, 2015

Solution:
We directly compute/expand
$$\begin{aligned} \operatorname{Re}\left(z^{p}-z\right) & = \operatorname{Re}\left((a+b i)^{p}-(a+b i)\right) \\ & = \left[a^{p} - \binom{p}{2} a^{p-2} b^{2} + \binom{p}{4} a^{p-4} b^{4} - \cdots \right] - a . \end{aligned}$$
Since $\binom{p}{i}$ is divisible by $p$ for all $i=2,4,6, \ldots$ (since $1 \leq i \leq p-1$), we have
$$\left[a^{p} - \binom{p}{2} a^{p-2} b^{2} + \binom{p}{4} a^{p-4} b^{4} - \cdots \right] - a \equiv a^{p} - a \equiv 0 \quad (\bmod p)$$
by Fermat's little theorem. Thus $p$ divides the real part of $z^{p}-z$.

Solution 2:
From the Frobenius endomorphism,
$$z^{p} = (a + b i)^{p} \equiv a^{p} + (b i)^{p} = a^{p} \pm b^{p} i \equiv a \pm b i \quad (\bmod p \cdot \mathbb{Z}*)$$
where we're using congruence of Gaussian integers (so that $u \equiv v (\bmod p)$ if and only if $\frac{u-v}{p}$ is a Gaussian integer). This is equivalent to the simultaneous congruence of the real and imaginary parts modulo $p$, so the real part of $z^{p}$ is congruent to $a$, the real part of $z$. So indeed $p$ divides the real part of $z^{p}-z$.