I want to find all functions $f:\mathbb{Z}\rightarrow\mathbb{Z}$ such that for all integers $n,m\in\mathbb{Z}$,


Obviously, any constant function is a solution and probably there are not any further solutions (I am not completely sure about that though). Unfortunately, I do not know how to solve the above problem and would like to ask you to help me.

Obviously, the function has to be an even function (put e.g. $m=0$). I tried some more substitutions to obtain further information. E.g. putting $n=\pm 1$ gives $f(m^2-1)=f(m^2+1)$ for all $m\in\mathbb{Z}$. etc.

I would appreciate any help.

  • $\begingroup$ I don't see how you get $f(m^2-1)=f(m^2+1)$. Is there a typo in the question or in this remark? Or am I just blind? $\endgroup$ – Arnaud Mortier Apr 3 '18 at 9:04
  • $\begingroup$ @ArnaudMortier Set $n=1$ to get $f(1+m) = f(1+m^2)$; set $n=-1$ to get $f(1+m) = f(-1+m^2)$. Both are equal to $f(1+m)$. $\endgroup$ – Patrick Stevens Apr 3 '18 at 9:05
  • $\begingroup$ @PatrickStevens Right! Thanks! $\endgroup$ – Arnaud Mortier Apr 3 '18 at 9:05
  • $\begingroup$ Maybe you can do something in this line: Take $n, m_1, m_2$. with $m_1 +m_2 = 0$. Then $n+m_1^2=n+m_2^2$, but $n^2+m_1 \neq n^2+m_2$ $\forall m_1 \neq 0$. $\endgroup$ – Botond Apr 3 '18 at 9:08
  • $\begingroup$ In fact $f(n^2+a)=f(n^2-a)$ for any $n$ and any $a$. $\endgroup$ – Arnaud Mortier Apr 3 '18 at 9:23

The answer is$$ f(x) = \begin{cases} a; & x \text{ odd}\\ b; & x \text{ even} \end{cases}, $$ where $a, b$ are constants.

First to verify that such $f$'s indeed satisfy the condition. For any $m, n \in \mathbb{Z}$, because $m^2 \equiv m \pmod{2}$ and $n^2 \equiv n \pmod{2}$, then either $m^2 + n$ and $m + n^2$ are both odd or both even. Thus either $f(m^2 + n) = a = f(m + n^2)$ or $f(m^2 + n) = b = f(m + n^2)$.

Next, suppose $f$ is any function satisfying the condition. For any $x \in \mathbb{Z}$, take $(m, n) = (x, 0)$, then $f(x) = f(x^2)$, which also implies $f(-x) = f(x^2) = f(x)$. Now take $(m, n) = (x - 1, 1)$, then $f(x) = f(x^2 - 2x + 2)$. Take $(m, n) = (-x + 1, 1)$, then $f(-x + 2) = f(x^2 - 2x + 2)$. Therefore,$$ f(x) = f(-x + 2) = f(x - 2). \quad \forall x \in \mathbb{Z} $$

| cite | improve this answer | |

Another way to arrive at Alex Francisco's answer:

Write $n\equiv m$ if $f(n)=f(m)$ is a consequence of the problem's conditions.

Then one easily sees that $n^2-a\equiv n^2+a$ for any integers $n$ and $a$. In other words an integer and its symmetric about $n^2$ are always equivalent.

Now take any $n$ and perform the symmetry about $1=1^2$ followed by the symmetry about $0=0^2$. The result is $n-2$. Therefore $f(n)=f(n-2)$ for any integer $n$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.