I'm trying to solve the functional equation $f(x+f(y)) = f(x)-y$ where $f : \mathbb{Z} \to \mathbb{Z}$. What I got so far is: $f$ is injective and $f(0) = 0$. Thanks in advance for your time.


As you have observed, $f$ must be injective and satisfy $f(0)=0$. Now for any fixed $a\in\mathbb{Z}$, setting $y=a$ and $x=0$ gives $f(f(a))=-a$, and then setting $x=f(a)$ gives $f(2f(a))=-2a$. You can continue similarly to find $f(nf(a))=-na$ for each $n\in\mathbb{N}$. You can also get $f(nf(a))=-na$ for negative integers $n$ by applying this argument backwards (for instance, for $n=-1$ set $x=-f(a)$).

Thus $f(nf(a))=-na$ for each $n\in\mathbb{Z}$. In particular, if $n=f(b)$ for some $b\in\mathbb{Z}$, this says $$f(f(a)f(b))=-f(b)a=-f(a)b,$$ since you can swap the roles of $a$ and $b$. Thus assuming $a,b\neq 0$, we find $f(a)/a=f(b)/b$. Since $a$ and $b$ were arbitrary, this means there is a constant $c\in\mathbb{Z}$ (the common value of $f(a)/a$) such that $f(x)=cx$ for all $x\in\mathbb{Z}$. Plugging $f(x)=cx$ back into the functional equation gives $$c(x+cy)=cx-y,$$ so $c^2=-1$. Since no integer can satisfy this, there are no functions which satisfy the functional equation.

  • $\begingroup$ Thanks you very much, got it :) $\endgroup$ – Topologicalife Apr 29 '16 at 22:34
  • $\begingroup$ Just one thing: I don't follow you when you say, setting $x = f(a)$ gives $f(2(f(a)) = -2a$. Can you give me any hint please? Thanks. $\endgroup$ – Topologicalife Apr 29 '16 at 22:41
  • $\begingroup$ Okay, I got it. We have to set $y=a$ and $x = f(a)$ so we have $f(f(a) + f(y)) = f(x) - y \iff f(2f(a)) = -2a$ thanks :) :) $\endgroup$ – Topologicalife Apr 29 '16 at 22:55
  • $\begingroup$ @EricWofsey Please explain why $f(0)=0$? $\endgroup$ – Hamid Reza Ebrahimi Sep 27 '17 at 15:39
  • 1
    $\begingroup$ We can also get $f(x)=cx$ by plugging first $x=-f(y)$ to get $f(-f(y))=y$ and then plugging $y=-f(y)$ to obtain Cauchy's functional equation $f(x+y)=f(x)+f(y)$, for which the result is known. $\endgroup$ – Sil May 12 at 10:47

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.