Find functions $f:\mathbb{N}\rightarrow \mathbb{N}$ such that $ f(m+n)=f(m)+f(n)+2mn$

Question

Find functions $f:\mathbb{N}\rightarrow \mathbb{N}$ such that $ f(m+n)=f(m)+f(n)+2mn$

Setting $m=n$ we see that: $$ f(2n)=2f(n)+2n^2$$

Setting $m=1$:

$$ f(n+1)=f(1)+f(n)+2n$$

Then, for example when $n=1$:

$$ f(2)=f(1)+f(1)+2=2f(1)+2$$

And we can observe that $f(n)=n^2$ satisfies the equations, as:

$$ f(m+n)=(m+n)^2=m^2+n^2+2mn=f(m)+f(n)+2mn$$

Then, if we assume that $f(n)=n^2+g(n)$, then we have:

$$ f(m+n)=(m+n)^2+g(m+n)=m^2+g(m)+n^2+g(n)+2mn $$ and so:

$$ g(m+n)=g(m)+g(n)$$ And that is Cauchy functional equation for natural numbers, so we see that we can only have $$g(n)=cn$$ for some $c$, so our solution would look like:

$$ f(n)=n^2+cn$$

And in this case:

$$ f(m+n)=(m+n)^2+c(m+n)$$ and $$f(m)+f(n)+2mn=m^2+cm+n^2+cn+2mn$$

so we see that $f(n)=n^2+cn$ is a solution. My question is:

is my solution correct? Does it suffice to do what I did to show that's the only one such function?

Answer 1

Hint Let $$g(n)=f(n)-n^2$$

Then $$g(m+n)=g(m)+g(n)$$

Then setting $g(1)=a$ what is $g(2), g(3),.., g(n)$?

Answer 2

Looks correct to me, wirh the addition of $f(1)$ being arbitrary and $f(n) = n^2+(f(1)-1)n$.