IMO ShortList 2002, algebra problem 4

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

3647 posts

orl

#1 h Sep 28, 2004, 1:20 PM • 12 Y

Y by amatysten, Davi-8191, tenplusten, e_plus_pi, mathematicsy, Adventure10, megarnie, ImSh95, Stuart111, Mango247, Rounak_iitr, cubres

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 $x,y,z,t$ .

This post has been edited 4 times. Last edited by orl, Sep 27, 2005, 4:57 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

inom

118 posts

inom

#2 h Jun 24, 2006, 3:20 PM • 5 Y

Y by Tawan, Adventure10, ImSh95, Mango247, cubres

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.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Albanian Eagle

1693 posts

Albanian Eagle

#3 h Jun 26, 2006, 9:24 PM • 7 Y

Y by vsathiam, Adventure10, ImSh95, Vladimir_Djurica, Mango247, cubres, and 1 other user

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

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

wojto111

42 posts

wojto111

#4 h Jan 22, 2007, 7:40 PM • 4 Y

Y by Adventure10, ImSh95, Mango247, cubres

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

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

olorin

588 posts

olorin

#5 h Jan 22, 2007, 11:58 PM • 7 Y

Y by Tawan, xdiegolazarox, Adventure10, ImSh95, Mango247, X.Luser, cubres

$x=z,y=t=0 \Rightarrow 2f(x)\cdot 2f(0)=2f(0)$ 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 $x\in\mathbb{R}$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

nealth

303 posts

nealth

#6 h May 19, 2009, 3:57 PM • 4 Y

Y by ImSh95, Adventure10, Mango247, cubres

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 $k = 1$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

MellowMelon

5850 posts

MellowMelon

#7 h May 19, 2009, 4:17 PM • 6 Y

Y by Jc426, ImSh95, Adventure10, Mango247, cubres, and 1 other user

Where did you show the function was differentiable?

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

nealth

303 posts

nealth

#8 h May 19, 2009, 4:52 PM • 4 Y

Y by ImSh95, Adventure10, Mango247, cubres

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?

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

MellowMelon

5850 posts

MellowMelon

#9 h May 19, 2009, 4:56 PM • 6 Y

Y by Jc426, ImSh95, Adventure10, Mango247, cubres, and 1 other user

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)

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

cnyd

394 posts

cnyd

#10 h Nov 30, 2009, 1:46 PM • 6 Y

Y by Tawan, vsathiam, ImSh95, Adventure10, Mango247, cubres

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

$\implies$ $f(x) = x^{2}$ $\forall\mathbb{R^{ + }}$

$f(x) = f( - x)$ $\implies$ $f(x) = x^{2}$

$\implies$ $\boxed{f(x) = 0},\boxed{f(x) = \frac {1}{2}},\boxed{f(x) = x^{2}}$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

cnyd

394 posts

cnyd

#11 h Dec 3, 2009, 6:26 PM • 4 Y

Y by ImSh95, Adventure10, Mango247, cubres

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

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

JuanOrtiz

366 posts

JuanOrtiz

#12 h May 26, 2014, 5:42 AM • 6 Y

Y by Tawan, Anar24, ImSh95, Adventure10, Mango247, cubres

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.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

fclvbfm934

759 posts

fclvbfm934

#13 h Aug 18, 2014, 4:58 AM • 5 Y

Y by Tawan, ImSh95, Adventure10, Mango247, cubres

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 $r < \sqrt{x^2 + \varepsilon}$ .

This post has been edited 1 time. Last edited by fclvbfm934, Sep 6, 2022, 1:33 AM
Reason: Fixing formatting

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Konigsberg

2225 posts

Konigsberg

#14 h Jul 7, 2015, 8:56 AM • 4 Y

Y by ImSh95, Adventure10, Mango247, cubres

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

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

mcdonalds106_7

1138 posts

mcdonalds106_7

#17 h May 19, 2016, 9:39 PM • 3 Y

Y by ImSh95, Adventure10, cubres

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 $f(x)=0, f(x)=\dfrac 12, f(x)=x^2$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

ZETA_in_olympiad

2211 posts

ZETA_in_olympiad

#54 h Oct 31, 2022, 8:57 AM • 1 Y

Y by ImSh95

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

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

SatisfiedMagma

458 posts

SatisfiedMagma

#55 h Dec 16, 2022, 7:44 PM • 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

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

YaoAOPS

1541 posts

YaoAOPS

#56 h May 10, 2023, 7:38 AM • 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 $f(x) = x^2$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

ezpotd

1266 posts

ezpotd

#57 h May 24, 2023, 4:45 AM • 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

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

F10tothepowerof34

195 posts

F10tothepowerof34

#58 h Jun 27, 2023, 10:11 PM • 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 $\boxed{f(x)=x^2, f(x)=\frac{1}{2}\text{ and }f(x)=0, \forall x\in\mathbb{R}}$ $\blacksquare$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

signifance

140 posts

signifance

#59 h Nov 1, 2023, 12:27 AM

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.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

cursed_tangent1434

624 posts

cursed_tangent1434

#60 h Dec 13, 2023, 6:18 AM

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.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

kamatadu

480 posts

kamatadu

#61 h Jan 1, 2024, 7:55 PM

Y by

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

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

KevinYang2.71

427 posts

KevinYang2.71

#62 h Apr 28, 2024, 7:43 AM • 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 $f(x)\equiv x^2$ . $\square$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

joshualiu315

2534 posts

joshualiu315

#63 h Aug 31, 2024, 5:35 AM

Y by

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.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

pie854

243 posts

pie854

#64 h Sep 8, 2024, 11:37 AM

Y by

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.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

OronSH

1739 posts

OronSH

#65 h Dec 3, 2024, 8:18 PM • 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.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Ilikeminecraft

623 posts

Ilikeminecraft

#66 h Feb 4, 2025, 1:13 AM

Y by

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 $f\equiv x^2.$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

cj13609517288

1915 posts

cj13609517288

#67 h Mar 3, 2025, 6:41 PM

Y by

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. $\blacksquare$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Ihatecombin

60 posts

Ihatecombin

#68 h Apr 10, 2025, 2:15 PM

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. $\blacksquare$

Z K Y