Iranian Mathematical Olympiad

$\boxed{f(n)=(n+2\ell {)}^2-2{\ell }^2,\forall n\in {\mathbb{Z}}^{+},\ell \ge 0}$.

We set $F(a,b)=f(a)+f(b)+2ab$ and we know that $\forall a,b\in {\mathbb{Z}}^{+}$, $F(a,b)$ is a perfect square.

$F(a,2)=f(a)+f(2)+4a=x^2$ $F(a,1)=f(a)+f(1)+2a=y^2$ $⟹2a+f(2)-f(1)=x^2-y^2$

If $f(2)-f(1)=2k$ $⟹2(a+k)=x^2-y^2$ $⟹2\mid x^2-y^2$ $⟹4\mid x^2-y^2$ $⟹4\mid 2(a+k)$ $⟹2\mid a+k$

This statement is right for any natural $a$ but we can choose $a$ such that $2\nmid a+k$ which is a contradiction.

So $f(2)-f(1)=2k+1$ $⟹2a+2k+1=x^2-y^2$

We choose an odd, prime number such as $p$ and we set $a=\frac{p-(2k+1)}{2}$ $⟹p=x^2-y^2=(x-y)(x+y)$ \implies \left\{ $$\begin{array}{l}x+y=p\\ x-y=1\end{array}$$ \right. $⟹x=\frac{p+1}{2}$ $⟹f\left(\frac{p-(2k+1)}{2}\right)+f(2)+4\left(\frac{p-(2k+1)}{2}\right)={\left(\frac{p+1}{2}\right)}^2$ $⟹f(a)+f(2)+4a={\left(\frac{2a+2k+2}{2}\right)}^2=(a+k+1{)}^2$ $⟹f(a)=a^2+2(k+1)a-4a+f(2)+(k+1{)}^2$ $=(a+(k-1){)}^2+(k+1{)}^2-(k-1{)}^2+f(2)$

We set $k-1=b$ and $(k+1{)}^2-(k-1{)}^2+f(2)=c$ so for any $a=\frac{p-(2k+1)}{2}$ we have $f(a)=(a+b{)}^2+c$

So for these kinds of $a$ $F(a,a)=2f(a)+2a^2=2((a+b{)}^2+c+a^2)=(2a+b{)}^2+b^2+2c$

So for infinite many $a$ we have $(2a+b{)}^2+b^2+2c$ is a perfect square but if $a$ is large enough so that $|b^2+2c|<2a$ we have either $$(2a+b-1{)}^2<(2a+b{)}^2+b^2+2c\le (2a+b{)}^2$$ or $$(2a+b{)}^2\le (2a+b{)}^2+b^2+2c<(2a+b+1{)}^2$$ so $$(2a+b{)}^2=(2a+b{)}^2+b^2+2c⟹b^2+2c=0⟹b=2\ell ,c=-2{\ell }^2$$

So for infinite many $a$ we have $$f(a)=(a+2\ell {)}^2-2{\ell }^2$$

We choose an arbitrary $b$ and $a$ such as chosen above we know $$f(a)+f(b)+2ab=(a+2\ell {)}^2-2{\ell }^2+f(b)+2ab=(a+b+2\ell {)}^2-b^2+f(b)-4b\ell -2{\ell }^2$$ is a perfect square. If we choose $a$ large enough we have $$(a+b+2\ell {)}^2\le (a+b+2\ell {)}^2-b^2+f(b)-4b\ell -2{\ell }^2<(a+b+2\ell +1{)}^2$$ or $$(a+b+2\ell -1{)}^2<(a+b+2\ell {)}^2-b^2+f(b)-4b\ell -2{\ell }^2\le (a+b+2\ell +1{)}^2$$ so $$-b^2+f(b)-4b\ell -2{\ell }^2=0⟹f(b)=(b+2\ell {)}^2-2{\ell }^2$$ so $$\forall n\in {\mathbb{Z}}^{+}:f(n)=(n+2\ell {)}^2-2{\ell }^2$$

If $\ell <0$ for $n=-2\ell$ we have $$f(-2\ell )\in {\mathbb{Z}}^{+}⟹-2{\ell }^2\in {\mathbb{Z}}^{+}$$ which is a contradiction. So $\ell \ge 0$ and we can check in the main equality that for any such $\ell$ $$f(a)+f(b)+2ab=(a+b+2\ell {)}^2.$$