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Let $\pi = a+bi \in \mathbb{Z}[i]$ and $q \equiv 3 \pmod{4}$ a rational prime. Show that $\pi^q \equiv \bar{\pi} \pmod{q}.$

It's a problem from chapter 9 "cubic and biquadratic reciprocity" of Rosen's classical introduction to modern number theory, and I have no idea how to solve, please helps.

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If you raise $a+bi$ to the $q$-th power and use binomial theorem all the terms are divisible by $q$ except the first and the last so $$ (a+bi)^q \equiv a^q + b^q i^q \pmod{q}$$ Now you can use Fermat's little theorem and the periodicity of the power's of $i$ to find the result.

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  • $\begingroup$ Very nice indeed. +1 $\endgroup$ – DonAntonio Mar 15 '14 at 13:15
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Here is a more conceptual explanation:

$\mathbb Z[i]/ (q)$ is a field of order $q^2$.

Since complex conjugation induces an automorphism of $\mathbb Z[i]$ which preserves the ideal $(q)$ (because $q = \overline{q}$) it induces a (non-trivial!) automorphism of $\mathbb Z[i]/ (q)$.

On the other hand, the general theory of finite fields tells us that the automorphism group of a field of order $q^2$ has order two, and that the non-trivial automorphism is given by raising to the $q$th power.

Combining the observations of the two paragraphs gives the desired result.

(It is a special case of the theory of Frobenius automorphisms in algebraic number theory.)

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