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jlacosta   0
May 1, 2025
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0 replies
jlacosta
May 1, 2025
0 replies
Kingdom of Anisotropy
v_Enhance   24
N 2 minutes ago by deduck
Source: IMO Shortlist 2021 C4
The kingdom of Anisotropy consists of $n$ cities. For every two cities there exists exactly one direct one-way road between them. We say that a path from $X$ to $Y$ is a sequence of roads such that one can move from $X$ to $Y$ along this sequence without returning to an already visited city. A collection of paths is called diverse if no road belongs to two or more paths in the collection.

Let $A$ and $B$ be two distinct cities in Anisotropy. Let $N_{AB}$ denote the maximal number of paths in a diverse collection of paths from $A$ to $B$. Similarly, let $N_{BA}$ denote the maximal number of paths in a diverse collection of paths from $B$ to $A$. Prove that the equality $N_{AB} = N_{BA}$ holds if and only if the number of roads going out from $A$ is the same as the number of roads going out from $B$.

Proposed by Warut Suksompong, Thailand
24 replies
v_Enhance
Jul 12, 2022
deduck
2 minutes ago
Incentre-excentre geometry
oVlad   2
N 11 minutes ago by Double07
Source: Romania Junior TST 2025 Day 2 P2
Consider a scalene triangle $ABC$ with incentre $I$ and excentres $I_a,I_b,$ and $I_c$, opposite the vertices $A,B,$ and $C$ respectively. The incircle touches $BC,CA,$ and $AB$ at $E,F,$ and $G$ respectively. Prove that the circles $IEI_a,IFI_b,$ and $IGI_c$ have a common point other than $I$.
2 replies
oVlad
Yesterday at 12:54 PM
Double07
11 minutes ago
Great similarity
steven_zhang123   4
N 33 minutes ago by khina
Source: a friend
As shown in the figure, there are two points $D$ and $E$ outside triangle $ABC$ such that $\angle DAB = \angle CAE$ and $\angle ABD + \angle ACE = 180^{\circ}$. Connect $BE$ and $DC$, which intersect at point $O$. Let $AO$ intersect $BC$ at point $F$. Prove that $\angle ACE = \angle AFC$.
4 replies
1 viewing
steven_zhang123
5 hours ago
khina
33 minutes ago
Unexpected FE
Taco12   18
N 39 minutes ago by lpieleanu
Source: 2023 Fall TJ Proof TST, Problem 3
Find all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ such that for all integers $x$ and $y$, \[ f(2x+f(y))+f(f(2x))=y. \]
Calvin Wang and Zani Xu
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Taco12
Oct 6, 2023
lpieleanu
39 minutes ago
No more topics!
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
458 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
1540 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
1263 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
Y by
"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
623 posts
#60
Y by
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
424 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
1731 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
616 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
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#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
60 posts
#68
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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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