$f(2k) = 2f(k) = 2f(N - k) = f(2N - 2k) = f(N - (2k - N)) = f(2k - N)$ Thus, by induction, $f(2k - rN) = f(2k)$ for all integers $r$. In particular, if we choose $r = 2s$, $f(2k - 2sN) =2f(k-sN)$ and $f(2k - 2sN) = f(2k) = 2f(k)$. Thus $f$ is periodic with period $N$. Now, suppose $N$ is even. Then, $2f(k) = f(2k) = f(2k + N) = f(2(k + N/2)) = 2f(k + N/2)$. Thus, $f$ is periodic with period $N/2$. By an easy induction, we get that $f$ is periodic with an odd period, say $b$. Then, $f(a2^\varphi^(^b^)) = f(a)$ by Euler's theorem and $f(a2^\varphi^(^b^)) = 2^\varphi^(^b^)f(a)$ by the given condition. Thus, $f(a) = 0$ for all integers $a$. *Typo edited

I do not see how, by induction, you then get $f(2k-rN) = f(2k)$; in particular, for $r=2$, you already have gotten in your first line that $f(2k) = f(2N-2k)$, so why should it be that $f(2N-2k) = f(2k-2N)$ ?

Sorry for the carelessness.....You are right, the argument above only works if $N$ is even. (Replace $k$ with $k - N/2$ in the first line for the induction.) I will try to fix it. Edit: Here is a fix: Let $N$ be odd. Then, for all $k$ odd, we have- $f(k) = f(N - k) = 2f((N - k)/2) = 2f(N - (N - k)/2) = 2f((N +k)/2) = f(N + k) = f(-k)$. Then, $f(k2^b) = 2^bf(k) = 2^bf(-k) = f(-k2^b)$. Thus $f(a) = f(-a)$ for all integers $a$. In particular, $f(k) = f(N - k) = f(k - N)$. Then, indeed, by induction (I hope), $f$ is periodic with period $N$.