Set $x=y=z=t=0$ then we will get $f(0)(2f(0)-1)=0$. So we will have two cases 1) $f(0)=o$ and 2)$f(0)=\frac{1}{2}$ 1) $f(0)=0$ Set $z=t=0$ then we will have $f(x)f(y)=f(xy)$ $(*)$ we know that $(*)$ is Cauchy's equation and its solutions are: (a) $f(x)=|x|^c$ (where $c$ is constant an we will look for it), (b) $f(x)=sgnx\times x$ (c) $f(x)=0$ in (a) put instead of each function which is given we will yield: $|xy|^c+|xt|^c+|zy|^c+|zt|^c=|xy-zt|^c+|xt+yz|^c$ now denote $|xy|=a, |xt|=b, |zy|=d$ and $|zt|=f$ $a^c+b^c+d^c+f^c=(a-f)^c+(d+b)^c$, so we have new function as example $f(a)=a^c$. Set $b=d=f=a$ then we will have $4a^c=(2a)^c$ take $c$ root from both sides, and have $4^{\frac{1}{c}} a =2a$ cancel $a$ an have $4^{\frac{1}{c}} =2$ from here $c=2$ only solution, so $f(x)=x^2$ (b) cannot be solution because it will give $f(x)=|x|$ (c) is obviously solution Now we will look in case 2) $f(0)=\frac{1}{2}$ Set $z=x$ and $y=t=0$ then $2f(x)=1\rightarrow f(x)=\frac{1}{2}$ So the solutions are: $f(x)=x^2$, $f(x)=0$, and $f(x)=\frac{1}{2}$, for all $x$ element of real numbers.

cauchy equation is valid only for rationals. we need an extra condition in this case (we can show that f is monotone)

can you show how to prove that $f(x)$ is monotone. I think that this is only dificulty of this problem.

$x=z,y=t=0 \Rightarrow 2f(x)\cdot 2f(0)=2f(0)$ for all $x\in\mathbb{R}$ $\Rightarrow f(0)=0$ or $f\equiv{1\over 2}$. Assume $f(0)=0$. $z=t=0 \Rightarrow f(xy)=f(x)f(y)$ for all $x,y\in\mathbb{R}$ (*) $x=y=0 \Rightarrow f(-zt)=f(z)f(t)=f(zt)$ for all $z,t\in\mathbb{R}$ So $f(-x)=f(x)$ for all $x\in\mathbb{R}$ (**) and $f(x)=f(\sqrt{x})^{2}\geq 0$ for all $x\geq 0$ (***) and $f(x)=f(x\cdot 1)=f(x)f(1)$ for all $x\in\mathbb{R}$ $\Rightarrow f(1)=1$ or $f\equiv 0$. Assume $f(1)=1$. $x=y=z=t=1\Rightarrow f(2)=4$ Then (*) gives $f(2^{r})=4^{r}=(2^{r})^{2}$ for all $r\in\mathbb{Q}$. (****) And now $x=t,y=z \Rightarrow (f(x)+f(y))^{2}=f(x^{2}+y^{2})=f(\sqrt{x^{2}+y^{2}})^{2}\\ \hspace*{0.9in}\Rightarrow f(y)\leq f(x)+f(y)=f(\sqrt{x^{2}+y^{2}})$ for all $x,y\geq 0$ using (***). So $f(x)=f(\sqrt{(\sqrt{x^{2}-y^{2}})^{2}+y^{2}})\geq f(y)$ for all $x\geq y\geq 0$. This and (****) give $f(x)=x^{2}$ for all $x>0$ and (**) and $f(0)=0$ give $f(x)=x^{2}$ for all $x\in\mathbb{R}$.

I'll skip $ f(0) = 0$. Set $ x = y, z = t$, we have $ (f(x) + f(z))^2 = f(x^2 - z^2) + f(2xz)$. Set $ x = t, z = y$, we have $ (f(x) + f(z))^2 = f(x^2 + z^2)$. Therefore $ f(x^2 + z^2) - f(x^2 - z^2) = f(2xz)$. \[ \lim_{z\rightarrow 0}\dfrac{f(x^2 + z^2) - f(x^2 - z^2)}{2z^2} = \lim_{z\rightarrow 0}\dfrac{f(2xz)}{2z^2} \] Let $ h(x) = \frac {f(x)}{x^2}$, then $ f'(x^2) = \lim_{z\rightarrow 0}h(2xz)\cdot 2x^2$. So $ \lim_{r\rightarrow 0}h(r) = \dfrac{f'(x^2)}{2x^2}$. This means $ \dfrac{f'(x^2)}{2x^2}$ is constant for all $ x$, therefore $ f(x)$ is constant or $ f(x) = kx^2$, and we can easily check that $ k = 1$.

Where did you show the function was differentiable?

MellowMelon wrote: Where did you show the function was differentiable? I just found out the same problem. But it can be fixed easily: $ \lim_{z\rightarrow 0}\frac{f(x^{2}+z^{2})-f(x^{2}-z^{2})}{2z^{2}}=\lim_{z\rightarrow 0}h(2xz)\cdot 2x^{2}$ Since the limit exist (we are fixing $ x$ and let $ z$ go to $ 0$), the funciton must be differentiable. Is that right?

That requires continuity at $ 0$, which you also have the burden of showing. (trying to use derivatives is almost certainly going to fail, by the way)

here is my solution; $ (f(x) + f(z))(f(y) + f(t)) = f(xy - zt) + f(xt + yz)$ if $ x = y = z = t = 0$ $ \implies$ $ 2f(0)2f(0) = 2f(0)$ $ \implies$ $ f(0) = \frac {1}{2}$ or $ f(0) = 0$ $ i)f(0) = \frac {1}{2}$ $ x = z = 0$ $ \implies$ $ 1[f(y) + f(t)] = 1$ $ \implies$ $ f(y) + f(t) = 1$ $ \implies$ $ f(x) = \frac {1}{2}$ $ ii)f(0) = 0$ $ \implies$ $ z = t = 0$ $ \implies$ $ f(x)f(y) = f(xy)$ $ \implies$ $ f(1)(f(1) - 1) = 0$ $ \implies$ $ f(1) = 0$ or $ f(1) = 1$ if $ f(1) = 0$,$ y = t = 1$ $ \implies$ $ f(x - z) + f(x + z) = 0$ if $ x = z$ $ \implies$ $ f(x) = 0$ $ f(1) = 1$ $ \implies$ $ f(x + z) + f(x - z) = 2f(x) + 2f(z)$ $ \implies$ $ f(2x) = 4f(x) = f(2)f(x)$ $ \implies$ $ f(2) = 4$ $ \implies$ $ 2f(a) + 2f(b) = f(a - b) + f(a + b)$ $ f(1) = 1,f(2) = 4$ ,Let $ f(n) = n^{2}$ $ a = nb$ $ \implies$ $ 2f(nb) + 2f(b) = f((n - 1)b) + f((n + 1)b)$ $ \implies$ $ (2n^{2} + 2)f(b) = [(n - 1)^{2} + f(n + 1)]f(b)$ $ \implies$ $ f(n + 1) = (n + 1)^{2}$ $ \implies$ $ f(x) = x^{2}$ $ \forall x\in\mathbb{N}$ $ 2f(a) + 2f(b) = f(a - b) + f(a + b)$ ,if $ a = 0,b = 1$ $ \implies$ $ f(1) = f( - 1) = 1$ $ \implies$ $ f(x) = f( - x)$ $ \implies$ $ f(x) = x^{2}$ $ \forall x\in\mathbb{Z}$ $ f(xy) = f(x)f(y)$ if $ y = \frac {1}{x}(x\in\mathbb{Z})$ $ \implies$ $ f(\frac {1}{x}) = \frac {1}{x^{2}}$ $ \implies$ $ f(\frac {p}{q}) = \frac {p^{2}}{q^{2}}$ $ \implies$ $ f(x) = x^{2}$ $ \forall x\in\mathbb{Q}$ We can analyze this function $ x > 0$ $ f(xy) = f(x)f(y)$, Let $ y = \frac {p}{q}$,$ p > q$ $ \implies$ $ f(x\frac {p}{q}) = f(x)\frac {p^{2}}{q^{2}}$ $ \implies$ $ f(x)$ is increasing function for $ x > 0$ Let $ \exists u\in\mathbb{R^{ + }}$,$ f(u) > u^{2}$ $ \implies$ $ u^{2} < r^{2} = f(r) < f(u),(r\in\mathbb{Q})$ $ \implies$ $ f(u) > f(r),r > u$ Contradiction! Let $ \exists u\in\mathbb{R^{ + }}$,$ f(u) < u^{2}$ $ \implies$ $ f(u) < f(r) = r^{2} < u^{2}$ $ \implies$ $ f(u) < f(r),u > r$ Contradiction $ \implies$ $ f(x) = x^{2}$ $ \forall\mathbb{R^{ + }}$ $ f(x) = f( - x)$ $ \implies$ $ f(x) = x^{2}$ $ \implies$ $ \boxed{f(x) = 0},\boxed{f(x) = \frac {1}{2}},\boxed{f(x) = x^{2}}$

my solution is wrongly for increasing $ x=t,y=z$ $ \implies$ $ [f(a) + f(c)]^{2} = f(a^{2} + c^{2})$ $ \implies$ $ x > 0$ $ \implies$ $ f(x)\geq 0$ $ \implies$ $ f(a^{2} + c^{2})\geq f(a^{2})$ $ \implies$ $ f$ is increasing for $ x > 0$ the rest is easy

By setting $x=0=z$ we see that for all $y,t$ we must have $2f(0)(f(y)+f(t))=2f(0)$. Therefore $f=1/2$ or $f(0)=0$. (Note that $f=1/2$ works, so WLOG $f(0)=0$). By setting $z=t=0$ we see that $f(x)f(y)=f(xy)$ for any $xy$. In particular we get that $f(p) \ge 0$ for any positive $p$, by setting $x=y=\sqrt{p}$. By setting $x=\frac{zt}{y}$ for any $z,t,y$ and by using multiplicity we get that $f(\frac{zt^2}{y}+yz)=(f(x)+f(z))(f(y)+f(t))=f(xy)+f(xt)+f(zy)+f(zt)=2f(zt)+f(\frac{zt^2}{y})+f(zy)$ and by multiplting by $f(y)/f(z)$ we get that $f(t^2+y^2)=(f(t)+f(y))^2$ by using multiplicity. Let $g=\sqrt{f}$ be defined on $\mathbb{R}^{+}$. Then for all $y,t$ positive, we get that $g(t^2+y^2)=g(t)^2+g(y)^2=g(t^2)+g(y^2)$ and so $g$ is additive. Since it's defined on the positives we get that $g(x)=kx$ for a constant $x$, and also $k$ must be positive (or $0$). So we have that $f(x)=Cx^2$ for positive $x$. Since $f$ is multiplicative we get that $C=1$ or $C=0$, since $C \ge 0$. If $C=0$ then because of multiplicity, $f=0$. Otherwise, $f(x)=x^2$ for positive $x$. Finally, in the original equation we have, by having $x,y,t$ very big positives and $z$ a small negative number such that $xt+yz > 0$, that $(x^2+f(z))(y^2+t^2) = (xy-zt)^2 + (xt+yz)^2=(xy)^2+(xt)^2+z^2(y^2+t^2)$ and so $f(z)(y^2+t^2)=z^2(t^2+y^2)$ and so $f(z)=z^2$. So we have $f=1/2$; $f=0$ or $f(x)=x^2$. Done.

The solutions are $f(x) = x^2$, $f(x) = 0$, and $f(x) = \frac{1}{2}$. Set all variables to $0$ and we get $f(0) = 0$ or $f(0) = \frac{1}{2}$. If we have that $f(0) = \frac{1}{2}$, letting $x = z = 0$ and $t = y$ gives us $2f(y) = 1$, so we see that $\boxed{f(y) = \frac{1}{2}}$ for all $y \in \mathbb{R}$. So now assume that $f(0) = 0$. Plugging in $y = t = 1, z = 0$, we have $$2f(1)f(x) = 2f(x).$$If $\boxed{f(x) = 0}$ for all real $x$, we have another solution. Otherwise, assume there exists an $x$ such that $f(x) \neq 0$, so we get $f(1) = 1$. Now letting $y = t = z = 1$, we get $$2f(x) + 2 = f(x+1) + f(x-1).$$This is a recurrence relation, and from induction, we can easily show that $f(n) = n^2$ for all $n \in \mathbb{Z}$. Now we prove multiplicity: set $z = t = 0$ and we get $$f(x)\cdot f(y) = f(xy).$$Setting $x = q$ and $y = \frac{p}{q}$ for integers $p$ and $q$ gives us $$f\left( \frac{p}{q} \right) = \frac{f(p)}{f(q)} = \frac{p^2}{q^2},$$so $f(x) = x^2$ for all $x \in \mathbb{Q}$. To show that $f$ is even: let $x = z$ and $y = 0$ in the original FE, which gives us $$2f(z)\cdot f(t) = 2f(zt) = f(-zt) + f(zt) \qquad \Rightarrow \qquad f(zt) = f(-zt)$$ Now we will show that $f$ is strictly increasing on the interval $[0, \infty)$. Let $x = y$ and $z = -t$. Then: \[f(x^2 + t^2)=(f(x) + f(t))^2 = f(x^2) + 2f(xt) + f(t)^2 \ge f(x^2)\]if $x \ge 0$ and $t > 0$. [We have $f(xt) \ge 0$, because $f(xt) = f(\sqrt{xt})f(\sqrt{xt}) \ge 0$]. So now we see that $f$ is strictly increasing. Because the rationals are dense within the reals, $\boxed{f(x) = x^2}$ for all $x \ge 0$, and because $f$ is even, we're done. So in conclusion, our answers are: $f(x) = x^2$, $f(x) = 0$, and $f(x) = \frac{1}{2}$. Footnote: In case the density argument wasn't entirely clear, suppose for some positive irrational number $x$, we have $f(x) = x^2 + \varepsilon$ for some $\varepsilon > 0$ (the proof for $\varepsilon < 0$ is analogous). Then, take a rational number $x < r < \sqrt{x^2 + \varepsilon}$, which is possible since between any two real numbers there is a rational number. Since we had shown $f(r) = r^2$, and we had shown $f$ is monotone, we must have $$f(r) \ge f(x) \qquad \Rightarrow \qquad r^2 \ge x^2 + \varepsilon$$which is a contradiction since $r < \sqrt{x^2 + \varepsilon}$.

Is Cauchy's equation's solution for monotone functions citable?

Let $P(x,y,z,t)$ be the given assertion. $P(0,0,0,0)$ implies that $f(0)=0$ or $f(0)=\dfrac 12$. If $f(0)=\dfrac 12$, then $P(0,0,0,t)$ gives us the solution $f(t)=\dfrac 12$. Now, we can assume that $f(0)=0$. $P(x,y,0,0)$ gives that $f(x)f(y)=f(xy)$, and $P(0,0,z,t)$ gives that $f(z)f(t)=f(-zt)$, so $f$ is multiplicative and even. Let $Q$ be the assertion that $f(x)f(y)=f(xy)$. Then, $Q(1,1)$ gives that $f(1)=0$ or $f(1)=1$. If $f(1)=0$, then $Q(1,y)$ gives the solution $f(y)=0$. Otherwise, assume $f(1)=1$. Then $Q\left(x,\dfrac 1x\right)$ gives that $f(x)\neq 0$ for all $x\neq 0$. Now, $P(x,1,1,1)$ gives that $2=f(x+1)-2f(x)+2f(x-1)$, so using the base cases $f(0)=0$ and $f(1)=1$, we can get that $f(n)=n^2$ for all integers $n$. Then, $Q\left(b, \dfrac ab\right)$ gives that $f(q)=q^2$ for all rationals $q$. Now, $Q(x,x)$ gives that $f(x^2)=f(x)^2>0$ for all $x\neq 0$. Furthermore, for any positive reals $x<w$, $P\left(x,\sqrt{w^2-x^2},\sqrt{w^2-x^2},x\right)$ gives that $\left[f(x)+f\left(\sqrt{w^2-x^2}\right)\right]^2=f(w)^2$, which implies that $f$ is strictly increasing over the positive reals. But since $f(q)=q^2$ over the positive rationals and $f$ is increasing over the positive reals, we can use a Cauchy-esque argument to show that $f(x)=x^2$ for all positive reals, and hence over all reals. Hence, the solutions are $f(x)=0, f(x)=\dfrac 12, f(x)=x^2$.

mcdonalds106_7 wrote: Let $P(x,y,z,t)$ be the given assertion. $P(0,0,0,0)$ implies that $f(0)=0$ or $f(0)=\dfrac 12$. If $f(0)=\dfrac 12$, then $P(0,0,0,t)$ gives us the solution $f(t)=\dfrac 12$. Now, we can assume that $f(0)=0$. $P(x,y,0,0)$ gives that $f(x)f(y)=f(xy)$, and $P(0,0,z,t)$ gives that $f(z)f(t)=f(-zt)$, so $f$ is multiplicative and even. Let $Q$ be the assertion that $f(x)f(y)=f(xy)$. Then, $Q(1,1)$ gives that $f(1)=0$ or $f(1)=1$. If $f(1)=0$, then $Q(1,y)$ gives the solution $f(y)=0$. Otherwise, assume $f(1)=1$. Then $Q\left(x,\dfrac 1x\right)$ gives that $f(x)\neq 0$ for all $x\neq 0$. Now, $P(x,1,1,1)$ gives that $2=f(x+1)-2f(x)+2f(x-1)$, so using the base cases $f(0)=0$ and $f(1)=1$, we can get that $f(n)=n^2$ for all integers $n$. Then, $Q\left(b, \dfrac ab\right)$ gives that $f(q)=q^2$ for all rationals $q$. Now, $Q(x,x)$ gives that $f(x^2)=f(x)^2>0$ for all $x\neq 0$. Furthermore, for any positive reals $x<w$, $P\left(x,\sqrt{w^2-x^2},\sqrt{w^2-x^2},x\right)$ gives that $\left[f(x)+f\left(\sqrt{w^2-x^2}\right)\right]^2=f(w)^2$, which implies that $f$ is strictly increasing over the positive reals. But since $f(q)=q^2$ over the positive rationals and $f$ is increasing over the positive reals, we can use a Cauchy-esque argument to show that $f(x)=x^2$ for all positive reals, and hence over all reals. Hence, the solutions are $f(x)=0, f(x)=\dfrac 12, f(x)=x^2$. What is a Cauchy-esque argument? Is it what cnyd used to extend from rationals to reals?

IMO ShortList 2002 A4 wrote: 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 $x,y,z,t$.

Let $P (x;y;u;v) $ the assertion of the functional equation. Skip case $f (x)=0$, $f (x)=\frac {1}{2} $ Easy to get cauchy's equation. To prove monotonicity put $P (x;y;y;x) $ we get: $(f (x)+f (y))^2 = f (x^2+y^2) $ sinnce $f $ is always positive, we get that $f (x^2 +y^2) \geq f (x^2) $ so $f $ is monotone on positive reals so $f (x)= x^a$ we get $f (x)=x^2$ on the postive reals and since $f $ is pair we get $f (x)=x^2$ for all $x $ real.

Let $P(t, x, y, z)$ refer to the assertion that \[\left(f(x)+f(z)\right)\left(f(y)+f(t)\right)=f(xy-zt)+f(xt+yz).\]$P(0,0,0,0)$ implies $f(0) \in \{0, \tfrac12\}$. If $f(0) = 0$, Claim 1: $f$ is multiplicative. \[P(0, x, y, 0) \implies f(x)f(y) = f(xy).\]Claim 2: $f$ is even. Proof. \[P(y, x, 0, x) \implies f(-xy) + f(xy) = 2f(x)f(y) = 2f(xy) \implies f(xy) = f(-xy).\]Claim 3: $f(x)$ is monotonically increasing for $x \ge 0$. Proof. Let $x, y \ge 0$, $f(x) = f(\sqrt{x})^2 \ge 0$. \[P(\sqrt{x},\sqrt{y},\sqrt{y},\sqrt{x}) \implies f(x+y) = (f(\sqrt{x}) + f(\sqrt{y}))^2.\]Therefore, $f(x+y) = (f(\sqrt{x}) + f(\sqrt{y}))^2 \ge f(\sqrt{x})^2 = f(x)$ which implies $f$ is monotonically increasing over positive reals. Thus, $f(x) = x^{2c}$ or $\boxed{f(x) = 0}$. Substituting the first solution into the original solution, we see that $\boxed{f(x) = x^2}$. If $f(0) = \tfrac12$, $P(0,x,0,x)$ implies $\boxed{f(x) = \tfrac12}$. Thus, the 3 solutions are \[f(x) = x^2\qquad f(x) = 0\qquad f(x) = \tfrac12.\]It's easy to verify that these indeed satisfy the original equation.

We claim that the only solutions are $f \equiv 0, \frac{1}{2}$ and $f(x)=x^2$ and they indeed work. Let $P(x, y, z, t)$ be the assertion. Now $P(0, 0, 0, 0)$ tells us that $f(0)$ is either $\frac{1}{2}$ or $0$. If $f(0)=\frac{1}{2}$, $P(y, 0, 0, 0) \rightarrow f(y) \equiv \frac{1}{2}$. So now let $f(0)=0$, now $P(x, y, 0, 0) \rightarrow f(xy)=f(x)f(y)$ so $f$ is multiplicative. Claim: $f$ is non-decreasing on $\mathbb{R}^+$ Proof: First, notice that $f(x^2)=f(x)^2 \geq 0$. Now $$P(x, y, y, x) \rightarrow f(x^2)+2f(xy)+f(y^2)=f(x^2+y^2)$$So $f(x^2) < f(x^2+y^2)$. $\blacksquare$ Now let $g(x)=\ln(f(a^x))$, notice that $g(x+y)=g(x)+g(y)$ and $g(x)=cx$ and this means $f(x)=x^c$, on plugging in the original equation, we see that only $f(x)=x^2$ works. $\blacksquare$

Le thonk Let $P(x,z,y,t)$ denote the assertion. $P(0,0,y,y)$ gives $4f(0)f(y)=2f(0)$ which shows either $f(y)=\frac{1}{2}$ for all $y$ or $f(0)=0.$ Note that $f(y)=\frac{1}{2}$ is a valid function. Now, suppose $f(0)=0,$ then $P(x,x,y,y)$ gives $4f(x)f(y)=f(2xy).$ Letting $x=y=\frac{1}{2}$ gives $4f(\frac{1}{2})^2=f(\frac{1}{2})$ so $f(\frac{1}{2})=0$ or $\frac{1}{4}.$ $~$ If $f(\frac{1}{2})=0$ then $P(0,\frac{1}{2},0,t)$ gives $0=f(-\frac{1}{2}t).$ Note that $-\frac{1}{2}t$ can be anything so $f(x)=0$ for all $x$ which is also a valid function, so suppose $f(\frac{1}{2})=\frac{1}{4}.$ $~$ Now, $P(x,x,y,-y)$ gives $2f(x)(f(y)+f(-y))=f(2xy)=4f(x)f(y).$ We can easily choose $f(x)\neq 0$ so $f(y)+f(-y)=2f(y),$ which implies that $f(y)=f(-y).$ Thus, we solve $f$ over only the positive numbers. We have $P(x,z,x,z)$ giving us $(f(x)+f(z))^2=f(x^2-z^2)+f(2xz)$ and $P(x,z,z,x)$ giving us $(f(x)+f(z))^2=f(x^2+z^2).$ Note that $(x^2-z^2)^2+(2xz)^2=(x^2+z^2)^2$ so $f(\sqrt{a})+f(\sqrt{b})=f(\sqrt{a+b}).$ $~$ Let $g:(0,\infty)\to \mathbb{R}$ such that $g(x)=f(\sqrt{x})$ which implies that $g(a)+g(b)=g(a+b).$ Thus, $g(2a)=2g(a)$ and $g(4a)=4g(a).$ Also, $4f(x)f(y)=f(2xy)$ so $4g(x^2)g(y^2)=g(4x^2y^2)=4g(x^2y^2)$ which implies $g(a)g(b)=g(ab).$ Thus, we have $g(x)=kx$ for some $k$ and so $f(x)=kx^2.$ Thus, plugging $x=\frac{1}{2}$ gives $f(x)=x^2$ which is indeed a valid function. Thus, our functions are $f(x)=0,\frac{1}{2},x^2.$

Similar problem from 2015 Korea https://artofproblemsolving.com/community/c6h1158064p5501261

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 $z$. $\blacksquare$

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 $f(x) = x^2$.

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 $f(x) = x^2$.

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 $\boxed{f(x)=x^2, f(x)=\frac{1}{2}\text{ and }f(x)=0, \forall x\in\mathbb{R}}$ $\blacksquare$.

"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.

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.

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 $P(x,0,x,0)\implies 2f(x) = 1 \implies f(x) = \dfrac{1}{2} \implies f\equiv \dfrac{1}{2}$.

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 $f(x)\equiv x^2$. $\square$

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.

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.

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.

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 $f\equiv x^2.$