37th Iranian Mathematical Olympiad

Claim i. $f(x)$ is injective. Proof. Assume there exist two natural numbers $x_1$ and $x_2$ such that $f(x_1)=f(x_2)$ and $x_1\ne x_2$. Now choose $y$ and $z$ such that $y+z=t^2-x_1$ then we have $$f(y)+f(z)+f(x_1)=a^2$$ for some natural number $a$. Therefore, $$f(y)+f(z)+f(x_1)=f(y)+f(z)+f(x_2)=a^2,$$ so $y+z+x_2$ must be square of some natural number $s$. Hence $$|x_1-x_2|=|t^2-s^2|\ne 0$$ and we have $$|t^2-s^2|\ge 2|t|-1.$$ If we choose $t$ large enough it gives us a contradiction.

Claim ii. For any $x,y\in \mathbb{N}$ we have $f(x+1)-f(x)=f(y+1)-f(y)$. Proof. Assume some arbitrary $x_0$ and $y_0$ with $x_0\ge 2$, and choose $z=t^2-x_0-y_0$ for some large $t$. Then $$f(x_0)+f(y_0)+f(z)=a^2,$$ and $$f(x_0-1)+f(y_0+1)+f(z)=b^2.$$ So we have $$|f(x_0)+f(y_0)-f(x_0-1)+f(y_0+1)|=|a^2-b^2|.$$ Assume that $f(x_0)+f(y_0)-f(x_0-1)+f(y_0+1)\ne 0$. If we choose $z$ large enough, $f(z)$ is a large number too (it's trivial because of the injectivity). But we know $$|f(x_0)+f(y_0)-f(x_0-1)+f(y_0+1)|=|a^2-b^2|\ge 2|a|-1.$$ That means $f(z)$ has an upper bound. Contradiction.

Claim ii implies that $f(x)$ is a linear function so $f(x)=ax+b$ and $$a(x+y+z)+3b,$$ becomes a perfect-square if and only if $$x+y+z,$$ does. That means $a{t}^2+3b$ is a perfect-square for any $t\in \mathbb{N}$. In particular $a(a{t}^2+3b)+3b=t^{\prime 2}$ hence $$3ba+a=t^{\prime 2}-t^2,$$ which is a contradiction unless $3ba+3b=0$ (because the left hand side is either 0 or very big for big $t$). Now obviously $a>0$ and we get $b=0$, $a=n^2$.