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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
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.

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0 replies
jlacosta
May 1, 2025
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
Inequality involving square root cube root and 8th root
bamboozled   3
N 3 minutes ago by Jackson0423
If $a,b,c,d,e,f,g,h,k\in R^+$ and $a+b+c=d+e+f=g+h+k=8$, then find the maximum value of $\sqrt{ad^3 g^4} +\sqrt[3]{be^3 h^4} + \sqrt[8]{cf^3 k^4}$
3 replies
bamboozled
Today at 4:46 AM
Jackson0423
3 minutes ago
Geometry
gggzul   3
N 4 minutes ago by nabodorbuco2
In trapezoid $ABCD$ segments $AB$ and $CD$ are parallel. Angle bisectors of $\angle A$ and $\angle C$ meet at $P$. Angle bisectors of $\angle B$ and $\angle D$ meet at $Q$. Prove that $ABPQ$ is cyclic
3 replies
gggzul
4 hours ago
nabodorbuco2
4 minutes ago
Geometry
Lukariman   0
4 minutes ago
Given circle (O) and point P outside (O). From P draw tangents PA and PB to (O) with contact points A, B. On the opposite ray of ray BP, take point M. The circle circumscribing triangle APM intersects (O) at the second point D. Let H be the projection of B on AM. Prove that <HDM = 2∠AMP.
0 replies
Lukariman
4 minutes ago
0 replies
Inequality
nguyentlauv   0
28 minutes ago
Source: Own
Let $a,b,c$ be positive real numbers such that $ab+bc+ca=3$ and $k\ge 0$, prove that
$$\frac{\sqrt{a+1}}{b+c+k}+\frac{\sqrt{b+1}}{c+a+k}+\frac{\sqrt{c+1}}{a+b+k} \geq \frac{3\sqrt{2}}{k+2}.$$
0 replies
nguyentlauv
28 minutes ago
0 replies
Geo metry
TUAN2k8   1
N 44 minutes ago by SimplisticFormulas
Help me plss!
Given an acute triangle $ABC$. Points $D$ and $E$ lie on segments $AB$ and $AC$, respectively. Lines $BD$ and $CE$ intersect at point $F$. The circumcircles of triangles $BDF$ and $CEF$ intersect at a second point $P$. The circumcircles of triangles $ABC$ and $ADE$ intersect at a second point $Q$. Point $K$ lies on segment $AP$ such that $KQ \perp AQ$. Prove that triangles $\triangle BKD$ and $\triangle CKE$ are similar.
1 reply
TUAN2k8
2 hours ago
SimplisticFormulas
44 minutes ago
Two equal angles
jayme   4
N an hour ago by jayme
Dear Mathlinkers,

1. ABCD a square
2. I the midpoint of AB
3. 1 the circle center at A passing through B
4. Q the point of intersection of 1 with the segment IC
5. X the foot of the perpendicular to BC from Q
6. Y the point of intersection of 1 with the segment AX
7. M the point of intersection of CY and AB.

Prove : <ACI = <IYM.

Sincerely
Jean-Louis
4 replies
jayme
May 2, 2025
jayme
an hour ago
PROVE THE STATEMENT
Butterfly   0
an hour ago
Given an infinite sequence $\{x_n\} \subseteq  [0,1]$, there exists some constant $C$, for any $r>0$, among the sequence $x_n$ and $x_m$ could be chosen to satisfy $|n-m|\ge r $ and $|x_n-x_m|<\frac{C}{|n-m|}$.
0 replies
Butterfly
an hour ago
0 replies
IMO Shortlist 2009 - Problem C5
April   38
N an hour ago by MathematicalArceus
Five identical empty buckets of $2$-liter capacity stand at the vertices of a regular pentagon. Cinderella and her wicked Stepmother go through a sequence of rounds: At the beginning of every round, the Stepmother takes one liter of water from the nearby river and distributes it arbitrarily over the five buckets. Then Cinderella chooses a pair of neighbouring buckets, empties them to the river and puts them back. Then the next round begins. The Stepmother goal's is to make one of these buckets overflow. Cinderella's goal is to prevent this. Can the wicked Stepmother enforce a bucket overflow?

Proposed by Gerhard Woeginger, Netherlands
38 replies
April
Jul 5, 2010
MathematicalArceus
an hour ago
Inspired by Austria 2025
sqing   4
N 2 hours ago by Tkn
Source: Own
Let $ a,b\geq 0 ,a,b\neq 1$ and $  a^2+b^2=1. $ Prove that$$   (a + b ) \left( \frac{a}{(b -1)^2} + \frac{b}{(a - 1)^2} \right) \geq 12+8\sqrt 2$$
4 replies
sqing
Today at 2:01 AM
Tkn
2 hours ago
Erasing the difference of two numbers
BR1F1SZ   1
N 2 hours ago by BR1F1SZ
Source: Austria National MO Part 1 Problem 3
Consider the following game for a positive integer $n$. Initially, the numbers $1, 2, \ldots, n$ are written on a board. In each move, two numbers are selected such that their difference is also present on the board. This difference is then erased from the board. (For example, if the numbers $3,6,11$ and $17$ are on the board, then $3$ can be erased as $6 - 3=3$, or $6$ as $17 - 11=6$, or $11$ as $17 - 6=11$.)

For which values of $n$ is it possible to end with only one number remaining on the board?

(Michael Reitmeir)
1 reply
BR1F1SZ
Yesterday at 9:48 PM
BR1F1SZ
2 hours ago
3-var inequality
sqing   1
N 2 hours ago by Natrium
Source: Own
Let $ a,b,c\geq 0 ,a+b+c =1. $ Prove that
$$\frac{ab}{2c+1} +\frac{bc}{2a+1} +\frac{ca}{2b+1}+\frac{27}{20} abc\leq \frac{1}{4} $$
1 reply
sqing
May 3, 2025
Natrium
2 hours ago
mathemetics
Pangbowen   0
2 hours ago
Let a,b,c≥0 and a+b+c=7. Prove that : a/b+b/c+c/a+abc≥ab+bc+ca-2
0 replies
Pangbowen
2 hours ago
0 replies
Property of a function
Ritangshu   1
N 3 hours ago by Natrium
Let \( f(x, y) = xy \), where \( x \geq 0 \) and \( y \geq 0 \).
Prove that the function \( f \) satisfies the following property:

\[
f\left( \lambda x + (1 - \lambda)x',\; \lambda y + (1 - \lambda)y' \right) > \min\{f(x, y),\; f(x', y')\}
\]
for all \( (x, y) \ne (x', y') \) and for all \( \lambda \in (0, 1) \).

1 reply
Ritangshu
May 3, 2025
Natrium
3 hours ago
max value
Bet667   2
N 3 hours ago by Natrium
Let $a,b$ be a real numbers such that $a^2+ab+b^2\ge a^3+b^3.$Then find maximum value of $a+b$
2 replies
Bet667
4 hours ago
Natrium
3 hours ago
Perfect square preserving polynomial
Omid Hatami   35
N Apr 27, 2025 by joshualiu315
Source: Iran TST 2008
Find all polynomials $ p$ of one variable with integer coefficients such that if $ a$ and $ b$ are natural numbers such that $ a + b$ is a perfect square, then $ p\left(a\right) + p\left(b\right)$ is also a perfect square.
35 replies
Omid Hatami
May 25, 2008
joshualiu315
Apr 27, 2025
Perfect square preserving polynomial
G H J
Source: Iran TST 2008
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Omid Hatami
1275 posts
#1 • 4 Y
Y by anantmudgal09, bumjoooon, Adventure10, PikaPika999
Find all polynomials $ p$ of one variable with integer coefficients such that if $ a$ and $ b$ are natural numbers such that $ a + b$ is a perfect square, then $ p\left(a\right) + p\left(b\right)$ is also a perfect square.
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Albanian Eagle
1693 posts
#2 • 3 Y
Y by Adventure10, Mango247, PikaPika999
very very nice problem. I don't have an elementary proof as of now so I would really like to see you solution.
I excluded P from having degree $ \geq 3$ so i'm assuming the only solutions are $ P(x)=k^2x$
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gagan_goku
18 posts
#3 • 2 Y
Y by Adventure10, PikaPika999
can we say that if a polynomial P(x) takes square values for all integers , then
P(x) = {f(x)}^2
where f(x) = polynomial with integer coefficients
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test20
988 posts
#4 • 3 Y
Y by Adventure10, PikaPika999, and 1 other user
gagan_goku wrote:
can we say that if a polynomial P(x) takes square values for all integers , then
P(x) = {f(x)}^2
where f(x) = polynomial with integer coefficients

Yes, that's a well-known theorem.

See for instance
H. Davenport, D.J. Lewis, A. Schinzel:
Polynomials of certain special types.
Acta Arithmetica 9, 1964, 107-116.
This post has been edited 1 time. Last edited by test20, May 26, 2008, 11:11 AM
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gagan_goku
18 posts
#5 • 2 Y
Y by Adventure10, PikaPika999
if thats true , then you can say that f(x,y) = P(x) + P(y^2 -x) = g(x,y) ^ 2
maybe this yields something
Z K Y
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fearailhold
24 posts
#6 • 2 Y
Y by Adventure10, PikaPika999
I've been spending a lot of time mulling this one over in my head. Can anyone solve it?
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Albanian Eagle
1693 posts
#7 • 3 Y
Y by Adventure10, Mango247, PikaPika999
If $ P$ is not an odd polynomial there exists an $ a$ so that $ P(x)+P(a)$ is not divisible by $ x+a$
$ P(x^2-a)+P(a)=Q(x)^2$ so $ P(x)+P(a)=Q(\sqrt{x+a})^2$ and $ P(x)+P(a)=R(x)^2$ also $ P(x)+P(b)=K(x)^2$
but $ R(x)^2-K(x)^2=c$ can not hold as an identity so $ c=0$ and the polynomial is a constant function $ P(x)=2m^2$
if P is an odd polynomial $ P(x)+P(a)=(x+a)R(x)^2$ and $ P(x)+P(b)=(x+b)K(x)^2$. how can we show that these last two equations imply some contradiction if the degree of $ P$ is too large?...
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fearailhold
24 posts
#8 • 3 Y
Y by Adventure10, Mango247, PikaPika999
Albanian Eagle wrote:
If $ P$ is not an odd polynomial there exists an $ a$ so that $ P(x) + P(a)$ is not divisible by $ x + a$

Do you have a proof of this?
Z K Y
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t0rajir0u
12167 posts
#9 • 3 Y
Y by Adventure10, Mango247, PikaPika999
Albanian Eagle wrote:
i'm assuming the only solutions are $ P(x) = k^2x$

Of course there's also $ P(x) = 2k^2$.
Z K Y
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ali666
352 posts
#10 • 3 Y
Y by Adventure10, Mango247, PikaPika999
Albanian Eagle wrote:
and $ P(x) + P(a) = R(x)^2$
why?...can you say more details please?
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harazi
5526 posts
#11 • 3 Y
Y by Adventure10, Mango247, PikaPika999
I will assume the answers to the preceeding questions have already been given, so let me try to answer Albanian Eagle's question. Suppose that $ (X+a)R(x)^2-(X+b)P(X)^2=c$, a nonzero constant (otherwise the conclusion is obvious...). Then obviously $ P$ and $ R$ are relatively prime and have the same degree. Differentiate the relation to obtain $ (X+a)2RR'+R^2-P^2-(X+b)2PP'=0$, thus $ R$ divides $ P+2(X-b)P'$ and thus must be a constant times this thing, by the remark on the degree. This gives you a differential equation on $ P$ and allows finding all solutions, but unfortunately it may very well happen that there are solutions of arbitrarily large degree (and it happens, actually...), but this imposes however very strong conditions on $ P$. Now, by choosing some other value $ d,e,..$ for $ a,b$, I think we can conclude. I'm still puzzled by this apparently extremely easy problem, which however does not seem to have a decent solution.
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Albanian Eagle
1693 posts
#12 • 2 Y
Y by Adventure10, PikaPika999
Oh nice..
so Harazi can substitute "a constant times" with "equal" by considering leading coefficients.
and so we have $ R=K+2(x+b)K'$ but also vice versa so
$ (x+b)K'+(x+a)R'=0$ and particularly
$ K'=R'=...$ by putting different choices So $ P(x)+P(m)=(x+m)(Q(x)+c_m)^2$ holds for some fixed $ Q$
this is possible only when $ Q$ is constant so $ P$ is linear and we can easily conclude
:lol: we don't even need the differential equations.
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harazi
5526 posts
#13 • 2 Y
Y by Adventure10, PikaPika999
But I love differential equations! Just kidding, I really hate them, thanks for pointing out this simplification, I had no paper with me to check if indeed it can be made easier. Anyway, great problem.
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ali666
352 posts
#14 • 2 Y
Y by Adventure10, PikaPika999
fearailhold wrote:
Albanian Eagle wrote:
If $ P$ is not an odd polynomial there exists an $ a$ so that $ P(x) + P(a)$ is not divisible by $ x + a$

Do you have a proof of this?
if for all $ a$, $ P(x)+P(a)$ is diviseble by $ x+a$ then $ G(x)=P(x)+P(-x)$ is zero for all $ a$....and P is odd...
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Albanian Eagle
1693 posts
#15 • 2 Y
Y by Adventure10, PikaPika999
:lol:
If the words Iran and Polynomial appear in the same thread it can never disappoint you :D
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gagan_goku
18 posts
#16 • 3 Y
Y by Adventure10, Mango247, PikaPika999
1 doubt
how does $ P(x) + P(a) = P(\sqrt{x+a})^2$ imply $ P(x) + P(a) = R(x)^2$
secondly , if you have $ P(x) + P(a) = R(x,a)^2$ , then i think we can conclude that the degree of $ P$ is 0
but then we dont get the solution $ P(x) = k^2x$
does this solution come from $ P$ being odd
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rchlch
35 posts
#17 • 3 Y
Y by Adventure10, Mango247, PikaPika999
gagan_goku wrote:
can we say that if a polynomial P(x) takes square values for all integers , then
P(x) = {f(x)}^2
where f(x) = polynomial with integer coefficients
This claim should be slightly modified, cause $ f(x)$ do not need to be with integer coefficients. For example $ P(x)=\frac{x^2(x-1)^2}{4}$.
Then I think I have a elementary proof to this fact.
The most important case is that the degree of $ P$ is even. In this case we let $ \deg P=2n$ and tha cofficent of the $ 2n-th$ power is $ A$.
Consider $ L=\lim_{x \rightarrow \infty}{(\sqrt{P(x+n-1)} -C_{n-1}^1 \sqrt{P(x+n-2)}+\ldots+(-1)^{n-1} \sqrt{P(x)})}$
Here is the most important result:
We claim that $ L$ is a constant.
In fact, we can find a polynomial $ Q(x)$ such that $ \deg Q=n$ and $ \deg(P-AQ^2) \leq n-1$ (This is true due to Newton's formula for symmitric sums.)
Then we have
${ \lim_{x \rightarrow \infty}{\sqrt{P(x)}-\sqrt{A}Q(x)}=\lim_{x \rightarrow \infty}{\frac{P(x)-AQ^2(x)}{\sqrt{P(x)}+\sqrt{A}Q(x)}}}=0$

So
$ L=\sqrt{A} \lim_{x \rightarrow \infty}{(Q(x+n-1) -C_{n-1}^1 Q(x+n-2)+\ldots+(-1)^{n-1} Q(x))}=  \sqrt{A} (n-1)!$
is a constant.
Finally, suppose $ P(x)=a_x^2$ for all $ x \in Z$. Then $ \lim_{x \rightarrow \infty}{(a_{x+n-1}-C_{n-1}^1 a_{x+n-2}+ \ldots +(-1)^{n-1} a_{x})}=L$. Therefore $ L$ is a integer, and for large enough $ x \in Z$, we have ${ a_{x+n-1}-C_{n-1}^1 a_{x+n-2}+ \ldots +(-1)^{n-1} a_{n}}=L$. Note that the sequence $ {a_n}$ has the fomula $ a_n=f(n)$ where $ \deg{f}=n$ for large enough $ n$. So we conclude $ P(x)=f^2(x)$ for large enough $ x \in Z$ and therefore $ P(x)=f^2(x)$ for all $ x$. And it's obvious that $ f(x) \in Q[x]$.

In the case that the degree of $ P$ is odd, let $ P'(x)=P(x^2)$ we do the same thing, and it's easy to claim that such $ P(x)$ don't exist.
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test20
988 posts
#18 • 3 Y
Y by Adventure10, Mango247, PikaPika999
rchlch wrote:
gagan_goku wrote:
can we say that if a polynomial P(x) takes square values for all integers , then
P(x) = {f(x)}^2 where f(x) = polynomial with integer coefficients
This claim should be slightly modified, cause $ f(x)$ do not need to be with integer coefficients. For example $ P(x) = \frac {x^2(x - 1)^2}{4}$.

Aaah, yes indeed. The correct statement is:

If a MONIC polynomial $ P(x)$ takes square values for all integers $ x$,
then $ P(x) = {f(x)}^2$ where $ f(x)$ is a polynomial with integer coefficients.
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Albanian Eagle
1693 posts
#19 • 2 Y
Y by Adventure10, PikaPika999
There are stronger lemmas out there
like http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1576618391&t=148696
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rchlch
35 posts
#20 • 3 Y
Y by Adventure10, Mango247, PikaPika999
Albanian Eagle wrote:
I think this lemma cannot be used in this problem because no direct evidence shows that $ P(x)$ is a monic polynomial.
And for your lemma, one of my classmates has given a beautiful solution.
Thank you all the same.
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Albanian Eagle
1693 posts
#21 • 3 Y
Y by Adventure10, Mango247, PikaPika999
rchlch wrote:
Albanian Eagle wrote:
I think this lemma cannot be used in this problem because no direct evidence shows that $ P(x)$ is a monic polynomial.
And for your lemma, one of my classmates has given a beautiful solution.
Thank you all the same.
It was just a reply to test20's post :lol:
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ith_power
233 posts
#22 • 6 Y
Y by toto1234567890, Adventure10, Mango247, Mango247, Mango247, PikaPika999
My solution was too big(6 pages). however here is an outline:
Click to reveal hidden text
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amir.sepehri
3 posts
#23 • 4 Y
Y by ali.agh, Adventure10, Mango247, PikaPika999
I give this exam and i couldn't solve this problem :(
in fact only one of students solved this problem, his main idea is : if a+b is square then a*x^2+b*x^2 is square too, so
p(ax^2)+p(bx^2) is perfect square for all integer x so there exist a g(x) such: p(ax^2)+p(bx^2)=g(x)^2
if p(x)=(a_n)*x^n+.... then (a_n)*(a^n+b^n)=u^2 (g(x)=u*x^k+...) for all a,b (a+b=perfect square) let a=b=2 so
(a_n)*2^(n+1)=u^2 and ... it's easy to see that p(x)=(k^2)*(x^2) or 2*(k^2), so we solve this hard problem
:wink:
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shekast-istadegi
81 posts
#24 • 2 Y
Y by Adventure10, PikaPika999
Theorem_ let $f$ be a integer polynomial.We have $f(a+tp^s)\equiv {}^{p{}^{s+1}}f(a)+tp^sf{}'(a)$such that $(t,p)=1$.So by above theorem we have for any integer polynomial $Q$ which has not any double root ,there are infinitely many prime $p$ which there exist some $n$ such that $p\mid Q(n)$ but $ Q(n)\not\equiv {}^p{}^{2}0$.now suppose $p(a)+p(x^2-a)=Q(x)$ .(step 1)if $Q$ has not any double root ,we have $p\mid Q(n)$ but $ Q(n)\not\equiv {}^p{}^{2}0$ a contraction.(step 2)if $Q$ has some double root, we have $Q(x)=R_{a}(x)T_{a}(x)^2$ such that $R_{a}(x)$ has not any double root .$Q(m)$ is a perfect square for any positive integer $m$ so $R_{a}(Y)$ is a perfect square for all $y> M$ and we have a contraction same way to step 1.
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toto1234567890
889 posts
#25 • 3 Y
Y by Adventure10, Mango247, PikaPika999
And how to solve the main problem?
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anantmudgal09
1980 posts
#26 • 13 Y
Y by rafayaashary1, v_Enhance, llpllp, Ankoganit, bumjoooon, mijail, Supercali, ywq233, centslordm, guptaamitu1, Adventure10, math_comb01, PikaPika999
Solution

Credits and remarks
This post has been edited 1 time. Last edited by anantmudgal09, Sep 23, 2016, 4:44 PM
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ywq233
38 posts
#28 • 2 Y
Y by Adventure10, PikaPika999
anantmudgal09 wrote:
and if $d$ is odd then we have $$\left((t^2-1)^{\frac{d-1}{2}}-1\right)^2<\left(\frac{g(t)}{tc}\right)^2<\left((t^2-1)^{\frac{d-1}{2}}\right)^2$$which is again a contradiction! Thus, we must have $d=1$.
How can you get the left side?
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bumjoooon
165 posts
#29 • 4 Y
Y by Mango247, Mango247, Mango247, PikaPika999
anantmudgal09 wrote:
Lemma 2. Let $f$ be a non constant polynomial with integer such that $\sqrt{f(n)}$ is an integer for all positive integers $n$. Then there exists a polynomial $g(x)$ with integer coefficients such that $f(x)=g(x)^2$.

I learned another simple way to prove this lemma from lectures on winter camp in Korea around 2018.

Proof) Think of $f(x+m)f(x)$ for some natural number $m$. If $m$ is sufficiently large, we can separate complex roots of $f(x)$ and $f(x+m)$ and therefore $f(x+m)$ and $f(x)$ is relatively prime. By assumption, $f(x+m)f(x)$ is even degree and the maximal coefficient is square. Therefore, $f(x+m)f(x)=h(x)^2$ for some $h(x) \in \mathbb{Z}[x]$. Since $f(x)$ and $f(x+m)$ is prime, you can easily see that $f(x)=g(x)^2$ for some $g(x) \in \mathbb{Z}[x]$.

Same idea is used in Proposition mentioned in https://artofproblemsolving.com/community/u314211h1985879p14247495
This post has been edited 2 times. Last edited by bumjoooon, Mar 10, 2020, 1:39 AM
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bumjoooon
165 posts
#30 • 1 Y
Y by PikaPika999
Another astonishing proof for Lemma by my student.

Proof) Let $f(x)=a_n x^n +....+a_0$. By assumption, $a_n >0$.
[Case 1] $n$ is odd.
Multiply the term $(a_n x +1)$. There are infinitely many $x$ such that $a_n x+1$ is perfect square. Therefore, infinitely many $x$ such that $(a_n x+1)f(x)$ is perfect square. However, there are also $x$ such that $a_n x +1$ is not a perfect square and contradiction.
[Case 2] $n$ is even, but $a_n$ is not a perfect square.
Multiply the term $(a_n x^2 +1)$. And same as above. Here we used the fact that pell equation $y^2 - a_n x^2 =1$ has infinitely many solution.

Therefore, $n$ is even and $a_n$ is perfect square and $f(x)=g(x)^2$ for some $g(x) \in \mathbb{Z}[x]$.
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primesarespecial
364 posts
#31 • 2 Y
Y by PRMOisTheHardestExam, PikaPika999
Here is a smol bren solution.
Write the polynomial,as $x^d.P(x)$, with $P(0)$ non zero.Assume $P$ non-constant.Then assuming,$P$ is not a square ,write it as a square times a bunch of irreducibles.Consider such a irreducible ,say $h(x)$,and then by Schur ,choose a very large prime factor of the polynomial $h(x^2)$.Then using Bezout,we can conclude that if $p|h(m^2)$,then $p$ does not divide any other irreducible factor of $P$ evaluated at $m^2$.Then,again using bezout but this time with $h'(x)$,we conclude that if $p$,is large enough ,then $p$ also doesn't divide $h'(m^2)$,thus from Hensel,we can uniquely lift the root corresponding to $m^2$ mod $p$ to mod $p^2$ .Due to this unique lift ,we can find such an $n$,such that $n \equiv m^2$ mod $p$ but $p^2$ does not divide $P(n)$.Thus we have that $v_p(P(n))$ is odd.Now choose a $b$ such that $v_p(b)=j$ is even and quite large.As $p$ is large (forgot to mention $p$ doesn't divide $P(0)$,so $p$ doesn't divide $P(b)$).
Now,we use hensel's again.Firstly note that $n$ is a quadratic residue mod $p$.So by hensel's on the polynomial $x^2-n$,we see that the derivative is non zero mod $p$.Thus,we can find a solution to this poly mod $p^{j+1}$,such that $v_p(x^2-n)$ is exactly $j$.Thus we have found $b+n$ is a perfect square but $v_p(n^d.P(n)+b^d.P(b))$ is odd which is a contradiction.Thus,$P(x)$ is a perfect square.

Using a very similar argument as above,we can prove that $d$ is even(Basically choosing $a,b$,such that $P(a)$ has a large $v_p$,which is obv
even,except here we choose $b$ to have small odd $v_p$,and obv $p$ doesn't divide $P(b)$.$a+b$ being a perfect square is ensured by the very similar hensel argument as above).
So we have that we can write $P(a)+P(b)$ as $a^{2d}Q(a)^2+b^{2d}Q(b)^2$.So if we fix $a$,we see that $a^dQ(a)$ is a part of infinitely many Pythagorean triples,which is impossible.
Observing the left out cases(which are not that hard),we get our solutions.
This post has been edited 2 times. Last edited by primesarespecial, Sep 2, 2022, 12:59 PM
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VicKmath7
1389 posts
#32 • 1 Y
Y by PikaPika999
As in #26, $f(ax^2)+f(bx^2)$ is always a square for a fixed pair $(a,b)$ with $a+b$ being a square, so $g(x)^2=f(ax^2)+f(bx^2)$ for some integer polynomial $g(x)$. In addition, we consider only the case $\deg(f) \geq 2$.

Next, we consider the leading coefficient of this polynomial; it is of the form $T(a^n+b^n)$ ($T$ is the leading coefficient of $f$), which shall be a square. If $n$ is even, plug $(1,p^2-1),(1,p^2+2p)$ for some prime $p$ to see that $2T, T$ are QR's modulo $p$ for all primes $p$, so $2$ is a QR for all primes $p$, absurd, since $(\frac{2}{p})=(-1)^{\frac{p^2-1}{8}}$. Now we can finish the other case with $(2,2),(3,1)$ : $2^{n+1}T$ is a square, hence $T$ is a square, so $3^n+1=x^2$ is a square, so $3^n=(x-1)(x+1)$ and thus $x-1=1$ and $n=1$, contradiction.
This post has been edited 2 times. Last edited by VicKmath7, Sep 25, 2022, 2:47 PM
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giglio
9 posts
#33 • 1 Y
Y by PikaPika999
I see pleny of solutions over here depending on this theorem about a polynomial being square too many times. I solved to problem using a totally different approach with pells equations, some estimations of the growth of some terms, and an unexpected use of Norms function at the end. I will write down the solution tomorrow.
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giglio
9 posts
#34 • 3 Y
Y by jrsbr, Dissonant, PikaPika999
Suppose $degP > 0$. If $P$ is constant so $2P(2)$ must be a square, so $P \equiv 2k^2$ for some integer $k$. Indeed all of these work.
Also suppose that the leading coefficient of $P$ is positive. If it's negative then $P(2N^2)$ is negative for big $N$, but is must also be twice a perfect square, contradiction.
Fix some positive integer $t$ such that the equation $b^2 - 2a^2=t$ has a solution.
We know there are infinetely many big $t$'s satisfying this (perfect squares for instance).
By pell's equation we know that, given one solution $b_0^2 - 2a_0^2$, we can find infinetely many solutions of the form
$$b_n + a_n \sqrt2= (b_0 + a_0 \sqrt2)(3+2sqrt2)^n.$$Just so I'll write quicker, set $ r = 3 + 2 \sqrt2$, we'll use it at lot.
Now we know that, as $2a_n^2 + 2a_n^2$ is the square of an integer, $2P(a_n^2)$ must be the square of an even integer, say, by definition, $4r_n^2$. So $P(a_n^2) = 2r_n^2$.
We also know that $b_n^2 - t + t$ is a perfect square, so $P(b_n^2 - t) = s_n^2 - P(t)$, for some integer $s_n$. But, by definition, $b_n^2 - t = 2a_n^2$, so $2r_n^2$ and $s_n^2 - P(t)$ must be equal. So we have that $s_n^2 - 2r_n^2 = P(t)$.
But this means that $s_n, r_n$ are pell's solutions for $x^2 - 2y^2 = P(t)$. This means that $$s_n + r_n \sqrt2 = (s_0 + r_0 \sqrt2)r^{t_n}$$for some minimal solution $(s_0,r_0)$ and some integer $t_n$. As there are only finitely many minimal solutions for pell's equation, by the pigeon hole principle one of them must appear infinetely many times, varying $n$. WLOG call it $(s_0,r_0)$.
In other words, we can say that
$$s_n = \frac{s_0 + r_0 \sqrt2}{2}r^{t_n} + \frac{s_0 - r_0 \sqrt2}{2}r^{-t_n}$$so
$$s_n^2 = \big( \frac{s_0 + r_0 \sqrt2}{2} \big)^2 r^{2t_n} + \big( \frac{s_0 - r_0 \sqrt2}{2} \big)^2 r^{-2t_n} + 2\frac{s_0^2 - 2r_0^2}{4} = (1+o(1))\big( \frac{s_0 + r_0 \sqrt2}{2} \big)^2 r^{2t_n}$$So this is one estimate for $s_n^2$.
Let $d = degP$ and $c$ the leading coefficient. Now we also know that
$$s_n^2 = P(b_n^2 - t) + P(t) = (1+o(1))P(b_n^2) = (1+o(1))c.b_n^{2d}.$$As $$b_n = (1+o(1))\frac{b_0+a_0 \sqrt2}{2}r^n,$$$$s_n^2 = (1+o(1))c.\big( \frac{b_0+a_0 \sqrt2}{2} \big) ^{2d}r^{2nd}$$Therefore the ratio between the $2$ estimates converges to $1$. So
$$lim_{n \to \infty} \frac{\big( \frac{s_0 + r_0 \sqrt2}{2} \big)^2 r^{2t_n}}{c.\big( \frac{b_0+a_0 \sqrt2}{2} \big) ^{2d}r^{2nd}} = 1.$$So
$$lim_{n \to \infty} \frac{\big( \frac{s_0 + r_0 \sqrt2}{2} \big)^2}{c.\big( \frac{b_0+a_0 \sqrt2}{2} \big) ^{2d}} r^{2t_n - 2dn} = 1.$$So, a constant times an integer power of $r$ converges to $1$. Due to the fact the the sequence $C.r^m$ is discrete, if it converges to $1$, it's eventually $1$. So there exists an integer $q$ such that
$$\frac{\big( \frac{s_0 + r_0 \sqrt2}{2} \big)^2}{c.\big( \frac{b_0+a_0 \sqrt2}{2} \big) ^{2d}} r^q = 1.$$Rearranging things and substituting $r$ back to $3+2 \sqrt2$, we have
$$(s_0 + r_0 \sqrt2)^2 (3+2 \sqrt2)^t = c(b_0 + a_0 \sqrt2)^{2d} 2^{2d-2}$$Now we are gonna make use of the Norm function defined by $N(m + n \sqrt2) = m^2 - 2n^2$, where $m,n$ are rationals. One can easily prove that this function is well defined and is multiplicative. So we have
$$(s_0^2 - 2r_0^2)^2 . 1^t = c^2(b_0^2 - 2a_0^2)^2 2^{4d-4}$$So
$$P(t)^2 = c^2 . t^{2d} . 2^{4d-4}$$As the leading coefficient of $P$ is positive, $P$ is eventually positive, so we can take the squareroot of both sides to conclude that
$$P(t) = c.t^d.2^{2d-2}$$But notice that this is true for infinetely many big integers $t$'s, so it must be true that
$$P(x) = x^d.c.2^{2d-2}$$But remember that $c$ is the leading coefficient of $P$, so $c = c.2^{2d-2}$, so $2^{2d-2}=1$, so $d$ must be $1$, and $P(x) = cx$.
Now, as $2P(2)$ is a square, so $4c$ is a square, so $c$ itself is a perfect square. Given that, clearly $a+b$ is a square if and only if ac+bc is a square.
Therefore all solutions are $P(x) = k^2x$ and $ P(x) = 2k^2$, where $k$ is an arbitrary integer.
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Pyramix
419 posts
#35 • 1 Y
Y by PikaPika999
We use a few well-known theorems to solve.

Theorem 1.
A polynomial is a perfect square at every integer iff it is a perfect square in $\mathbb{Z}[x]$

Theorem 2. (Mihailescu)
For $a,b,c,d>1$ being positive integers, the Diophantine $a^b-c^d=1$ has a unique solution, which is $(a,b,c,d)=(3,2,2,3)$.

For any $x$, we have $2P(2x^2)$ is a perfect square and $P(x^2)+P(3x^2)$ is also a perfect square.

So, from theorem 1 there are polynomials $A(x),B(x)\in\mathbb{Z}[x]$ such that
\[2P(2x^2)=A(x)^2,\ \ P(x^2)+P(3x^2)=B(x)^2\]Let $k$ be the leading coefficient of $P$ and $d=\deg P>0$.
Then we have $2^{d+1}k=u^2$ and $(3^d+1)k=v^2$ for some integers $u,v$.

If $d$ is even then $k=2w^2$ for some $w$, which means $3^d+1=2y^2$ for some $y$, which is a contradiction $\pmod{3}$.
So, $d$ is odd. But then $k$ is a perfect square from first equation, which means $3^d+1$ is a perfect square. From theorem 2, $d=1$ is forced.

Finally, note that for $d=1$ only $P(x)=c^2x$ works for some $c$ and for $d=0$ only $P(x)=2c^2$ works. These are the only solutions.
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HamstPan38825
8859 posts
#36 • 1 Y
Y by PikaPika999
The key lemma is the following:

Lemma: Let $P(x)$ be a polynomial such that $P(n)$ is a perfect square for all sufficiently large positive integers $n \geq 1$. Then there exists a polynomial $Q(x)$ such that $P(x) = Q(x)^2$ identically.

Proof: This is a special case of Steps 3 and 4 to this proof to USEMO 2019/2.

Now we just kind of destroy the problem. Observe that for each fixed positive integer $k$, we have the numerical equality \[P\left(n^2 - k\right) + P(k) = P_k(n)^2\]for some function $P_k(n)$ for all sufficiently large $n$. Thus the polynomial $P\left(n^2-k\right) + P(k)$ in $n$ takes only perfect square values, and by the lemma, it follows that it must be identically equal to $P_k(n)^2$ for some polynomial $P_k$.

On the other hand, the expression on the LHS is even, hence $P_k(n)$ is either even or odd. (Otherwise, it is easy to construct an $x$ such that $|P_k(x)| \neq |P_k(-x)|$.) We make the following claim:

Claim: Either $P_k(x)$ is even for all $k \geq 1$ or odd for all $k \geq 1$.

Proof: Suppose there existed a $k$ such that $P_k(x)$ was odd and $P_{k+1}$ was even. Then we can write
\begin{align*}
P\left(x^2-k\right) + P(k) = x^2 Q_k\left(x^2\right)^2 &\implies P(x-k) + P(k) = xQ_k(x)^2 \\
P\left(x^2-k-1\right) + P(k+1) = Q_{k+1}\left(x^2\right)^2 &\implies P(x-k-1) + P(k+1) = Q_{k+1}(x)^2
\end{align*}for some polynomials $Q_k(x)$ and $Q_{k+1}(x)$. So in particular, we have the equality \[Q_{k+1}(x+1)^2 - P(k+1) = P(x-k) = xQ_k(x)^2 - P(k).\]However, taking $x$ to be a sufficiently large negative number yields a negative RHS and positive LHS, hence contradiction. The symmetric case works out analogously. $\blacksquare$

So now we have to resolve two cases.

First Case: Suppose that $P_k(x)$ is even for all $x$. Then the same equation in the previous display reads \[Q_{k+1}(x+1)^2 - P(k+1) = Q_k(x)^2 - P(k).\]So for $Q_{k+1}'(x) = Q_{k+1}(x+1)$ and $Q_k(x)$, the coefficients of the squares of these polynomials agree everywhere except the constant term. It follows easily that their constant terms must also agree, thus $P(k) = P(k+1)$ for all positive integers $k$, and $P$ is constant. We can verify $P \equiv 2c^2$ in this case.

Second Case: Suppose that $P_k(x)$ is odd for all $x$. Then by setting $x=0$ in the polynomial equality \[P(x-k) + P(k) = xQ_1(x)^2\]for all positive integers $k \geq 1$, we get that $P(k) + P(-k) = 0$, i.e. $P$ is odd. Then setting $x=1$ in this sequence of equalities, note that $P(k)-P(k-1)= Q_k(1)^2$ is always a perfect square. So reading as a polynomial, $P(k) - P(k-1) = R(k)^2$ is the perfect square of a polynomial. By applying the same argument with $x=4$, we obtain that $R(x)^2+R(x+1)^2+R(x+2)^2+R(x+3)^2 = S(x)^2$ for some polynomial $S(x)$, as the numerical equality holds for all sufficiently large $x$. [EDIT: the rest of this solution ended up being a fakesolve, so I would greatly appreciate anyone who could finish it off from here.]

Remark: This problem is a nightmare.
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joshualiu315
2534 posts
#37 • 1 Y
Y by PikaPika999
The answer is $P(x) = \boxed{2c^2, c^2x}$, which can be manually confirmed to be the only valid polynomials with degree less than $2$. Let $P(x) = a_0+a_1x+\dots+a_dx^d$ so that it has degree $d$.

Notice that plugging in $(a,b) = (2x^2, 2x^2)$ and $(a,b) = (x^2, 3x^2)$ gives

\[2P(2x^2) = 2a_0+4a_1x^2+\dots+2^{d+1}a_dx^{2d} = [Q(x)]^2\]\[P(x^2)+P(3x^2) = 2a_0+4a_1x^2+\dots+[1+3^d]a_dx^{2d} = [R(x)]^2,\]
where $Q, R \in \mathbb{Z}[x]$ and both have degree $d$. Let the leading coefficients of $Q(x)$ and $R(x)$ be $m$ and $n$, respectively. Then,

\[m^2 = 2^{d+1}a_d\]\[n^2 = a_d(1+3^d),\]
where $m$ and $n$ are positive integers.

If $d$ is even, then $a_d = 2c^2$ for some integer $c$, so

\[\left(\frac{n}{c} \right)^2 = 2(1+3^d) \equiv 2 \pmod{3},\]
when $d \neq 0$. Thus, there are no solutions for even $d$ except for $d=0$.

If $d$ is odd, then $a_d = c^2$ for some integer $c$, so

\[\left(\frac{n}{c} \right)^2 - 3^d = 1,\]
which by Mihailescu only has $d=1$ as a solution. Hence, there are no solutions when $d > 1$.
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