Let $f, g:\mathbb{N_0} \mapsto \mathbb{N_0} $ be such that

  1. $f(1) >0, g(1)>0 $
  2. $f(g(n)) = g(f(n))$ $ \forall n\in \mathbb{N_0}$
  3. $f(m^2+g(n))=f(m)^2+g(n)$ $\forall m, n \in \mathbb{N_0}$
  4. $g(m^2+f(n))=g(m)^2+f(n)$ $\forall m, n \in \mathbb{N_0}$

Prove that $f(n)=n$ $\forall n \in \mathbb{N_0}$

What have I done so far:

Plugging in $m=n=0$ and using $(3)$ and $(4)$, we see that $$f(g(0))=f(0)^2+g(0)$$ and $$g(f(0))=g(0)^2+f(0)$$ Now using the commutativity of the functions we see that $$ g(0)^2+f(0)=f(0)^2+g(0)$$ The hint in the book I'm referring says to use this and conclude $2g(0)^2=g(g(0))^2$ and therefore $g(0)=0$. Similarly $f(0)=0$. Now this is the part where I'm confused at. Please explain how I should proceed in this step, as the other parts of the proof are clear to me.

The book I am using is Functional Equations: A Problem Solving Approach by B.J. Venkatachala.

Thank you.

  • $\begingroup$ Someone please help $\endgroup$
    – QFTheorist
    Feb 13, 2018 at 11:20

1 Answer 1


Note that $$g(g(f(0)))=g(g(0)^2+f(0))=g(g(0))^2+f(0).$$ On the other hand, $$g(f(g(0)))=g(0^2+f(g(0)))=g(0)^2+f(g(0))=g(0)^2+g(f(0))=2g(0)^2+f(0).$$ Since $g$ and $f$ commute, these two expressions are actually equal, and so $g(g(0))^2=2g(0)^2$.

  • $\begingroup$ Thank you for your help. I certainly didn't see that :) $\endgroup$
    – QFTheorist
    Feb 13, 2018 at 17:22
  • 1
    $\begingroup$ Yeah, it wasn't easy to find! I actually was stumped and was just looking for a different way to prove $g(0)=0$ when I finally stumbled across it. $\endgroup$ Feb 14, 2018 at 2:18

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