Find all functions for $f:\Bbb{N}\to\Bbb{N}$ such that $f\left(m^2+f(n)\right)=f\left(m^2\right) +n$

Question

I would have given my approach but i didnt get anywhere. I just substituted zeroes and got $f(f(n)) =n$ and I'm just lost. Any help would be appreciated

Answer 1

Assume $0 $ is an element of $\mathbb N$. Set $f(0) = N$. Taking $m=0$ we see $$ ff(n) = N + n $$ Now take $n=0$ in the original equation, we get a second identity $$f(m^2 + N) = f(m^2)$$

Apply $f$ to both sides to see (using first identity), $$ m^2 + 2N = N + m^2$$ Taking $m=0$ implies $N=0$. So now we know $ff(n) = n$. This implies $f$ is bijective. Suppose $f(M)=1$. Then $$ f(m^2 + 1) = f(m^2) + M$$ so we have $$ f(0)=0\\ f(M) = 1\\ f(1) = M $$ Now $f(2) = f(1^2+f(M))=f(1^2)+M = 2M$, and therefore $$ f(2) = 2M \\ f(2M)=2 \\ $$ Now $f(3)=f(1^2+f(2M))=f(1^2) + 2M = 3M$, and therefore $$ f(3)= 3M\\ f(3M)=3.$$ Inductively we obtain $$ f(M)=M^2 \\ f(M^2)=M$$ so $M$ solves $M^2=1$ i.e. $M=1$, and $f(n)=n$ for all $n$.

Answer 2

Solution 1. assuming $0\in\mathbb{N}$

Note that $f$ is clearly injective, so if we substitute $n=0$ we find that $f(m^2+f(0))=f(m^2)$, from which we see that by injectivity that $f(0)=0$. Letting $m=0$ in the original equation, we see that $f$ is indeed an involution as OP found.

Then if we set $m=1$ and $n=f(n)$, we find that $f(n+1)=f(n)+f(1)$. A simple induction shows that $f(n)=nf(1)$ for all $n\in\mathbb{N}$, and so $f(n)=nc$ for some constant $c$.

Substituting this into our original equation, we find $c^2=1$, and after checking, we see that $f(x)=x$ is the only solution to the functional equation. ($c\in\mathbb{N}$, and so is not negative)

Solution 2. assuming $0\notin \mathbb{N}$

Note that when we put in $m=1$, we find that $$f(f(n)+1)=n+1\tag{1}$$ Setting $n=f(n)+1$ in $(1)$ shows that $$\begin{align*} f(f(f(n)+1)+1)&=f(n)+2 \\ \implies f(n+2)=f(n)+2 \tag{2}\end{align*}$$We now prove that $f(1)=1$. Suppose not, i.e. $f(1)=k$ for some integer $k>1$. Then $k-1\in\mathbb{N^+}$. So putting $n=k-1$ in $(2)$ shows that $f(k+1)=f(k-1)+2$. But if we put $n=1$ in $(1)$, we find that $$f(k+1)=f(f(1)+1)=2$$ which implies that $f(k-1)=0$, a contradiction. Hence $k=1\implies f(1)=1$. Hence $f(2)=2$, and a simple induction using $(2)$ shows that $f(n)=n$ for all $n\in\mathbb{N^+}$.