parmenides51 wrote: Find all functions ${ f : N \rightarrow N}$ (where ${N}$ is the set of the natural numbers and is assumed to contain ${0}$), such that ${f(x^2) - f(y^2) = f(x + y)f(x - y)}$ for all ${x, y \in N}$ with ${x \ge y}$. Let $P(x,y)$ be the assertion $f(x^2)-f(y^2)=f(x+y)f(x-y)$ $P(0,0)$ $\implies$ $f(0)=0$ $P(x,0)$ $\implies$ $f(x^2)=f(x)^2$ So $P(x,y)$ may be written as $Q(x,y)$ : $f(x)^2-f(y)^2=f(x+y)f(x-y)$ This implies that $f(x)\ge f(y)$ $\forall x\ge y\ge 0$ and so $f(x)$ is nondecreasing If $f(u)=0$ for some $u>0$, then $Q(x+u,x)$ $\implies$ $f(x+u)=f(x)$ so $f(x)$ is periodic and so, since nondecreasing, constant. And so $\boxed{\text{S1 : }f(x)=0\quad\forall x\in\mathbb Z_{\ge 0}}$ which is the only constant solution. So let us from now consider $f(x)>0$ $\forall x>0$ $P(1,0)$ $\implies$ $f(1)=1$ Subtracting $Q(x+1,1)$ from $Q(x+2,1)$, we get : $\frac{f(x)+f(x+2)}{f(x+1)})=\frac{f(x+1)+f(x+3)}{f(x+2)}$ $\forall x\ge 0$ And so $\frac{f(x)+f(x+2)}{f(x+1)}=a$ constant $\forall x\ge 0$ Which is $f(x+2)=af(x+1)-f(x)$ $\forall x\ge 0$ and for some $a$ This gives : $f(0)=0$ $f(1)=1$ $f(2)=a$ $f(3)=a^2-1$ $f(4)=a^3-2a$ Since $f(4)=f(2^2)=f(2)^2$, we get $a^3-2a=a^2$ and so $a\in\{-1,0,2\}$ and so $a=2$ and $f(x+2)=2f(x+1)-f(x)$ Which immediately gives $\boxed{\text{S2 : }f(x)=x\quad\forall x\in\mathbb Z_{\ge 0}}$ which indeed is a solution.

Let $P(x,y)$ be the given assertion. $P(0,0)\implies f(0)=0.$ $P(x,0)\implies f(x^2)=(f(x))^2.$ $P(1,0)\implies f(1)=0$ or $f(1)=1.$ Case 1: $f(1)=0.$ $P(x+1,x)\implies f((x+1)^2)-f(x^2)=f(2x+1)f(1)=(f(x+1))^-(f(x))^2=0$ $\implies f(x+1)=f(x) \implies \boxed{f(x)=0},$ which obviously works. Case 2: $f(1)=1.$ $P(2,1)\implies f(2^2)-f(1)=f(1)f(3)=(f(2))^2-1 \implies f(3)=(f(2))^2-1.$ $P(3,1)\implies ((f(2))^2-1)^2-1=(f(2))^3,$ it follows that $$f(2)=0\text{ or } f(2)=2.$$Subcase 1: $f(2)=0.$ Then $f(3)=-1,$ a contradiction. Subcase 2: $f(2)=2.$ Then by induction, $\boxed{f(n)=n},$ for all natural $n$ works.