Find the function $f$ such that $f(m^2 + f(n))=(f(m))^2 + n$

Question

Find all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ such that $f(m^2 + f(n))=(f(m))^2 + n, \forall m,n \in \mathbb{Z} \tag 1$

---

If $m=0$ then $f(f(n))=f^2(0) + n \tag 2, \forall n$ From (2) $f \circ f$ is injective, therefore $f$ is injective.

Replacing $n=0, m:=-m$ in (1) we get $f(m^2 + f(0))=f^2(-m)$ therefore $f^2(m)=f^2(-m), \forall m \tag3$ From (3), using $f$ injectivity we get $f(-m)=-f(m), \forall m \ne 0 \tag 4$

Replacing $n:=-n$ in (2) we get $f(f(-n))=(f(0))^2 - n \tag 5$ and, from (4) and (5) $-f(f(n))=f^2(0) - n \tag 6$ Now, from (2) and (6) we get $f(0)=0$ therefore: $f(f(n))=n, \forall n \tag 7$ The last one proves $f$ is also surjective, therefore bijective.

It's also easy to prove $f(1)=1$. I have strong feelings that $f(n)=n, \forall n$ is the only solution, but I cannot prove it.

Any help is appreciated.

Answer 1

We prove by induction that $f(n)=n$ for all $n \in \mathbb N_0$:

The cases $n=0$ and $n=1$ are clear.

Now suppose $f(n)=n$ for some $n$. Then, use $f(m^2 + f(n))=(f(m))^2 + n$ with $m=1$,

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

From $(4)$ you get

$f(m)=m$ for all $m \in \mathbb Z$.

Answer 2

You can do this by induction. You have already done the base case i.e $n=1$.
Let $f(n)=n$, then, $$ f(n+1)=f(1^2+f(n))=f(1)^2+n=n+1.$$

Answer 3

From (1) we have $$ f(1+f(n))= f(1)^2 + n = 1+n. $$

applying $f(\cdot)$ in this equation and using (7) we have $$ 1+f(n) = f(f(1+f(n))) = f(1+n). $$

Thus your conjecture follows by induction

Answer 4

Well, if it's really easy to prove $f(1)=1$: $m=1$ gives $f(f(n)+1)=n+1,$ replacing $n$ by $f(n)$ and using $f(f(n))=n$ gives $f(n+1)=f(n)+1.$ The rest is obvious.