This is also usajmo problem 3.

This was also JMO #3. Also, I hear many people saying the answers are $ f(x) = 0 $ and $ f(x) = x^2 $. How did you prove that we can't have a function $ f $ such that $ f(x) = 0 $ for some integers and $ f(x) = x^2 $ for all other integers?

Oops did cases of $|f(0)|=2$ and somehow got no solutions and missed case of $f(0)$

Here's a quick way to get $f(0) = 0$. So plug in $y = 0$. This gives $xf(2f(0)-x) = \frac{f(x)^2}{x}+f(0)$ Now take $x$ to be a prime $p$. Then $p | f(p)^2 \implies p|f(p)$. So $p| \frac{f(x)^2}{x}$, so $p | f(0)$. So $f(0) = 0$. Then we can plug in $f(0) = 0$ to get $x^2f(-x) = f(x)^2$, Plug in $-x$ to get $x^2f(x) = f(-x)^2$. Therefore, $f(-x) = f(x) = x^2$ or $0$ for all $x$. Now by dumb stuff (bashy too) we show that its either always $0$ or always $x^2$. My sketch. Say $f(c) = 0$ for some $c \not= 0$. Then we prove that $f(2x) = 0$ always. Now assume some $f(d) = d^2$ for $d$ odd. We prove that $f(4k+3) = 0$ always. Then either $-d$ or $d$ is $3 \pmod{4}$, so it equals 0 too.

Question: if I somehow forgot the f(0)=0 case and disproved the +/- 2 case will I salvage 1 point?

In order to plug in x= 2f(0) into the equation doesn't it require f(0) to not be 0 because x can't be 0 in the equation.

calculatorwiz wrote: In order to plug in x= 2f(0) into the equation doesn't it require f(0) to not be 0 because x can't be 0 in the equation. Exactly, that's how you prove f(0)=0

How many points do I get for showing that f(x)=x^2 works but saying that it works where x is not 0 and that I conjectured it was the only function?

What I wrote seems different from most, which makes me doubt it. Could someone please check?

How many points would I get if I forgot the case f(x)=0?

borntobewild: What if $a=1$?

Ughh... forgot that. It can be $-1$ as well... How many points do you think I'll lose?

infiniteturtle wrote: borntobewild: What if $a=1$? oh, that's how everyone was saying it's 2 or -2...

BOGTRO wrote: Then $P(c_2, c_1)$ gives us $c_2f(c_2)+c_1^2f(2c_2)=\frac{f(c_2)^2}{c_2} \implies c_1^2f(2c_2)=2c_2^3$, This isn't true. You get $f(2c_2) = 0$, which tells you all even guys die. Then you show all guys 1 mod 4 die and then 3 mod 4 die.

You can skip the 1mod4 / 3mod4 step by plugging in $x = \frac{y^2+y}{2}$ or $x=\frac{y^2-y}{2}$ -- whichever one is even.

fermat007 wrote: BOGTRO wrote: Then $P(c_2, c_1)$ gives us $c_2f(c_2)+c_1^2f(2c_2)=\frac{f(c_2)^2}{c_2} \implies c_1^2f(2c_2)=2c_2^3$, This isn't true. You get $f(2c_2) = 0$, which tells you all even guys die. Then you show all guys 1 mod 4 die and then 3 mod 4 die. Oops that's what I get for doing things in my head on the car ride back lol

Claim: $f(0)=0$. Proof: Suppose not. $P(2f(0),0): 2f(0)^2=\frac{f(2f(0))^2}{2f(0)}+f(0)\implies 4f(0)^3=f(2f(0))f^2+2f(0)^2\implies 4f(0)^3-2f(0)^2=f(0)^2(4f(0)-2)=f(2f(0))^2$. This implies $4f(0)-2$ is a perfect square, which is a contradiction as it's $2\pmod4$. $P(x,0): xf(-x)=\frac{f(x)^2}{x}\implies x^2f(-x)=f(x)^2$. $P(-x,0): x^2f(x)=f(-x)^2$. If one of the two equations is $0$, then the other is also zero, which implies $f(x)=f(-x)=0$. If none of the two equations is $0$, then we divide them. \[\frac{f(-x)}{(f(x))}=(\frac{f(x)}{f(-x})^2\] Setting $\frac{f(-x)}{f(x)}=k$ gives $k=\frac{1}{k^2}\implies k^3=1\implies k=1$. So $f(x)=f(-x)$. Thus, we can conclude that $f$ is even. So $x^2f(x)=f(x)^2\implies f(x)=0\text{ or }f(x)=x^2$, which we obtain from dividing both sides by $f(x)$. Suppose there exist $a,b\ne0$ so that $f(a)=0$ and $f(b)=b^2$. $P(b,a): bf(-b)+a^2f(2b)=b^3+a^2f(2b)=\frac{b^4}{b}=b^3\implies a^2f(2b)=0\implies f(2b)=0$. $P(a,a): a^2f(2a)=0\implies f(2a)=0$. Thus, $f(x)=0$ if $x$ is even. Now suppose $b$ is odd and $f(b)=b^2$. $P(2a,b): 2af(2b^2-2a)+b^2f(4a-b^2)=b^2f(4a-b^2)=f(b^3)$. Case 1: $f(b^3)=b^6$. Then $f(4a-b^2)=b^4$. So $16a^2-8ab^2+b^4=b^4\implies 16a^2-8ab^2=0\implies a(16a-8b^2)=0\implies 16a=8b^2$, which implies $b$ is even, a contradiction. Case 2: $f(b^3)=0$. Then $f(4a-b^2)=0$. This is already a contradiction if $b=\pm1$ as $b^3=b$, which implies $f(b)=0$. Now we have $4a-b^2$ covers all $q$ that are $3\pmod4$, and by taking the negative, it covers all $q\equiv1\pmod4$, which implies that $f(b)=0$, a cntradiction. So we can never have $a,b\ne0$ so that $f(a)=0$ and $f(b)=b^2$, which implies the only solutions are $\boxed{f\equiv0}$ and $\boxed{f(x)=x^2\forall x\in\mathbb{Z}}$.

Answer is $f(x)=0$ or $x^2$. Claim: $f(0)=0.$ Proof: Let $P(x,y)$ be assertion. Take a prime $p$ that does not divide $f(0)$. Plugging in $x=p$, we have that $\frac{f(x)^2}{x}$ which means that $p | \frac{f(p)^2}{p}$ which gives the contradiction that $p | f(0).$ Thus, the claim has been proven. Claim: $f$ is an even function. Proof: $P(-x,0) $ and $P(x,0)$ implies the claim. Not only this; it implies further that $f(x)=f(-x)=0$ or $x^2$. Now, we show that the solutions that we have are the only ones, and resolve the pointwise trap. We do this by showing that if there exists an $n$ such that $f(n)=0,$ then in fact $f\equiv 0$. $P(x,n)$ gives \[xf(x)+n^2f(2x)=\frac{f(x)^2}x.\]From this, we know that for both $f(x)=0$ and $x^2$, all evens are then $0$. For some $m$ such that $f(m)=m^2$, consider $P($even,$m)$. [Denote the even number by $x$, as usual.] Then we have $m^2f(2x-m^2)=f(m^3)$. Now, we consider 2 cases: if $f(m^3)=(m^3)^2,$ then we have that $m^4$ is either $0$ or $(2x-m^2)^2$ for all $x$, which is an obvious contradiction. If $f(m^3)=0$, then $f(4k+3)=0$ besides $\pm m^2.$ Then contradiction $f(m)=0$ iff $m\neq 1$, but we get contradiction at $P(5,1)$ so no sols in this case, so only answers are the ones at beginning.

The solutions are $\boxed{f(x) \equiv x^2}$ and $\boxed{f(x) \equiv 0}$ for all $x \in \mathbb{Z}.$ Let $P$ be the given. If $f(0) \ne 0,$ $P(2f(0), 0)$ gives \[2f(0)^2=\frac{f(2f(0))^2}{2f(0)}+f(0) \implies 4f(0)^3-2f(0)^2 =f(2f(0))^2 \implies f(0)^2(4f(0) - 2) = f(2f(0))^2 \]thus $4f(0)-2$ is a perfect square, but this is contradiction by mod $4$ reasons. So $f(0) = 0$ and $P(x,0)$ gives \[xf(-x)=\frac{f(x)^2}{x} \implies x^2f(-x)=f(x)^2.\]But also $x^2f(x) = f(-x)^2,$ so for any $x,$ $f(x) = x^2$ or $0.$ Claim: If $f(a) = 0$ for a nonzero $a,$ then $f$ evaluates to $0$ everywhere. Proof: $P(x,a)$ gives $xf(-x) + a^2f(2x) = \frac{f(x)^2}{x}$ for all nonzero $x.$ Since $xf(-x) = \frac{f(x)^2}{x},$ $f$ evaluates to zero for all nonzero even numbers $2x.$ For any odd $b,$ $P(2x,b)$ implies $b^2f(4x-b^2)=f(b^3)$ for any nonzero integers $x.$ If $f(b^3) = b^6,$ then $f(4x-b^2) = b^4$ for all $x,$ contradiction. So $f(b^3) =0,$ and $f$ evaluates to zero for all inputs that equalling $ \pm (4x-b^2)$ for some nonzero integer $x,$ i.e. all odd integers except for $\pm b^2.$ This will force $f(b)=0$ unless $b = \pm 1,$ but in that case $f(b) = f(b^3) = 0$ which finishes. $\square$ Thus either $f(x) \equiv x^2$ for all $x,$ or $f(x) \equiv 0$ for all $x$ as needed. $\blacksquare$

Missed the quick $f(0)=0$ conclusion. The answers are $f \equiv 0$ and $f(x)=x^2$, which clearly work. We now prove that these are the only solutions; denote the assertion as $P(x,y)$. First, I claim that $f(0)=0$. Suppose otherwise, so from $P(f(0),0)$, we have $$f(0)f(f(0))=\tfrac{f(f(0))^2}{f(0)}+f(0) \implies f(f(0))^2+f(0)^2=f(0)^2f(f(0)).$$Let $f(0)=a,f(f(0))=b$, so $a^2+b^2=a^2b$. Then taking $\pmod{a^2}$ we have $a \mid b$, so let $b=ak$. Substituting, the equation becomes $$a^2(k^2+1)=a^3k \implies k^2+1=ak,$$whence taking $\pmod{k}$ implies $k \mid 1$, i.e. $k=1,-1$. This then means that $(a,b)=(-2,2),(2,2)$. Then from $f(2f(0),0)$ we have $$2f(0)^2=\frac{f(2f(0))^2}{2f(0)}+f(0) \implies 4f(0)^3-2f(0) \text{ is a square},$$but substituting $f(0)=2,-2$ both fail (in hindsight, one can just use the fact that $4x^3-2x=x^2(4x-2)$ is never a square for nonzero $x$ by mod 4, but I didn't realize this), hence $f(0)=0$. Now, from $P(x,0)$, we obtain $$xf(-x)=\frac{f(x)^2}{x} \implies x^2f(-x)=f(x)^2,$$and from $P(-x,0)$ we obtain $x^2f(x)=f(-x)^2$. Solving this system of equations for $f(x),f(-x)$ yields $f(x) \in \{0,x^2\}$ for all $x$, and $f(x)=f(-x)$ hence $f$ is even. We now deal with the pointwise trap. I claim that if there exists $a \neq 0$ with $f(a)=0$, then $f \equiv 0$. First, by comparing $P(x,0)$ with $P(x,a)$, we obtain $$a^2f(2x)=0 \implies f(2x)=0$$for all $x \in \mathbb{Z}$. Now from $P(2x,y)$ for $y$ odd and $x \neq 0$, we have $$2xf(2(f(y)-x))+y^2f(4x-f(y))=\frac{f(2x)^2}{x}+f(yf(y)) \implies f(4x-f(y))=\frac{f(yf(y))}{y^2}.$$Suppose that there exists some choice of $(x,y)$ such that $f(4x-f(y)) \neq 0$. Then we must have $f(yf(y)) \neq 0 \implies f(yf(y))=y^2f(y)^2=y^6$, so $$(4x-f(y))^2=f(4x-f(y))=y^4 \implies 4x-f(y)=y^2 \implies 4 \mid y^2+f(y),$$but we have $y^2+f(y) \in \{y^2,2y^2\}$, the former of which is odd and the latter of which is $2 \pmod{4}$, so this is a contradiction. Thus, $f(4x-f(y))=0$ for all odd $y$, hence $f(yf(y))=0$. If $f(b) \neq 0$ for some odd $b$ and thus $f(b)=b^2$, then $f(b^3)=0$. If $b \equiv 1 \pmod{4}$, then letting $(x,y)=(\tfrac{b(b-1)}{4},b)$, we have $$f(b^2-b-b^2)=f(-b)=0=f(b)$$as $f$ is even, which is a contradiction. Otherwise, if $b \equiv 3 \pmod{4}$, then let $(x,y)=(\tfrac{b(b+1)}{4},b)$, so $$f(b^2+b-b^2)=f(b)=0,$$which is also a contradiction. Thus no such $b$ exists, hence $f(x)=0$ for all odd $x$, which implies that $f(x)=0$ for all $x$. This eliminates the pointwise trap, so we must have $f(x)=x^2$ or $f \equiv 0$ as desired. $\blacksquare$

Let $P(x,y)$ denote the given assertion. We claim that $f\equiv 0$ and $f\equiv x^2$ are the only answers, these clearly work. Claim 1: $f(0)=0.$ Proof. Assume the contrary. Then $P(2f(0),0)$ yields $4f(0)-2$ is perfect square, absurd since it is $2$ modulo 4. $\blacksquare$ Claim 2: $f(x)\in \{0,x^2\} ~\forall x.$ Proof. Follows from $P(x,0)$ and $P(-x,0).$ $\blacksquare$ Claim 3: The solution set mentioned. Proof. Take $f(u)=0$ for some $u\neq 0.$ Then by $P(x,u)$ we get $f(n)=0$ for all even $n.$ Take an odd $v$ such that $f(v)\neq 0$ so $f(v)=v^2.$ Then by $P(x,v)$ we get $f(v^3)=v^2f(2x-v^2).$ Note that the LHS is zero (trivially) Hence $f(2x-v^2)=0$ but since $P(x,0)$ implies $x^2f(-x)=f(x)^2$ it actually implies $f(n)=0$ for all $n$ of the form $4k+1.$ A contradiction since now $f(m)=0$ for all odd $m.$ $\blacksquare$

FEs are nice Set $y=0$ we can get $$x^2f(2f(0)-x)=f(x)^2+xf(0)$$, which we call eqn (1). Assuming $f(0)\neq0$ in (1) and set $a=f(0),b=f(f(0))$ we can get $a^2+b^2=a^2b$ Taking $\mod a^2$ and $\mod b^2$ to this we get $a^2|b^2,b^2|a^2(b-1)$, which gives $a=b=2$. Substituting $f(0)=2$ back into (1) and setting $x=4$ gives $f(4)^2=24$, but $24$ is not a perfect square, a contradiction, so $f(0)=2$ is impossible. Thus we must have $f(0)=0$, in which case we can substitute into (1) and get $f(x)=x^2\forall x$ or $f(x)=0\forall x$, both of which work Full proof here: https://infinityintegral.substack.com/p/usajmo-2014-contest-review

Denote the assertion as $P(x, y)$. Claim: We have $f(0) = 0$. Proof: Note that \[ P(x, 0) \implies xf(2f(0)-x) = \frac{f(x)^2}{x} + f(0) \]Now, suppose $x$ is a prime $p$. Then $p$ divides the RHS, but the RHS must bve the integer as the LHS is an integer. So $p \mid f(p)^2$ implying $p \mid f(p)$, so $p^2 \mid f(p)^2$, implying $p \mid f(0)$. This is true for infinitely many primes, so $f(0) = 0$. $\square$ Then, we now have \[ xf(-x) = \frac{f(x)^2}{x} \implies x^2f(-x) = f(x)^2 \] Claim: $f$ is even, and either $f(x) = 0$ or $f(x) = x^2$. Proof: If we use $-x$ then \[ x^2f(x) = f(-x)^2 \]If $f(x) \neq 0$ then $f(-x) \neq 0$ and dividing the two equations implies $f(x) = f(-x)$; If $f(x) = 0$ then $f(-x) = 0$. Thus, $f$ is even, and either $f(x) = x^2$ or $f(x) = 0$. $\square$ Now, if $f(a) = 0$ and $f(b) = b^2$ where $a, b \neq 0$, then \[ P(x, a) \implies xf(x) + b^2f(2x) = \frac{f(x)^2}{x} \]Note that $xf(x) = \frac{f(x)^2}{x}$ for all $x$, so then we must have that $f(2x) = 0$. Thus, all evens give zeros. Hence, $b$ is odd. Then, \[ P(2x, b) \implies b^2f(4x - b^2) = f(b^3) \]This holds for all $x \neq 0$. If $f(b^3)$ is nonzero then we find $f(4x - b^2) = b^4$, implying $x = 0$, contradiction. Thus, $f(b^3) = 0$ and $f(4x - b^2) = 0$ for all $x \neq 0$. But then $f(b^2 - 4x) = 0$ as well. This implies that all numbers give outputs at zero except for $f(b^2)$, $f(-b^2)$ and $f(b)$. This means $b = 1, -1$. Checking the function which gives $f(1) = 1$, $f(-1) = 1$ and $f(x) = 0$ otherwise, we find that it fails. Thus, we must have either $f(x) = 0$ or $f(x) = x^2$, both of which work. $\blacksquare$

Claim 1: $f(0)=0$. Assume otherwise. Take the least prime $p$ not dividing $f(0)$. Then \[A(p,0) \implies xf(2f(0)) = \frac{f(p)^2}{p}+f(0) \implies \frac{f(p)^2}{p} \in \mathbb{Z},\] so $p \mid f(p)^2$. But this also suggests that $p^2 \mid f(p)^2$, so $p \mid f(0)$, contradiction. ${\color{blue} \Box}$ Claim 2: $f$ is even. Assume otherwise with the existence of $t$ such that $f(t) \ne f(-t)$. Subtracting our equations for $A(t,0)$ and $A(-t,0)$, we find $f(t) + f(-t) = -t^2$. Then we can get a contradiction by bounding \[0 > - \frac{3t^4}{4} = t^2 \left(f(t)+f(-t)\right) + \frac{t^4}{4} = f(t)^2 + t^2f(t) + \frac{t^4}{4} = \left(f(t)+\frac 12 t^2\right)^2. \quad {\color{blue} \Box}\] Hence $A(x,0)$ can be rewritten as $f(x)=0,x^2$ for all $x$, so our final step is to combat this pointwise trap one case at a time. Suppose by contradiction the existence of $m$, $n$ such that $f(m)=0$ and $f(n)=n^2$, where $m, n \neq 0$. $A(x,m) \implies f(\text{even})=0$, so $n$ must be odd. $A(2x,n) \implies n^2 \cdot f(4k-n^2) = f(n^3)$. If the RHS is nonzero, we have \[n^2 \cdot (4k-n^2)^2 = n^6 \implies 2 \mid n \text{ and } k=0, \text{contradiction.}\] If the RHS is 0, then $f(4k-n^2)=f(n^2-4k)=0$, which covers all odd integers except $\pm n^2$, which includes $n$. This is a contradiction, unless $n = \pm n^2 \implies n = \pm 1$, which is still a contradiction as $f(n) = f(n^3) = 0$ in this subcase. Finally, we find that our solutions are just $\boxed{f(x)=0, \quad f(x)=x^2}$. $\blacksquare$

pov: this took you 2.5 hours even you you are '''''good''''' at FEs Let the assertion be $P(x,y)$. Let $f(0)=c$. Setting $P(2c, 0)$ gives \[2c^2 = \frac{f(x)^2}{2c}+c \implies f(x)^2 = c^2(4c-2).\] Since $4c-2$ is never a perfect square, this is a contradiction. This means that dividing by $2c$ must be invalid, or that $c=0$. Note that $P(x,0)$ and $P(-x,0)$ give \[xf(-x)=\frac{f(x)^2}{x}\]\[-xf(x)=-\frac{f(x)^2}{x},\] whence substituting yields $f(x)=x^2$. A side note of this is that $f$ is even. Before we are done, we must check for piecewise functions. Suppose there are integers $a$ and $b$ not equal to $0$ such that $f(a)=0$ and $f(b)=b^2$. The assertions $P(a,a)$ and $P(b,a)$ yield \[a^2f(2a)=0\]\[a^2f(2b)=0,\] from which we are able to conclude that $f(x)=0$ for any even $x$. Finally, notice that $P(2a,b)$ for an odd $b$ gives \[2af(2f(b)-2a) + b^2f(4a-b^2) = \frac{f(2a)^2}{2a}+f(b^3)\]\[\implies b^2f(4a-b^2) = f(b^3).\] If $f(b^3)$ is nonzero, we get $f(4a-b^2)=f(b^2-4a) = b^4$, which implies $a=0$, a contradiction. Thus, $f(b^3)=0$, which means $f(b^2-4a)=0$. This means every value of $x$ except for $f(b), f(b^2), f(-b^2)$ is zero. Since $f(-b)$ is covered in the "zero" set, we must have $b= \pm b^2$, which gives $b = \pm 1$. Checking the desired piecewise function shows us it fails, so our solution set is $\boxed{f(x) \equiv 0, x^2}$.

fes are very fun Let $P(x, y)$ be the assertion. Claim 1: $f(0) = 0$: Say for the sake of contradiction, $f(0) \neq 0$. $P(2f(0), 0)$ yields: $$2f(0)^2 = \frac{f(2f(0))^2}{2f(0)} + f(0) \implies 4f(0)^3 = f(2f(0))^2 + 2f(0)^2$$This gives $f(0)^2(4f(0) - 2) = f(2f(0))^2$, which means $4f(0) - 2$ is a perfect square, however $4f(0) - 2 \equiv 2\mod{4}$, contradiction, meaning $f(0) = 0$. Claim 2: $f(x) = 0, x^2$: $P(x, 0)$ gives $x^2f(-x) = f(x)^2$, $P(-x, 0)$ gives $x^2f(x) = f(-x)^2$, which means $f(-x) = f(x) \implies x^2f(x) = f(x)^2 \implies f(x) = x^2, 0$. Now, all we have to do is avoid the pointwise trap. Say there exists non-zero $a, b$, such that $f(a) = 0, f(b) = b^2$. $P(x, a)$ gives $a^2f(2x) = 0$ (since $xf(x) = \frac{f(x)^2}{x}$), which means $f(2x) = 0$. $P(2x, b)$ gives: $$2xf(2b^2 - 2x) + b^2f(4x - b^2) = f(b^3)$$Since $2b^2 - 2x$ is even, $f(2b^2 - 2x) = 0 \implies b^2f(4x - b^2)$. If $f(b^3) \neq 0, f(b^3) = b^6 \implies f(4x - b^2) = b^4 \implies x = 0$, impossible. So $f(b^3) = 0$, which means $f(4x - b^2) = f(b^2 - 4x) = 0$, which makes $f(x) = 0$ except for $f(-b^2)$, $f(b^2)$, $f(b)$, so $b = 1, - 1$. The function which maps $1, -1$ to $1$ and rest $x$ to $0$, fails by $P(x, 1)$. So either $f(x) = x^2$ or $f(x) = 0$.

Love love love I claim that the answer is $f(x)=0$ and $f(x)=x^2$ only. These two solutions can be checked by plugging and chugging C: First, I claim that $f(0)=0$. FTSOC, assume that $f(0)\neq 0$. Then, if we plug in $x=f(0)$ and $y=0$, we get that \[f(0)f(f(0))=\frac{f(f(0))^2}{f(0)}+f(0),\]and since $f(0)\neq 0$, if we multiply both sides by $f(0)$, we get \[f(0)^2f(f(0))=f(0)^2+f(f(0))^2.\]Let $f(0)^2=a$ and let $f(f(0))=b$. We now get that \[b^2-ab+a=0,\]\[\iff b=\frac{a\pm \sqrt{a^2-4a}}{2},\]by the quadratic formula. However, since the function is from $\mathbb{Z}$ to $\mathbb{Z}$, note that $a$ and $\sqrt{a^2-4a}$ must then both be integers. Now, let $a^2-4a=n^2$ and let $m=a-2$. Note that \[m^2-n^2=4,\]\[\iff (m,n)=(\pm 2,0),\]by using the difference of squares and factorizing (since $m$ and $n$ must both be integers). Since $m=-2$ or $2$, this means that $a=2$ or $a=0$. However, since $a=f(0)^2$, and we already established that $f(0)\neq 0$, we get that \[f(0)^2=a=2,\]which is a contradiction since by problem conditions, $f(0)$ must be an integer. Therefore we must have that $f(0)=0$. -- Now plugging in $(x,0)$ gives \[f(-x)=\frac{f(x)^2}{x^2},\]and similarly plugging in $(-x,0)$ gives that \[f(x)=\frac{f(-x)^2}{x^2}.\]Now using these equations, we have that \[f(-x)^2=\frac{f(x)^4}{x^4},\]\[\iff f(x)=\frac{\left(\frac{f(x)^4}{x^4}\right)}{x^2},\]\[\iff f(x)=\frac{f(x)^4}{x^6},\]which gives that either $f(x)=0$ or $f(x)=x^2$. -- However, there's still a pointwise trap. I claim that there do not exist DISTINCT nonzero integers $a$, $b$ such that $f(a)=a^2$ and $f(b)=0$. Assume FTSOC that there do exist such $a$, $b$. Then plugging $(b,a)$ gives \[af(-a)+b^2f(2a)=a^3,\]and we now have two cases. C1: $f(-a)=0$. Now, plugging in $(-a,a)$ gives that \[af(-a)+a^2f(2a)=a^3,\]\[\iff a^2f(2a)=a^3,\]which means that $f(2a)=a$, a contradiction. This is because we already established beforehand that $f(x)=0$ or $f(x)=x^2$. Since $a\neq 0$, it cannot be the former, therefore we must have that $f(2a)=a=4a^2$, which is a contradiction, since there does not exist an integer $a$ such that $4a^2=a$. Therefore, $f(-a)=0$, meaning that we must have $f(-a)=a^2$. -- C2: $f(-a)=a^2$. Then, plugging in $(b,a)$, we have that \[af(-a)+b^2f(2a)=a^3,\]\[\iff b^2f(2a)=0,\]and since $b\neq 0$, this implies that $f(2a)=0$. Now plugging in $(2a,a)$ gives \[2af(-a)+4a^2f(2a)=a^3,\]\[\iff 2a^3+4a^2f(2a)=a^3,\]\[\iff f(2a)=-\frac{a^3}{4a^2}=-\frac{a}{4}.\]However, we already know that $f(2a)=0$, meaning that we must have $a=0$, a clear contradiction! Therefore, there cannot exist such $a$, $b$ as stated above. Finally, this means that our only solutions are $f(x)=0$ and $f(x)=x^2$, as desired, finishing the problem.

$P(x,0)\implies xf(2f(0)-x)=\frac{f(x)^2}{x}+f(0)$. Let $x=2f(0), 2f(0)^2(2f(0)-1)=f(2f(0))^2$. Since the function is defined in integers, we have to make sure $2(2f(0)-1)$ is an integer, but $4f(0)-2\equiv 2\pmod{4}$, which is impossible to be a perfect square. Thus, we could conclude $f(0)=0$. Now, we can have $xf(-x)=\frac{f(x)^2}{x}, x^2f(-x)=f(x)^2$, which could conclude $f(x)=0$. Assertion $P(-x,0)\implies x^2f(x)=f(-x)^2$. Thus, we have $\frac{f(x)^2}{f(-x)}=\frac{f(-x)^2}{f(x)}\implies f(x)=f(-x)$. As a result, we could conclude $\boxed{x^2=f(x)}$

Bluesoul wrote: Now, we can have $xf(-x)=\frac{f(x)^2}{x}, x^2f(-x)=f(x)^2$, which could conclude $f(x)=0$. Assertion $P(-x,0)\implies x^2f(x)=f(-x)^2$. Thus, we have $\frac{f(x)^2}{f(-x)}=\frac{f(-x)^2}{f(x)}\implies f(x)=f(-x)$. As a result, we could conclude $\boxed{x^2=f(x)}$ Second half here is not correct (poinwise trap), read some of the solutions above.

First I will show that $f(0)=0$, let $P(x, y)$ denote assertion. $P(p, 0)$ where $p$ is prime. \[pf(-p)=\frac{f(p)^2}{p}+f(0)\]Clearly we have that $p\mid f(p)$, thus we get that $p\mid \frac{f(p)^2}{p}+f(0)$ as $p\mid \frac{f(p)^2}{p}$ we get $p\mid f(0)$ for all $p$ which suffices. $P(x, 0)$ \[x^2f(-x)=f(x)^2\]Thus we get that $f(x)=f(-x)\in {0, x^2}$ for all $x$, now suppose that there exists some $k\neq 0$ such that $f(k)=0$ $P(x, k)$ \[xf(-x)+k^2f(2x)=\frac{f(x)^2}{x}\]Thus we get that $f(2x)=0$ for all $x$. Suppose there exists some $i$ such that $f(i)=i^2$ and $i\neq 0$. $P(2x, i)$ \[i^2(4x-i)=f(i^3)\]Thus $f(i^3)=0$ for all $i$ and $f(4x-i)=0$ for all $x$. Thus $f(1)=0$. As we have $f(-i)=i^2$ we also get $f(4x+i)=0$ for all $x$ which suffices to prove that $f(x)=0$ for all $x$, or $f(x)=x^2$ for all $x$.