USJL wrote: Find all function $f:\mathbb{Z}\rightarrow\mathbb{Z}$ such that $f(f(x)+f(y))+f(x)f(y)=f(x+y)f(x-y)$ for all integer $x,y$ Let $P(x,y)$ be the assertion $f(f(x)+f(y))+f(x)f(y)=f(x+y)f(x-y)$ Let $a=f(0)$ $P(0,0)$ $\implies$ $f(2a)=0$ $P(0,2a)$ $\implies$ $f(a)=0$ $P(2a,a)$ $\implies$ $a=0$ 1) Let $x$ such that $f(x)\ne 0$ $P(x,0)$ $\implies$ $f(f(x))=f(x)^2$ $P(f(x),0)$ $\implies$ $f(f(x)^2)=f(x)^4$ $P(x,x)$ $\implies$ $f(2f(x)^2)=-f(x)^4$ $P(f(x),f(x))$ $\implies$ $f(2f(x))=-f(x)^2$ $P(2f(x),f(x))$ $\implies$ $f(3f(x))=-f(x)^2$ $P(3f(x),f(x))$ $\implies$ $f(4f(x))=f(x)^2$ $P(2f(x),2f(x))$ $\implies$ $f(-2f(x)^2)=-f(x)^4$ $P(4f(x),f(x))$ $\implies$ $f(5f(x))=0$ $P(5f(x),f(x))$ $\implies$ $f(6f(x))=f(x)^2$ $P(4f(x),3f(x))$ $\implies$ $f(7f(x))=-f(x)^2$ $P(5f(x),2f(x))$ $\implies$ $f(-f(x)^2)=f(x)^4$ From there, it is not very difficult to prove that $f(kf(x))=h(k)f(x)^2$ $\forall k\in\mathbb Z$ where : $h(\{0,1,2,3,4,5,6,7,8,9,10,11...\})=\{0,1,-1,-1,1,\overline{0,1,-1,-1,1,},0,1,...\}$ and $h(-x)=h(x)$ Then $f(x)^4=f(f(x)^2)=h(f(x))f(x)^2$ and so $h(f(x))=f(x)^2$ and so $f(x)\in\{-1,1\}$ If $f(x)=1$, this implies $f(k)=h(k)$ If $f(x)=-1$, this implies $f(-k)=h(k)$ and so $f(k)=h(-k)=h(k)$ 2) Solution. We obviously have a first solution which is $\boxed{\text{S1 : }f(x)=0\text{ }\forall x\in\mathbb Z}$ If $f(x)$ is not allzero, then previous paragraph implies $f(x)=h(x)$ and so $\boxed{\text{S2 : }f(\{0,1,2,3,4\}=\{0,1,-1,-1,1\}\text{ and }f(x+5)=f(x)\text{ }\forall x\in\mathbb Z}$ which indeed is a solution.

A way to clarify that $f(x)=(\frac{x}{5})$ is a solution is to notice that $f(x)\equiv x^2\mod 5$ and $(x^2+y^2)^2=(x+y)^2(x-y)^2+4x^2y^2$

This problem is amazing. The answer is $f\equiv0$ and $f(x)=\left(\tfrac x5\right)$. The former obviously works; to check the latter, note that both sides of the functional equation are in $[-2,2]$, so it suffices to show they are congruent modulo $5$. Since $\left(\tfrac x5\right)\equiv x^2\pmod5$, just note that \begin{align*} \left(x^2+y^2\right)^2+x^2y^2&\equiv\left(x^2+y^2\right)^2-4x^2y^2\\ &\equiv(x+y)^2(x-y)^2\pmod5. \end{align*}We will now show these are the only solutions. Let $P(x,y)$ denote the assertion. Claim 1. $f(0)=0$. Proof. Check that $P(0,0)\implies f(2f(0))=0$, $P(0,2f(0))\implies f(f(0))=0$, and $P(f(0),f(0))\implies f(0)=0$, as claimed. $\blacksquare$ Claim 2. $f$ is even. Proof. Note that $P(x,y)$ and $P(y,x)$ give $f(x+y)f(x-y)=f(x+y)f(y-x)$. This means that if $f(n)\ne f(-n)$, for every choice of $x$, $y$ with $x-y=n$, we have $f(x+y)=0$; in other words, $f(m)=0$ for all $m\equiv n\pmod2$. Then $f(n)=f(-n)=0$, contradiction. $\blacksquare$ Claim 3. For integers $n$, we have $f(nf(x))=f(x)^2\left(\tfrac n5\right)$. Proof. This is clearly true for $n=0$, and also when $f(x)=0$ --- henceforth assume $f(x)\ne0$. We prove the hypothesis only for positive integers $n$; this is sufficient since $f$ is even. First $P(x,0)$ and $P(x,x)$ give $f(f(x))=f(x)^2$ and $f(2f(x))=-f(x)^2$ respectively, proving the hypothesis for $n=1$ and $n=2$. Now, we prove the hypothesis inductively for $5\nmid n$ as follows. (Here, $k$ is a nonnegative integer.) For $n=5k+3$, $P(f(x),(5k+2)f(x))\implies-f(x)^4=f(x)^2f( (5k+3)f(x))$. For $n=5k+4$, $P(f(x),(5k+3)f(x))\implies-f(x)^4=-f(x)^2f( (5k+4)f(x))$. For $n=5k+6$, $P(2f(x),(5k+4)f(x))\implies-f(x)^4=-f(x)^2f( (5k+6)f(x))$. For $n=5k+7$, $P(3f(x),(5k+4)f(x))\implies-f(x)^4=f(x)^2f( (5k+7)f(x))$. To prove the hypothesis for $5\mid n$, we first check that \[f(2f(x)^2)=f(2f(f(x)))=-f(f(x))^2=-f(x)^4,\]so $P(f(x),(5k+4)f(x))\implies0=-f(x)^2f( (5k+5)f(x))$, and the claim has been proven. $\blacksquare$ Finally by Claim 3, whenever $f(x),f(y)\ne0$, \[f(x)^2\left(\frac{f(y)}5\right)=f(f(x)f(y))=f(y)^2\left(\frac{f(x)}5\right)\implies\frac{\left(\frac{f(x)}5\right)}{f(x)^2}=\frac{\left(\frac{f(y)}5\right)}{f(y)^2}\]is fixed. Since the numerator of each fraction is in $\{0,1,-1\}$, we know $f(x)\ne0$ implies $|f(x)|$ is fixed. Recalling that $f(f(x))=f(x)^2$ shows that this fixed value is $1$, so if $f\not\equiv0$, then at least one of $1$ or $-1$ is in the range of $f$. Claim 3 with $x$ obeying $|f(x)|=1$ completes the proof.

I claim that the answers are $f(x)\equiv 0$ and $f(x)=\begin{cases}0&x\equiv 0\pmod 5\\1&x\equiv 1\pmod 5\\-1&x\equiv 2\pmod 5\\-1&x\equiv 3\pmod 5\\1&x\equiv 4\pmod 5\end{cases}=\left(\frac{x}{5}\right)$. To show that the latter works, note that $f(x)\equiv x^2$. Using this in the initial assertion, we get $LHS\equiv RHS\pmod 5$. However, as their ranges are between $-2$ and $2$, we get that the assertion is satisfied. Now, denote the assertion as $P(x,y)$. If $f(x)\not\equiv 0$, then choose $c$ such that $f(c)\neq 0$. $P\left(\frac{c+x}{2},\frac{c-x}{2}\right)$ and $P\left(\frac{c-x}{2},\frac{c+x}{2}\right)$ yield $f(c)f(x)=f(c)f(-x)\implies f(x)=f(-x)$. So, $f$ is even. $P(0,0)$ yields $f(2f(0))=0$, so $P(x,2f(0))$ gives $f(f(x))=f(x+2f(0))f(x-2f(0))$. Plugging in $x=0$ gives $f(f(0))=0$, and plugging in $f(0)$ gives $f(f(f(0)))=0\implies f(0)=0$. Now, $P(x,0)$ gives $f(f(x))=f(x)^2$ and $P(x,x)$ gives $f(2f(x))=-f(x)^2$. Using these two equations, we will inductively show that $f(kf(x))=\left(\frac{k}{5}\right)f(x)^{2}$. Of course, our base cases are good from the two equations above. For the inductive step, we split into the cases when $k$ is even or odd. If $k=2n$, consider $P((n-1)f(x),(n+1)f(x))$. We get $$f(kf(x))f(2f(x))=f((n-1)f(x))f((n+1)f(x))+f(f((n-1)f(x))+f((n+1)f(x)))$$$$\implies -f(x)^2f(kf(x))=\left(\frac{n-1}{5}\right)\left(\frac{n+1}{5}\right) f(x)^4+f\left(\left(\left(\frac{n-1}{5}\right)+\left(\frac{n+1}{5}\right)\right)f(x)^2\right)$$Now, I claim that the second term on the RHS is just $\left(\frac{\left(\frac{n-1}{5}\right)+\left(\frac{n+1}{5}\right)}{5}\right)f(x)^4$. To show this, note that we must have $\left(\frac{n-1}{5}\right)+\left(\frac{n+1}{5}\right)\in\{-2,-1,0,1,2\}$. If the sum is $2$ or $-2$, we have $f(2f(x)^2)=f(2f(f(x)))=-f(f(x))^2=-f(x)^4=\left(\frac{\pm 2}{5}\right) f(x)^4$. If the sum is $1$ or $-1$, we have $f(f(x)^2)=f(f(f(x)))=f(f(x))^2=f(x)^4=\left(\frac{\pm 1}{5}\right)f(x)^4$. And, if it is $0$, we have $f(0)=0=\left(\frac{0}{5}\right) f(x)^4$, as desired. So, our equation becomes $$-f(x)^2f(kf(x))=f(x)^4\left(\left(\frac{n-1}{5}\right)+\left(\frac{n+1}{5}\right)+\left(\frac{\left(\frac{n-1}{5}\right)+\left(\frac{n+1}{5}\right)}{5}\right)\right)=\left(\frac{k}{5}\right)\left(\frac{2}{5}\right)f(x)^4=-f(x)^4\left(\frac{k}{5}\right)$$where the simplification is true because $f(x)=\left(\frac{x}{5}\right)$ is a solution to the FE. So, we get $f(kf(x))=\left(\frac{k}{5}\right)f(x)^2$ as desired. When $k=2n+1$, we instead use $P(nf(x),(n+1)f(x))$, and we get the same result, so the claim is proven. (Note that our induction only goes over the positive integers. However, $k\to-k$ easily shows that it extends to negatives as well.) First, suppose that there exists $x$ such that $f(x)\equiv 0\pmod 5$, $f(x)\neq 0$. Note that this means $\nu_5(f(2f(x))>2\nu_5(f(x))$. Now, consider $f(\text{lcm}(f(x),f(2f(x))))$. On one hand, we have that $f\left(\frac{\text{lcm}(f(x),f(2f(x)))}{f(x)}\cdot f(x)\right)=\left(\frac{\frac{\text{lcm}(f(x),f(2f(x)))}{f(x)}}{5}\right) f(x)^2=0$ because $5\bigg | \frac{\text{lcm}(f(x),f(2f(x)))}{f(x)}$. However, we also have $f\left(\frac{\text{lcm}(f(x),f(2f(x)))}{f(2f(x))}\cdot f(2f(x))\right)=\left(\frac{\frac{\text{lcm}(f(x),f(2f(x)))}{f(2f(x))}}{5}\right) f(2f(x))\neq 0$, which is a contradiction. So, if $f(x)\neq 0$, it cannot be divisible by $5$. Now, suppose we have $f(x)=a,f(y)=b$ with $a,b\neq 0$. Then, we have that $f(ab)=f(af(y))=\left(\frac{a}{5}\right) b^2$ and $f(ab)=f(bf(x))=\left(\frac{b}{5}\right)a^2$. These expressions are both nonzero because $5\nmid a,b$. Comparing the magnitudes of the two expressions, we must have $a^2=b^2$. So, if $f$ takes on a value which is not $0\pmod 5$, then all nonzero values have the same magnitude. However, if $f(x)$ is a possible magnitude, then $f(x)^2$ is as well. So, we must have $f(x)\in \{-1,0,1\}$ If $f(1)=-1$, we get $f(-1)=f(1)^2=1$, which is a contradiction to evenness. If $f(1)=0$, $P(x,1)$ yields $f(f(x))=f(x+1)f(x-1)=f(x)^2$, so $f$ is a geometric series. However, as $f(0)=f(1)=0$, we get that $f(x)\equiv 0$. Finally, if $f(1)=1$, our above claim gives $f(k)=\left(\frac{k}{5}\right)(1)^2=\left(\frac{k}{5}\right)$, which is indeed a solution. So, we have shown that our only solutions are $0$ and $\left(\frac{k}{5}\right)$, as desired.

Since zero function works, assume $f$ is not identically zero. Let $P(x,y)$ be the assertion $f(f(x)+f(y))+f(x)f(y)=f(x+y)f(x-y)$. $P(0,0)$ implies $f(2f(0))=0$ and $P(2f(0),0)$ implies $f(f(0))=0$. Now $P(2f(0),f(0))$ implies $f(0)=0$. Assume $f(a)\neq 0$. Comparing $P(\tfrac{n+a}2,\tfrac{a-n}2)$ and $P(\tfrac{a-n}2,\tfrac{n+a}2)$ shows $f(n)=f(-n)$. \begin{align*} P(a,0) &\implies f(f(a))=f(a)^2\\ P(a,a) &\implies f(2f(a))=-f(a)^2 \\ P(f(a),f(a)) &\implies f(2f(a)^2)=-f(a)^4\\ P(2f(a),f(a)) &\implies f(3f(a))=-f(a)^2 \\ P(3f(a),f(a)) &\implies f(4f(a))=f(a)^2 \\ P(4f(a),f(a)) &\implies f(5f(a))=0 \\ P(4f(a),2f(a)) &\implies f(6f(a))=f(a)^2 \\ P(4f(a),3f(a)) &\implies f(7f(a))=-f(a)^2\end{align*}By induction, $f(nf(a))=f(a)^2\left (\frac n5\right)$ [legendre symbol]. For $n=f(a)$, (using $P(f(a),0)$), $f(a)^2=\left (\frac{f(a)}5\right )$ implying $f(a)\in \{1,-1\}$. In any case, $f(n)=\left (\frac n5\right)$. It is not very hard to see this work.

Denote the assertion with $P(x, y)$. Claim: $f$ is even. Proof. Comparing $P(x, y)$ and $P(y, x)$ gives that \[ f(x + y)f(x - y) = f(x + y)f(y - x) \]so for each $x$ and $y$ either $f(x - y) = f(y - x)$ or $f(x + y) = 0$. If $f$ is nonzero at some $a$, then we can fix $x - y$ to any value of the same parity and vary such $x + y = a$, giving that $f$ is even on integers with parity $a$. Else, $f$ is uniformly zero on that parity, in which case the result still holds. $\blacksquare$ Claim: $f(0) = 0$. Proof. Note that by $P(0, 0)$ it follows that $f(2f(0)) = 0$. Then, by $P(2f(0), 0)$ it follows that $f(f(0)) = 0$. Then, by $P(2f(0), f(0))$ it follows that $f(f(2f(0))) = 0$ so $f(0) = 0$. $\blacksquare$ Claim: $f(nf(x)) = f(x)^2 \cdot \left(\frac{n}{5}\right)$ where the last term represents the Legendre symbol. Proof. Skimming over the details, \begin{align*} P(x, 0) \implies f(f(x)) &= f(x)^2 \\ P(x, x) \implies f(2f(x)) &= -f(x)^2 \\ P(2f(x), f(x)) \implies f(3f(x)) &= -f(x)^2 \\ P(3f(x), f(x)) \implies f(4f(x)) &= f(x)^2 \\ P(4f(x), f(x)) \implies f(5f(x)) &= 0 \\ \end{align*}As such, by $P(5f(x), af(x))$ \[ f(af(x))^2 = f((5+a)f(x)) \cdot f((5-a)f(x)) \]which allows us to inductively establish the result. $\blacksquare$ Claim: If $f$ is not uniformly $0$, then $1$ is in the image of $f$. Proof. Suppose that $f(c)$ is nonzero for some $c$. Then, $5 \not\mid f(c)$ as $f(c)^2$ is not in the kernel. Since $f^k(c) = f(c)^{2^{k-1}}$, it follows that there exists a $d \equiv 1 \pmod{5}$ such that $f(d) = d^2$. Then, by $P(5kd, d)$ \[ d^4 = f(d^2) = f(5kd + d)f(5kd - d) \]We can take $k$ such $d^2 = (1 + 5k)d$ and cancel to get $f(5kd - d) = 1$. $\blacksquare$ Thus, if $f$ is not uniformly $0$, then it follows that $f(n) = \left(\frac{n}{5}\right)$, which can be verified to work.