Find all injections $f: \mathbb N \to \mathbb N$ such that:
$f(n + m) + f(n - m) = f(n) - f(m) + f(f(m) + n) $
I have an idea to substitute $n=m$, since if $f(n)=f(m)$, then $n=m$ (the rule for injective functions).
If I do that, I get $f(2m)+f(0)=f(f(m)+m)$ and after that I have no idea what to do.
Any help?
Setting $m=0$ we get:
$f(n)+f(n)=f(n)-f(0)+f(f(0)+n)$, this implies $f(n)=f(f(0)+n)-f(0)$.
This gives us: $f(n+f(0))=f(n)+f(0)$.
Now, recall that we have $f(2m)+f(0)=f(f(m)+m)$, but the left side is equal to $f(2m+f(0))$.
So we have $f(2m+f(0))=f(f(m)+m)$, we conclude $2m+f(0)=f(m)+m$ with injectivity.
and so $f(m)=m+f(0)$.