Suppose I have a problem involving functional equations, such as the one below ($0 \not \in \mathbb{N}$):
Find all functions $f:\mathbb{N} \rightarrow\mathbb{N}$ such that
- $f(n)$ is a square for each $n\in \mathbb{N}$;
- $f(m+n) = f(m) + f(n) + 2mn \forall m,n \in \mathbb{N}$
By sheer inspection, we see that $f(n)=n^2$ might be a solution. Suppose we prove by induction that if $f(n)=n^2$ for $n=1,2,\ldots,k$, then $f(k+1)=(k+1)^2$ (it's not hard).
Does this prove that $f(n)=n^2$ is the only solution to the given functional equation?