,f(a)-\ell)$ to get that $\ell^2-\ell f(2a)+f(a)f(2a)$ is a perfect square and for fixed $a$ this means that by some trivial size argument we have that when $f(a)f(2a)=d^2$ then $f(2a)=2d$ this means $d=2d_1^2$ but also $f(2a)^2=4f(a)f(2a)$ so $f(2a)=4f(a)$. Thus giving $f(2\ell)=4\ell^2$.
Also notice from $f(a)-2a+1$ being a perfect square we must have that $0 \le f(a) \le a^2$ for all $a \in \mathbb Z$ by simple size checking and therefore now if we considered $P(\ell, \ell^2-a)$ we get that $f(a)+4(\ell^2-a)\ell^2$ is always a perfect square, sorting this we have $(2\ell^2)^2-2a(2\ell^2)+f(a)$ being a perfect square for infinitely many $\ell$ and thus again from a size argument this implies that $4f(a)=(2a)^2$ and thus $f(a)=a^2$ for all $a \in \mathbb Z$ is the second solution which indeed works, thus we are done :cool:.
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