I like this - you really just need to calculate... The functional equation $f\left(m-n\right)f\left(m+n\right)=f\left(m^2\right)$ allows us to recurrently compute all values of the function f in terms of the first three values, which we denote by f(1) = a, f(2) = b and f(3) = c. Note that these numbers a, b, c are naturals, so we can divide by them (this is important, since we will often do this in future). Then, $f\left(4\right)=f\left(2^2\right)=f\left(2-1\right)f\left(2+1\right)=f\left(1\right)f\left(3\right)=ac$. Now, $af\left(5\right)=f\left(1\right)f\left(5\right)=f\left(3-2\right)f\left(3+2\right)=f\left(3^2\right)$ $=f\left(3-1\right)f\left(3+1\right)=f\left(2\right)f\left(4\right)=b\cdot ac$. Division by a yields f(5) = bc. Furthermore, $bf\left(6\right)=f\left(2\right)f\left(6\right)=f\left(4-2\right)f\left(4+2\right)=f\left(4^2\right)$ $=f\left(4-1\right)f\left(4+1\right)=f\left(3\right)f\left(5\right)=c\cdot bc=bc^2$. Division by b yields $f\left(6\right)=c^2$. Continuing the game, $cf\left(7\right)=f\left(3\right)f\left(7\right)=f\left(5-2\right)f\left(5+2\right)=f\left(5^2\right)$ $=f\left(5-1\right)f\left(5+1\right)=f\left(4\right)f\left(6\right)=ac\cdot c^2=ac^3$. Division by c yields $f\left(7\right)=ac^2$. Hence, $a^2c^2=a\cdot ac^2=f\left(1\right)f\left(7\right)=f\left(4-3\right)f\left(4+3\right)=f\left(4^2\right)$ $=f\left(4-1\right)f\left(4+1\right)=f\left(3\right)f\left(5\right)=c\cdot bc=bc^2$. Division by $c^2$ yields $a^2=b$. Thus, f(2) = b becomes $f\left(2\right)=a^2$, and f(5) = bc becomes $f\left(5\right)=a^2c$. So we have kicked out b from the parameters a, b, c - our first progress. Let's continue the computing spree: $ac\cdot f\left(8\right)=f\left(4\right)f\left(8\right)=f\left(6-2\right)f\left(6+2\right)=f\left(6^2\right)$ $=f\left(6-1\right)f\left(6+1\right)=f\left(5\right)f\left(7\right)=a^2c\cdot ac^2=a^3c^3$. Division by ac yields $f\left(8\right)=a^2c^2$. Thus, $a^4c^2=a^2\cdot a^2c^2=f\left(2\right)f\left(8\right)=f\left(5-3\right)f\left(5+3\right)=f\left(5^2\right)$ $=f\left(5-1\right)f\left(5+1\right)=f\left(4\right)f\left(6\right)=ac\cdot c^2=ac^3$. Division by $ac^2$ transforms this into $a^3=c$. Thus, the variable c is unnecessary and we are left with a only. Let's first combine: $f\left(1\right)=a$; $f\left(2\right)=a^2$; $f\left(3\right)=c=a^3$; $f\left(4\right)=ac=aa^3=a^4$; $f\left(5\right)=a^2c=a^2a^3=a^5$; $f\left(6\right)=c^2=\left(a^3\right)^2=a^6$; $f\left(7\right)=ac^2=a\left(a^3\right)^2=a^7$; $f\left(8\right)=a^2c^2=a^2\left(a^3\right)^2=a^8$. Reminds you of something? Well, then you might be disappointed in a few moments. We note that $af\left(9\right)=f\left(1\right)f\left(9\right)=f\left(5-4\right)f\left(5+4\right)=f\left(5^2\right)$ $=f\left(5-1\right)f\left(5+1\right)=f\left(4\right)f\left(6\right)=a^4\cdot a^6=a^{10}$, so that, upon division by a, we get $f\left(9\right)=a^9$. But this leads to $a^9=f\left(9\right)=f\left(3^2\right)=f\left(3-1\right)f\left(3+1\right)=f\left(2\right)f\left(4\right)=a^2a^4=a^6$, and thus to $a^3=1$, so that a = 1. How boring! Anyway, the problem is now almost solved. We will show that f(n) = 1 holds for every natural n. This will be proven by induction over n. In fact, for n = 1, n = 2, n = 3, n = 4 and n = 5, this follows from the above formulas f(1) = a, $f\left(2\right)=a^2$, $f\left(3\right)=a^3$, $f\left(4\right)=a^4$ and $f\left(5\right)=a^5$ combined with the killer equation a = 1. Now, assume that N is a natural number $\geq 5$, and the identity f(n) = 1 holds for all naturals n < N. Then, particularly, f(N - 1) = 1, f(N - 3) = 1 and f(N - 4) = 1. Thus, $f\left(N\right)=1\cdot f\left(N\right)=f\left(N-4\right)f\left(N\right)=f\left(\left(N-2\right)-2\right)f\left(\left(N-2\right)+2\right)$ $=f\left(\left(N-2\right)^2\right)=f\left(\left(N-2\right)-1\right)f\left(\left(N-2\right)+1\right)=f\left(N-3\right)f\left(N-1\right)$ $=1\cdot 1$, so that f(N) = 1, and the induction step is done. Thus, f(n) = 1 holds for every natural n. So the only solution of our functional equation is the function f(x) = 1. It is indeed a solution, as a reader could easily verify. Darij

^^^^ Great solution.

what about f(x)=0

$0 \not\in \mathbb{N}$ as long as nothing else is said.

why do you exlude 0 it's a positive and negative integer

In most parts of the world $0$ is considered neither positive nor negative, but thats not important here (did anybody say anything about that¿). $0 \not\in \mathbb{N}$ for me and a lot of other peoples, and it's the problem of the problem-poster when it's not well defined what he wants (and there are good reasons here that $0$ is excluded, like that then also $f(0)$ is defined and the problem gets a at least not harder).

Yes 0 is not natural number in Bulgaria, too And I think that this problem is too easy to be in TST

Here is my approach: By induction we obtain that :$f(m+2)=cf(m)$ for all $m\geq 3$ where $c$ is constant.So : $F(2n)=c^{n-1}f(2)$ and $f(2n+1)=c^nf(1)$ This one is really classic !

EDIT: whoops did some stuff wrong, should be fixed now

If $n\ge 3$, then $f(2n+2)f(2)=f(2n)f(4)=f((n+2)^2)$, so $\frac{f(2n+2)}{f(2n)}=\frac{f(4)}{f(2)}$. Thus, for $n\ge 3$, $f(2n)=AB^n$, for some $A,B \ge 0$, and $f(4n)f(6)=f((2n+3)^2)$, or $A^2B^{4n+6}=AB^{(2n+3)^2}$. This is true as $n \to \infty$, so $A=B=1$, and $f(2n)=1$ for $n\ge 3$ . Then for any $k \ge 1$, take $N \ge k+3$. $f(4N-k)f(k)=f(4N^2)=1, \Rightarrow f(k)=1$.

It's easy to see $\frac {f(m+2)} {f(m)}=\frac {f((n+1)^2)} {f(n^2)}$ So $f(2n)=k^{n-1}f(2)$ and $f(2n+1)=k^nf(1)$ now it's obvious that $\frac {f((m+2)^2)} {f(m^2)}=k^{2m+2}$, from another way it's also $k^2$ so $k=1$,so $f(1)=f(2)=1$ so $f(n)=1$ for all $n$.

EDIT: I mixed up $m$ and $n.$

Could someone check if my solution is correct? Thanks in advance!

Note that for positive $x$ and $d$ such that all terms are positive, we have \begin{align*} f(x-3d)f(x-d)f(x+d)f(x+3d)=(f(x-3d)f(x-d))(f(x+d)f(x+3d))&=(f(x-3d)f(x+d))(f(x-d)f(x+3d)) \\ \implies f((x-2d)^2)f((x+2d)^2)&=f((x-d)^2)f((x+d)^2) \\ \implies f(x^2+4d^2)&=f(x^2+d^2) \end{align*}and \begin{align*} f(x-5d)f(x-d)f(x+d)f(x+5d)=(f(x-5d)f(x-d))(f(x+d)f(x+5d))&=(f(x-5d)f(x+d))(f(x-d)f(x+5d)) \\ \implies f((x-3d)^2)f((x+3d)^2)&=f((x-2d)^2)f((x+2d)^2) \\ \implies f(x^2+9d^2)&=f(x^2+4d^2) \end{align*} In particular, this means that $f(17)=f(20)$ and $f(45)=f(40)$, by setting $(x,d)$ to be $(4,1)$ and $(6, 1)$, respectively. Now $f(17)=f(20)\implies f(16)=f(19)$ because $f(18^2)=f(17)f(19)=f(20)f(16)$. This actually implies for all $x$, we know $f(x)=f(x+3)$ because depending on whether $x$ is odd or even, we can multiply both sides by $f(16)$ or $f(17)$ to get an equality we know is true. Similarly, from $f(45)=f(40)$, we prove that $\forall x, f(x)=f(x+5)$. Now, since $3$ and $5$ are relatively prime, this implies that $\forall x, y\in\mathbb{N}, f(x)=f(y)=k$, for some positive integer $k$. Then, the original equation becomes $k^2=k$, which implies $k=1$. Thus, we've concluded that the only solution is $f(x)=1$.