certain polynomials in $\mathbb{Z}[x]$ have the property that their evaluations at different integers all lie in a proper ideal of $\mathbb{Z}$.
for example, this is trivially true if the coefficients of the polynomial themselves lie in the same ideal.
non-trivial examples can be found as a consequence of Fermat's Little Theorem. If $p$ is prime then: $$ \forall n \in \mathbb{Z} \; n^p-n \in (p) $$ stated differently: the polynomial $x^p -x$ annihilates the finite ring $\mathbb{Z}/(p)$
my question concerns the proof (or disproof) of a generalization of this idea, which can be expressed as follows:
for each positive integer $N$ there is a polynomial of degree $N$, with all non-zero coefficients in $\{-1,1\}$, which annihilates $\mathbb{Z}/(N)$
given $N$, the relevant polynomial is defined by: $$ f_N(x) = \sum_{d|N} \mu\bigg(\frac{N}d\bigg) x^d $$ where $\mu$ denotes the Mobius function familiar from elementary number theory. When $N$ is a prime then $f_N(x)$ reduces to $x^N - x$
Now assume $N$ is quadratfrei, so that $(N)$ is a radical ideal of $\mathbb{Z}$. Let $p$ be a prime dividing $N$. For any $t \in \mathbb{Z}$ we have: $$ f_N(t) = f_{\frac{N}p}(t^p) - f_{\frac{N}p}(t) $$ and the RHS lies in $(p)$ by Fermat's Little Theorem. an example should make this clear. Take $N=30$ and $p=5$: $$\begin{align} f_{30}(t) &= t^{30} -t^{15} - t^{10} -t^6 +t^5+t^3 +t^2 -t \\ &= (t^{30} - t^{15}-t^{10} +t^5) -(t^6 -t^3-t^2+t)\\ &= ((t^6)^5-t^6) -((t^3)^5-t^3) -((t^2)^5-t^2) +(t^5-t) \end{align} $$ similar re-arrangements show that if $p$ is any prime dividing the quadratfrei $N$, then $p|f_N(t)$ for any integer $t$. Hence $N|f_N(t)$. If $N$ is not quadratfrei this last assertion does not follow, and the argument shows only that $f_N(t) \in \sqrt{(N)}$ (the radical of $(N)$).
In the general case, for a non-quadratfrei number $M$ we have $M=Nm$ where $N=\sqrt{M}$ and $m \gt 1$. So $$ f_M(t) = f_N(t^m) $$ my question is, can we prove for all $t \in \mathbb{Z}$ that $f_M(t) \in M$?
