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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.

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.

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[*]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
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 em981
oldbeginner   15
N 2 minutes ago by sqing
Source: Own
Let $a, b, c>0, a+b+c=3$. Prove that
\[\sqrt{a+\frac{9}{b+2c}}+\sqrt{b+\frac{9}{c+2a}}+\sqrt{c+\frac{9}{a+2b}}+\frac{2(ab+bc+ca)}{9}\ge\frac{20}{3}\]
15 replies
1 viewing
oldbeginner
Sep 22, 2016
sqing
2 minutes ago
Functional equation
shobber   19
N an hour ago by Unique_solver
Source: Canada 2002
Let $\mathbb N = \{0,1,2,\ldots\}$. Determine all functions $f: \mathbb N \to \mathbb N$ such that
\[ xf(y) + yf(x) = (x+y) f(x^2+y^2)  \]
for all $x$ and $y$ in $\mathbb N$.
19 replies
shobber
Mar 5, 2006
Unique_solver
an hour ago
Prove the inequality
Butterfly   0
an hour ago
Let $a,b,c$ be real numbers such that $a+b+c=3$. Prove $$a^3b+b^3c+c^3a\le \frac{9}{32}(63+5\sqrt{105}).$$
0 replies
Butterfly
an hour ago
0 replies
Functional equation
shactal   1
N an hour ago by ariopro1387
Let $f:\mathbb R\to \mathbb R$ a function satifying $$f(x+2xy) = f(x) + 2f(xy)$$for all $x,y\in \mathbb R$.
If $f(1991)=a$, then what is $f(1992)$, the answer is in terms of $a$.
1 reply
shactal
3 hours ago
ariopro1387
an hour ago
interesting diophantiic fe in natural numbers
skellyrah   5
N an hour ago by skellyrah
Find all functions \( f : \mathbb{N} \to \mathbb{N} \) such that for all \( m, n \in \mathbb{N} \),
\[
mn + f(n!) = f(f(n))! + n \cdot \gcd(f(m), m!).
\]
5 replies
skellyrah
Yesterday at 8:01 AM
skellyrah
an hour ago
Non-linear Recursive Sequence
amogususususus   3
N 2 hours ago by SunnyEvan
Given $a_1=1$ and the recursive relation
$$a_{i+1}=a_i+\frac{1}{a_i}$$for all natural number $i$. Find the general form of $a_n$.

Is there any way to solve this problem and similar ones?
3 replies
amogususususus
Jan 24, 2025
SunnyEvan
2 hours ago
Inspired by 2025 Beijing
sqing   6
N 2 hours ago by sqing
Source: Own
Let $ a,b,c,d >0  $ and $ (a^2+b^2+c^2)(b^2+c^2+d^2)=36. $ Prove that
$$ab^2c^2d \leq 8$$$$a^2bcd^2 \leq 16$$$$ ab^3c^3d \leq \frac{2187}{128}$$$$ a^3bcd^3 \leq \frac{2187}{32}$$
6 replies
sqing
Yesterday at 4:56 PM
sqing
2 hours ago
Serbian selection contest for the IMO 2025 - P4
OgnjenTesic   2
N 2 hours ago by sqing-inequality-BUST
Source: Serbian selection contest for the IMO 2025
For a permutation $\pi$ of the set $A = \{1, 2, \ldots, 2025\}$, define its colorfulness as the greatest natural number $k$ such that:
- For all $1 \le i, j \le 2025$, $i \ne j$, if $|i - j| < k$, then $|\pi(i) - \pi(j)| \ge k$.
What is the maximum possible colorfulness of a permutation of the set $A$? Determine how many such permutations have maximal colorfulness.

Proposed by Pavle Martinović
2 replies
OgnjenTesic
May 22, 2025
sqing-inequality-BUST
2 hours ago
Nice "if and only if" function problem
ICE_CNME_4   14
N 2 hours ago by wh0nix
Let $f : [0, \infty) \to [0, \infty)$, $f(x) = \dfrac{ax + b}{cx + d}$, with $a, d \in (0, \infty)$, $b, c \in [0, \infty)$. Prove that there exists $n \in \mathbb{N}^*$ such that for every $x \geq 0$
\[
f_n(x) = \frac{x}{1 + nx}, \quad \text{if and only if } f(x) = \frac{x}{1 + x}, \quad \forall x \geq 0.
\](For $n \in \mathbb{N}^*$ and $x \geq 0$, the notation $f_n(x)$ represents $\underbrace{(f \circ f \circ \dots \circ f)}_{n \text{ times}}(x)$. )

Please do it at 9th grade level. Thank you!
14 replies
ICE_CNME_4
Friday at 7:23 PM
wh0nix
2 hours ago
2-var inequality
sqing   1
N 2 hours ago by sqing
Source: Own
Let $ a,b> 0 , ab(a+b+1) =3.$ Prove that$$\frac{1}{a^2}+\frac{1}{b^2}+\frac{24}{(a+b)^2} \geq 8$$$$ \frac{a}{b^2}+\frac{b}{a^2}+\frac{49}{(a+  b)^2} \geq \frac{57}{4}$$Let $ a,b> 0 ,  (a+b)(ab+1) =4.$ Prove that$$\frac{1}{a^2}+\frac{1}{b^2}+\frac{40}{(a+b)^2} \geq 12$$$$\frac{a}{b^2}+\frac{b}{a^2}+\frac{76}{(a+ b)^2}  \geq 21$$
1 reply
sqing
3 hours ago
sqing
2 hours ago
Balkan Mathematical Olympiad
ABCD1728   1
N 3 hours ago by ABCD1728
Can anyone provide the PDF version of the book "Balkan Mathematical Olympiads" by Mircea Becheanu and Bogdan Enescu (published by XYZ press in 2014), thanks!
1 reply
1 viewing
ABCD1728
Yesterday at 11:27 PM
ABCD1728
3 hours ago
area of quadrilateral
AlanLG   1
N 3 hours ago by Altronrren
Source: 3rd National Women´s Contest of Mexican Mathematics Olympiad 2024 , level 1+2 p5
Consider the acute-angled triangle \(ABC\). The segment \(BC\) measures 40 units. Let \(H\) be the orthocenter of triangle \(ABC\) and \(O\) its circumcenter. Let \(D\) be the foot of the altitude from \(A\) and \(E\) the foot of the altitude from \(B\). Additionally, point \(D\) divides the segment \(BC\) such that \(\frac{BD}{DC} = \frac{3}{5}\). If the perpendicular bisector of segment \(AC\) passes through point \(D\), calculate the area of quadrilateral \(DHEO\).
1 reply
AlanLG
Jun 14, 2024
Altronrren
3 hours ago
Inspired by 2025 SXTB
sqing   1
N 3 hours ago by sqing
Source: Own
Let $ a,b  $ be real number such that $ a^2+b^2=\frac12. $ Prove that
$$-\frac{\sqrt{9+6\sqrt 3}}{2}\leq(a+1)^2- (b-1)^2\leq\frac{\sqrt{9+6\sqrt 3}}{2}$$Let $ x $ be real number . Prove that
$$-\frac{2\sqrt 2}{3}\leq \frac{x}{x^2+1}+ \frac{ 2x}{x^2+4} \leq\frac{2\sqrt 2}{3}$$
1 reply
sqing
3 hours ago
sqing
3 hours ago
IMO Shortlist 2014 G2
hajimbrak   14
N 3 hours ago by ezpotd
Let $ABC$ be a triangle. The points $K, L,$ and $M$ lie on the segments $BC, CA,$ and $AB,$ respectively, such that the lines $AK, BL,$ and $CM$ intersect in a common point. Prove that it is possible to choose two of the triangles $ALM, BMK,$ and $CKL$ whose inradii sum up to at least the inradius of the triangle $ABC$.

Proposed by Estonia
14 replies
hajimbrak
Jul 11, 2015
ezpotd
3 hours ago
IMO ShortList 2002, algebra problem 4
orl   62
N Apr 10, 2025 by Ihatecombin
Source: IMO ShortList 2002, algebra problem 4
Find all functions $f$ from the reals to the reals such that \[ \left(f(x)+f(z)\right)\left(f(y)+f(t)\right)=f(xy-zt)+f(xt+yz)  \] for all real $x,y,z,t$.
62 replies
orl
Sep 28, 2004
Ihatecombin
Apr 10, 2025
IMO ShortList 2002, algebra problem 4
G H J
Source: IMO ShortList 2002, algebra problem 4
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ZETA_in_olympiad
2211 posts
#54 • 1 Y
Y by ImSh95
Similar problem from 2015 Korea https://artofproblemsolving.com/community/c6h1158064p5501261
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SatisfiedMagma
461 posts
#55 • 3 Y
Y by Pranav1056, ImSh95, Mango247
Solved with Pranav1056 and proxima1681.

Solution: We will show that $f \equiv 0, \frac{1}{2}$ and $f(x) = x^2$ for all $x \in \mathbb{R}$ are the only possible solutions. It is easy to check all of these works. Denote $P(x,z,y,t)$ as the assertion to the functional equation.

Throughout the solution, assume that $f$ is a non-constant function since the constant ones are fairly trivial. We will prove $f(x) = x^2$ for all $x$.

From $P(0,0,0,0)$ we get $f(0) \in \left\{0,\frac{1}{2}\right\}$. If $f(0) = \frac{1}{2}$, then $P(x,0,0,0)$ gives $f \equiv \frac{1}{2}$ which is a contradiction since we have assumed $f$ as non-constant. This gives $f(0) = 0$.


Claim: $f$ is multiplicative and strictly increasing over $\mathbb{R}^+$.

Proof: $P(x,0,y,0)$ gives
\[f(x)f(y) = f(xy)\]as desired. Also note that this gives $f(x) = 0 \iff x = 0$ for non-constant $f$. For proving increasing nature over $\mathbb{R}^+$, we consider the substitution $P(x,y,y,x)$ with $x,y>0$.
\[f(x^2 + y^2) = f(x^2) + f(y^2) + 2f(xy) > f(x^2)\]which proves the claim. $\square$
By a well-known result of Cauchy, we get to know that $f(x) = x^n$ for all $x \in \mathbb{R}^+$. $P(1,1,1,1)$ would yield $n = 2$. Since we already have $f(0) = 0^2$, we need to show $f(x) = x^2$ for negative $x$ as well.

This part is easy to see. Take $P(x,z,1,1)$ with $z$ some arbitrary negative $z$ and sufficiently large positive $x$ such that $x-z>0$. This would give $f(z) = z^2$ for all negative $z$. $\blacksquare$
This post has been edited 1 time. Last edited by SatisfiedMagma, Jan 17, 2023, 6:05 PM
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YaoAOPS
1541 posts
#56 • 1 Y
Y by ImSh95
Denote the assertion with $P(x, z, y, t)$.
Note that the constant solutions of $f(x) = \frac{1}{2}$ and $f(x) = 0$ work.
By $P(x, z, 0, 0)$ it follows that \[ (f(x) + f(z))2f(0) = 2f(0) \]so either $f(x)$ is constant for all $x$ or $f(0) = 0$. Assume the latter.

Claim: $f(x) \ge 0$ for all $x$ and $f$ is even.
Proof. By $P(x, x, x, x)$ it follows that \[ 4f(x)^2 = f(0) + f(2x^2) \]so $f(x) \ge 0$ for $x \ge 0$ and $f(-x) = \pm f(x)$.
Then, if $f(x) = -f(-x)$ for $x \ne 0$ then by $P(x, -x, t, t)$ it follows that \[ f(2xt) = 0 \]and thus $f(x)$ is uniformly $0$. $\blacksquare$

Claim: If there is a nonzero $x$ such $f(x) = 0$ then $f(x)$ is uniformly $0$. As such, we can take the codomain to be ${\mathbb R}^+$ over the domain ${\mathbb R}^+$.
Proof. Substitute $P(x, x, y, t)$ to get \[ f(xy - xt) + f(xy + xt) = 0 \]which implies that $f(x(y-t)) = f(x(y+t)) = 0$. $\blacksquare$
Assume that $0$ is the only root. Now, by $P(x, z, y, t)$ and $P(z, x, y, t)$ it follows that \[ f(yz - xt) + f(xy + zt) = f(yz + xt) + f(xy - zt) \]If we substitute it follows that for $ad = bc$ it follows that \[ f(a - d) + f(b + c) = f(a + d) + f(b - c) \]Over the positive reals, this reduces to ELMO 2011/4 which has the solution set of $f(x) = ax^2 + b$ for $a$ and $b$ both $\ge 0$ where equality does not hold for both.
Since $f$ is even, this holds for all $x \ne 0$. Then, it follows that since \[ f(x - x) + f(x^2 + 1) = f(x + x) + f(x^2 - 1), \]$b = 0$ and thus $f(x) = ax^2$ for $a > 0$ holds for all $x$.
It can be seen that by substituting in the original equation that only one value of $a$ can possible work.
Furthermore, \[ (x^2 + z^2)(y^2 + t^2) = (xy - zt)^2 + (xt + yz)^2 \]holds so $a = 1$ and thus $f(x) = x^2$.
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ezpotd
1289 posts
#57 • 1 Y
Y by ImSh95
I claim that the only solutions are $f(x) = x^2, f(x) = 0, f(x) = \frac 12$, and it is obvious that all of these work.

Let $P(x,z,y,t)$ denote the assertion, then $P(x,z,0,0)$ gives $2f(0)(f(x) + f(z)) = 2f(0)$, so either $f(x) + f(z) = 1$ for all $x$, or $f(0) = 0$. The former case gives $f(x) = \frac 12$ for all $x$. In the latter case, consider $P(x,0,0,t)$, which gives $f(x)f(t) = f(xt)$. This then immediately forces $f(x) = 0$ or $f(x)$ nonzero for $x$ nonzero. We assume the latter case. Then $P(1,0,1,0)$ combined with the condition before gives $f(1) = 1$. Then plugging in $P(1,1,1,1)$ gives $f(2) = 2$. Then we use induction to prove $f(n) = n^2$ for all positive integers. We have already proved the base cases. Now assume the claim is true for all $k \le n$. Notice $P(n,1,1,1)$ gives $2(n^2 + 1) = (n-1)^2 + f(n+1)$, so we have $f(n + 1) = (n+1)^2$. Then by multiplicity, it is easy to get $f(\frac 1n) = \frac{1}{n^2}$, and the claim for all positive rationals is easy. Then notice that $P(x,x,x,x)$ gives $4f(x)^2 = f(2x^2)$, so $f(x) = f(-x)$, so this claim is in fact true for all rationals.

Now we extend this to the reals. Firstly, notice that $f$ is always nonnegative. We first show that $f(x) \ge x^2$ for all $x$. Consider $P(x,r,r,x)$ for real $x$ and rational $r$, with $x>r$. This then gives $(f(x)+r^2)^2 = f(x^2 + r^2)$. We can write this as $f(x^2) = (f(\sqrt{x^2 - r^2}) + r^2)^2$. Now assume $f(x) < x^2$ for some real $x$. We can then find some rational $r^4$ between $f(x),x^2$, such that $r^4 > f(x) = (f(\sqrt{x - r^2}) + r^2)^2 > r^4$, contradiction. A flipped argument can show the other side, so $f(x) = x^2$.
This post has been edited 2 times. Last edited by ezpotd, May 24, 2023, 4:47 AM
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F10tothepowerof34
195 posts
#58 • 1 Y
Y by ImSh95
Let $P(x,y,z,t):=(f(x)+f(z))(f(y)+f(t))=f(xy-zt)+f(xt+yz)$

$P(0,0,0,0)$ yields $4f(0)^2=2f(0)\Longrightarrow 2f(0)(2f(0)-1)=0$ thus $f(0)=0$ or $f(0)=\frac{1}{2}$

Case 1: $f(0)=0$
$P(x,y,0,0)$ yields $f(x)f(y)=f(xy)$ therefore $f$ is a multiplicative function, and $f(x^2)=f(x)^2$ for $x\neq0$, however we can extend this for all $x$ as $f(0)=0$
$P(x,x,x,x)$ yields $4f(x)^2=f(2x^2)$ thus $f(x)\ge0\text{ for }x\ge0$
$P(y,0,x,y)$ yields $f(y)^2+f(xy)=f(-xy)+f(y^2)\Longrightarrow f(xy)=f(-xy)$ therefore $f$ is even.
Furthermore let $x,y\ge0$, thus $P(\sqrt{x},\sqrt{y},\sqrt{y},\sqrt{x})$ yields $f(x+y)=\left(f(x)+f(y)\right)^2\ge f(\sqrt{x})^2=f(x)$ thus $f(x+y)\ge f(x)$ for $x,y\ge0$ therefore $f$ is monotonically increasing over positive reals. Thus $f(x)=x^c$, and by plugging into our original $fE$ we obtain that $c=2$ and $f(x)=x^2, \forall x\in\mathbb{R}$

Case 2: $f(0)=\frac{1}{2}$
$P(0,x,0,x)$ yields $2f(x)=1\Longrightarrow f(x)=\frac{1}{2}, \forall x\in \mathbb{R}$

Case 3: $f(x)=0$
Assume that $f$ is constant and not equal to $\frac{1}{2}$, when we plug in $f(x)=c$ into our original $fE$ we obtain $4c^2=2c\Longrightarrow 2c(2c-1)=0$, this forces $c=0$
Therefore $f(x)=0, \forall x\in\mathbb{R}$

So, to sum up $\boxed{f(x)=x^2, f(x)=\frac{1}{2}\text{ and }f(x)=0, \forall x\in\mathbb{R}}$ $\blacksquare$.
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signifance
140 posts
#59
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"good easy problem. solved in my head in a few minutes."

$x=z=0\implies 2f(0)(f(y)+f(t))=2f(0)\implies\{f\equiv\frac12,f(0)=0\}$. In the latter, we have f(-xy)=f(x)f(y)=f(xy) by setting two variables equal to 0 (ill omit which ones). With $x=t,y=z\implies(f(x)+f(y))^2=f(x^2+y^2)$ so f is bounded over $x\ge0$, whence multiplicity with bounded with f(-xy)=f(xy) implies $f(x)=x^2$. Otherwise, for any weird edge cases check that the other constant sol is f(x)=0.
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cursed_tangent1434
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#60
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The answers are

\[f\equiv 0,f\equiv \frac{1}{2}  \text{ and } f(x)=x^2 \text{ for all}x\in \mathbb{R}\]
It’s easy to see that these functions satisfy the given equation. We now show these are the only solutions.

$P(0,0,0,0)$ gives us that
\[4f(0)^2 = f(0)\]Thus, $f(0)=0$ or $f(0)=\frac{1}{2}$. We then break into two cases.

\textbf{Case 1 : }$f(0)=0$. Note that now, $f(x,y,0,0)$ gives,
\[f(x)f(y)=f(xy)\]for all $x,y \in \mathbb{R}$. Thus, $f(1)=0$ or $f(1)=1$. If $f(1)=0$, we know that the above equation gives $f\equiv 0$ and if $f(1)=1$ then $f(x)=x^k$ for all $x \in \mathbb{R}$ for some constant $k \in \mathbb{R}$. Consider $P(a,a,a,a)$ for some $a>1 \in \mathbb{R}$. This gives,
\[4f(a)^2=f(2a^2)\]plugging in $f(x)=x^k$ gives us that $4a^{2k}=2^ka^{2k}$ from which it is quite clear that we must have $k=2$. Thus, the only solutions from this case are $f\equiv 0$ and $f(x)=x^2$ for all $x\in \mathbb{R}$.

\textbf{Case 2 : } $f(0) = \frac{1}{2}$. Plugging in $y=z=t$ gives us that,
\[f(x)+\frac{1}{2} = \frac{1}{2} +\frac{1}{2} \]for all $x\in \mathbb{R}$. Thus, $f\equiv \frac{1}{2}$ in this case.

This implies that indeed all solutions to the given functional equation are of the claimed forms and we are done.
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kamatadu
480 posts
#61
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The solutions are $f\equiv 0$, $f\equiv \dfrac{1}{2}$, $f\equiv x^2$.

$P(0,0,0,0)\implies 4f(0)^2 = 2f(0) \implies f(0) \in \left\{0,\frac{1}{2}\right\}$.

$\textbf{Case 1:}$ If $f(0) = 0$. Now if $f\not\equiv 0$, let $f(p) \neq 0$.

$P(0,y,z,0) \implies f(z) f(y) = f(zy)$.

Put $y=z=1$ to get $f(1) \in \left\{0,1\right\}$.
$\textbf{Case 1.1:}$ If $f(1) = 1$.

\begin{align*}
    P(0,y,p,t)&\implies f(p)(f(y) + f(t)) = f(-pt) + f(yp)\\
    &\implies f(p)(f(y) + f(t)) = f(p) (f(y) + f(-t))\\
    &\implies f(t) = f(-t)
.\end{align*}
Now, $P(1,1,z,1) \implies (1+f(z))(2) = f(1-z) + f(1+z)$.

Now by induction, $f(n) = n^2$ and by using multiplicativity, $f(q) = q^2$ for all $q\in \mathbb Q$.

Also, $P(x,x,z,z) \implies f(2xz) = (f(x) + f(z))^2 \implies f \ge 0$.

Then for any $t\neq 0$,
\begin{align*}
    P\left(x,y,\frac{xy}{24},t\right)\implies f(xy)+f(xt) + f\left(\frac{xy}{2t} \cdot y\right) + f\left(\frac{xy}{2}\right) = f\left(\frac{xy}{2}\right) + f\left(xt + \frac{xy^2}{2t}\right)\\
    \implies f(2t^2 + y^2) = f(2t^2) + f(2yt) \ge f(2t^2) \\
    \implies f(x + \varepsilon) \ge f(x) \;\forall\; x,\varepsilon \in\mathbb R^+
.\end{align*}
Now using multiplicativity and using some $\log$ stuff, we can convert it into the multiplicativity into Cauchy and then use the monotone fact to show that $f=x^k$ and $f(2) = 4 \implies k = 2 \implies f\equiv x^2$.
$\textbf{Case 1.2:}$ Otherwise if $f(1) = 0$.

Then $f(z) = f(z) \cdot f(1) = 0 \implies f\equiv 0$.
$\textbf{Case 2:}$ Finally, if otherwise at the beginning we had $f(0) = \dfrac{1}{2}$, then $P(x,0,x,0)\implies 2f(x) = 1 \implies f(x) = \dfrac{1}{2} \implies f\equiv \dfrac{1}{2}$. :yoda:
This post has been edited 2 times. Last edited by kamatadu, Jan 1, 2024, 7:56 PM
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KevinYang2.71
428 posts
#62 • 1 Y
Y by deduck
Solved with the help of megarnie.

We claim the only functions are $\boxed{f(x)\equiv 0}$, $\boxed{f(x)\equiv\frac{1}{2}}$, and $\boxed{f(x)\equiv x^2}$. It is easy to check that these all work.

Let $P(x,y,z,t)$ denote the given assertion. $P(0,0,0,0)$ yields $2f(0)^2=f(0)$ so $f(0)\in\left\{0,\frac{1}{2}\right\}$. If $f(0)=\frac{1}{2}$ then $P(x,0,0,0)$ gives $f(x)=\frac{1}{2}$ so $f(x)\equiv\frac{1}{2}$.

So assume $f(0)=0$. $P(x,y,0,0)$ gives $f(x)f(y)=f(xy)$. It follows that $f(1)^2=f(1)$ so $f(1)\in\{0,1\}$. If $f(1)=0$ then $f(x\cdot 1)=f(x)f(1)=0$ so $f(x)\equiv 0$. So we may assume $f(1)=1$. $P(0,0,x,1)$ gives $f(x)=f(-x)$. We also have $f(x^2)=f(x)^2\geq 0$ for all $x$. If $f(a)=0$ for some $a\neq 0$ then
\[
f(1)=f\left(\frac{1}{a}\cdot a\right)=f\left(\frac{1}{a}\right)f(a)=0,
\]a contradiction. Thus $f(x)>0$ for all $x\neq 0$.

From $P(x,y,y,x)$ we get $(f(x)+f(y))^2=f(x^2+y^2)$ so $f(x^2+y^2)=f(x^2)+f(y^2)+f(xy)>f(x^2)$. Thus for all $x>y>0$, we have
\[
f(x)=f\left(\sqrt{y}^2+\sqrt{x-y}^2\right)>f(y)
\]so $f$ is increasing in $\mathbb{R}^+$. Let $g(x):=\log f(e^x)$, which is defined since $f(e^x)$ is always positive. Note that $g$ is increasing. Since $g(x+y)=g(x)+g(y)$ and $g$ is increasing, it follows that $g(x)=cx$ for some constant $c$. Thus $f(e^x)=(e^x)^c$ so $f(x)=x^c$ for all $x>0$. $P(1,1,1,1)$ gives $f(2)=4$ so $c=2$. Thus $f(x)=x^2$ for all $x>0$. Since $f(-x)=f(x)$, it follows that $f(x)\equiv x^2$. $\square$
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joshualiu315
2534 posts
#63
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The answer is $f(x) = \boxed{0, \tfrac{1}{2},x^2}$, which works. Denote the given assertion as $P(x,y,z,t)$.

Plugging in $P(0,0,0,0)$ yields

\[4f(0)^2 = 2f(0),\]
which means $f(0) = 0$ or $f(0) = \tfrac{1}{2}$; we will tackle the latter case first. Note that $P(x,0,x,0)$ yields $f \equiv \tfrac{1}{2}$.

Now, suppose that $f(0)=0$. Then, $P(x,y,0,0)$ yields

\[f(x)f(y) = f(xy),\]
upon which we see that $f(x) = 0 \iff x = 0$ unless $f \equiv 0$, which is indeed a solution.

Plugging in $P(x,y,y,x)$ gives

\[ f(x^2+y^2) = (f(x)+f(y))^2 > f(x^2),\]
when considered over $\mathbb{R}^+$. Hence, $f$ is monotonic over $\mathbb{R}^+$, implying it is of the form $f(x) = x^n$. Simply setting $P(x,x,x,x)$ shows $n=2$, giving us our final solution set.
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pie854
243 posts
#64
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Let $P(x,y,z,t)$ denote the given functional equation. We claim that the solutions are $f(x)=0$ for all $x$ and $f(x)=x$ for all $x$ and $f(x)=1/2$ for all $x$. All of them work, as we can easily verify.

We have $$P(0,0,0,0) \Rightarrow 4f(0)^2=2f(0)\Rightarrow f(0)\in \{0,1/2\}.$$If $f(0)=1/2$, then $P(x,0,0,0)$ implies that $f(x)=1/2$ for all $x$. So we get one of our claimed functions.

Now assume $f(0)=0$. Then $$P(x,y,0,0)\Rightarrow \ f(x)f(y)=f(xy)$$so $f$ is a multiplicative function. So our equation can also take the form $$f(xy)+f(xt)+f(yz)+f(zt)=f(xy-zt)+f(xy+yz).$$On this form, $P(0,1,1,z)$ gives us $f(z)=f(-z)$. So $f$ is an even function. Also we have $$P(\sqrt x, \sqrt y, \sqrt y, \sqrt x)\Rightarrow f(x+y)=\left (f\left (\sqrt x\right )+f\left (\sqrt y\right )\right)^2=\left (\sqrt{f(x)}+\sqrt{f(y)}\right )^2 \qquad (1)$$This implies $f(x)\geq 0$ if $x\geq 0$. But since $f$ is even, $f(x)\geq 0$ for all $x$. So we can define the function $g:\mathbb R\to \mathbb R_{\geq 0}$ such that $g(x)=\sqrt{f(x)}$. Note that $g$ is multiplicative, since $f$ is. From $(1)$ we get $g(x+y)=g(x)+g(y)$, so $g$ is additive. It is well know that the only additive and multiplicative functions are the identity function and zero function. So either $g\equiv 0$, in which case $f\equiv 0$. Or, $g(x)\equiv x$, in which case $f(x)\equiv x^2$. And these are all the solutions we claimed.
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OronSH
1748 posts
#65 • 1 Y
Y by megarnie
We claim $f(x)=0,\tfrac12,x^2$ are the solutions.

First $P(0,0,x,x)$ gives $f(x)=\tfrac12$ or $f(0)=0$.

If $f(0)=0$ then $P(x,y,x,y),P(x,y,y,x)$ gives $(f(x)+f(y))^2=f(x^2-y^2)+f(2xy)=f(x^2+y^2)$. If $a,b$ are nonnegative reals then $x=\sqrt{\frac{\sqrt{a+b}+\sqrt a}2},y=\sqrt{\frac{\sqrt{a+b}-\sqrt a}2}$ gives $f(\sqrt a)+f(\sqrt b)=f(\sqrt{a+b})$. We also have $f(x)\ge 0$ for $x\ge 0$, and $f$ is even by swapping $x,y$.

Define $g$ such that $g(x)=f(\sqrt x)$ for $x\ge 0$ and $g$ is odd. Then $g$ is additive and $g(x)\ge 0$ for $x\ge 0$, so $g(x)=cx$ and $f(x)=cx^2$. Checking, $c=0,1$ work.
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Ilikeminecraft
658 posts
#66
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The solution set is $f\equiv 0, \frac12, x^2.$

$(0, y, 0, t) \implies 2f(0)(f(y) + f(t)) = 2f(0).$ Thus, either $f\equiv \frac12$ or $f(0) = 0.$

$(0, 0, z, t) \implies f(z)f(t) = f(-zt)$ while $(0, y, z, 0) \implies f(y)f(z) = f(yz).$ Thus, $f$ is both multiplicative and even.

$(x, y, y, x) \implies (f(x) + f(y))^2 = f(x^2 + y^2).$ Thus, if $x\geq0,$ then $f\geq0.$ Sub $g \equiv \ln f(e^x)$ for $x>0$ to get $g(x + y) = g(x) + g(y),$ and $g$ is increasing over an interval. Thus, $g\equiv \alpha x.$ Thus, $f\equiv |x|^\alpha, x^\alpha.$ Simple testing gives us $f\equiv x^2.$
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cj13609517288
1922 posts
#67
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The answer is $\boxed{f(x)=x^2}$, $\boxed{f(x)=0}$, and $\boxed{f(x)=\frac12}$.

Comparing $P(x,y,z,t)$ and $P(z,x,t,y)$ gives that $f$ is even. Then comparing $P(x,y,z,t)$ and $P(x,t,z,y)$ gives
\[f(xy+zt)-f(xy-zt)=f(xt+yz)-f(xt-yz).\]Now define $g(x)=f(\sqrt{x})$ as a function for $x\in\mathbb{R}^{\ge 0}$. Note that if we set a positive constant $c=xyzt$, then we may choose $xy$ and $xt$ to be whatever values we want. Thus we may choose some $xy,xt\ge\sqrt{c}$ then get
\[f\left(x+\frac cx\right)-f\left(x-\frac cx\right)\]is independent of $x$, as long as $x\ge\sqrt{c}$. So then
\[g(x^2+4c)-g(x^2)\]is independent of $x$. At this point we have gotten Jensen's FE. But also $P(x,x,x,x)$ gives $f(2x^2)$ is bounded below, so $g(x)$ is bounded below. So $g$ is linear, so $f(x)=ax^2+b$ (recalling that $f$ is even). Now it is simple to check that only $f(x)=x^2$, $f(x)=0$, and $f(x)=\frac12$ work. $\blacksquare$
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Ihatecombin
66 posts
#68
Y by
Great problem, I really enjoyed it. The solutions are \(\boxed{f(x) \equiv 0,\frac{1}{2},x^2}\), these clearly work, we shall show that they are the only solutions.
Claim 1: \(f(0) = 0\) or \(f(x) \equiv \frac{1}{2}\).

Proof: We can just substitute \(x=y=t=z=0\), this gives us
\[4{f(0)}^2 = 2f(0)\]which gives us \(f(0) = 0\) or \(f(0) = \frac{1}{2}\), if \(f(0) = 0\) we are done, otherwise assume \(f(0) = \frac{1}{2}\).
We can just substitute \(x=z=t=0\), this gives us
\[(1)(f(y) + f(0)) = 2f(0) \Longrightarrow f(y) = \frac{1}{2}\]$\blacksquare$
Hence assume \(f(0) = 0\), a very important property arises from this
Claim 2: \(f(x)\) is multiplicative i.e \(f(xy) = f(x)f(y)\), and \(f(x)\) is even.

Proof: We just substitute \(x=t=0\), this gives us
\[f(z)f(y) = f(yz)\]The even condition comes from subbing \(x=y=0\), this gives us
\[f(z)f(t) = f(-zt)\]since \(f(x)\) is multiplicative, it then follows that
\[f(-zt) = f(zt)\]$\blacksquare$
Claim 3: \(f(x) \equiv 0\) or \(f(1) = 1\)

Proof: Since \(f(x)\) is multiplicative, we know that
\[f(1)f(1) = f(1\cdot 1) = f(1)\]This gives us \(f(1) = 0\) or \(f(1) = 1\), if \(f(1) = 0\), then we can substitute \(x=1\), \(z=0\), \(t=0\) to obtain
\[f(y) = 0\]$\blacksquare$
Hence we shall assume \(f(1) = 1\)
Claim 4: \(f(x) = x^2\) for all \(x \in \mathbb{Q}\)

Proof: Notice that by the multiplicative property and due to the fact that \(f(1) = 1\) we know that
\[f\left(\frac{1}{b}\right) = \frac{1}{f(b)} \Longrightarrow f\left(\frac{a}{b}\right) = \frac{f(a)}{f(b)}\]Hence it suffices to show that \(f(x) = x^2\) for all \(x \in \mathbb{Z}\). Due to the fact that \(f(x)\) is even, it suffices to show that \(f(x) = x^2\) for the natural numbers.
However we can substitute \(x=z\) and \(t = 1\) to obtain
\[2(f(x))(f(y) + 1) = f(x(y-1)) + f(x(1+y))\]thus we obtain
\[2f(x)(f(y) + 1) = f(x)f(y-1) + f(x)f(1+y) \Longrightarrow 2(f(y) + 1) = f(y-1) + f(y+1)\]which gives
\[f(y+1) = 2f(y) - f(y-1) + 2\]We use induction. Due to the fact that \(f(0) = 0\) and \(f(1) = 1\), our base case is done. We can just finish the induction step by noting that the induction hypothesis implies
\[f(y+1) = 2y^2 - (y^2-2y+1) + 2 = y^2 + 2y + 1 = {(y+1)}^2\]hence we are done. $\blacksquare$
Claim 5: \({(f(a) + f(b))}^2 = f(a^2 + b^2)\), \(f(x) \geq 0\) for all \(x\), and \(f(x) < 1\) for all \(0 \leq x< 1\).

Proof: We can just substitute \(x = y\) and \(z=-t\) to obtain
\[{(f(x) + f(t))}^2 = f(x^2 + t^2)\]Notice that since \(f(x)\) is even, the fact that \(f(x) \geq 0\) for all \(x\) immediately follows.
To show that \(f(x) < 1\) for all \(0 \leq x< 1\), we can just assume otherwise that there exists some \(x\) such that \(f(x) > 1\), substituting \(t = \sqrt{1-x^2}\) gives us a contradiction. $\blacksquare$

Now we shall show that \(f(x) = x^2\) for all \(0 \leq x \leq 1\), notice that this will finish the problem due to the fact that
if \(f(a) \neq a^2\) and \(a>1\), we can just find some rational \(b\) such that \(b < \frac{1}{a}\), and since \(f(x) = x^2\) for the rationals we obtain
\[{(ab)}^2 = f(ab) = f(a)f(b) = f(a)b^2\]which is a contradiction.
Claim 6: \(f(x) = x^2\) for all \(0 \leq x \leq 1\).

Proof: Claim \(5\) is the key, we can just substitute \(a = \sin(\theta)\) and \(b = \cos(\theta)\) (the trig is just to avoid having to write the awkward \(\sqrt{1-a^2}\)).
Notice that
\[f(\sin(\theta)) + f(\cos(\theta)) = 1\]Suppose there exists some \(\theta\), such that \(f(\sin(\theta)) \neq \sin^2(\theta)\). Without loss of generality assume that
\(f(\sin(\theta)) > \sin^2(\theta)\) (since otherwise \(f(\cos(\theta)) > \cos^2(\theta)\) and we would be able to make the same argument that we are about to make).
Thus there exists some \(\delta\), such that \(f(\sin(\theta)) = \sin^2(\theta) + \delta\). Now since the rationals are dense in the reals, we can find some number \(a \in \mathbb{Q}\) such that
\(a\sin(\theta)\) is arbitrarily close to \(1\). Now notice that
\[f(a\sin(\theta)) = f(a)f(\sin(\theta)) = a^2(\sin^2(\theta) + \delta)\]Notice that this implies
\[\lim_{a \to \frac{1}{\sin(\theta)}} f(a\sin(\theta)) = 1 + \frac{\delta}{\sin^2(\theta)} > 0\]hence by the delta-epsilon definition of a limit, we can find some \(a\) such that \(a\sin(\theta) < 1\) and \(f(a\sin(\theta)) > 1\), which contradicts claim \(5\), thus we are done since we have shown that this claim is sufficient. $\blacksquare$
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