Find all functions $ f : \mathbb N \to \mathbb N $ such that
(a) $ f ( n ) $ is a square for each $ n \in \mathbb N $;
(b) $ f ( m + n ) = f ( m ) + f ( n ) + 2 m n $, for all $ m , n \in \mathbb N $.
Clearly $f(x)=x^2$ is one such function, but is it the only one? If yes, then how can I prove it?
It appeals the "Taiwanese Transformation".
As pointed by Markus Zetto, we will let $f(x)=g(x)^2$ and then "translate by the obvious solution" $f(x)=x^2$, i.e., set $h(x)=g(x)^2-x^2$. You can rewrie the equation as $g(m+n)^2-(m+n)^2=(g(m)^2-m^2)+(g(n^2)-n^2)$ which thus returns $h(m+n)=h(m)+h(n)$ and note that its a Cauchy Equation which is bounded, and hence it is linear, thus $h(x)=ax+c$ for some $c\in \mathbb{R}$, but substituiting this in the original equation clearly gives $c=0$, thus $f(x)=x^2+ax$ for some $a\in \mathbb{N}_0$ and note that $x^2+ax$ cannot be a square for every $x$ as $a$ is fixed, and thus the only solution is $f(x)=x^2$ which works.