Re: Find all primes $p$ such that $\phi(x)=x^{13}$ is a homomorphism $\Bbb Z/p\Bbb Z \rightarrow \Bbb Z/p\Bbb Z$.

Question

Note:

This is NOT a duplicate of this question.

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The Question:

Find all primes $p$ such that $\phi(x)=x^{13}$ is a homomorphism $\Bbb Z/p\Bbb Z \rightarrow \Bbb Z/p\Bbb Z$

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My Thoughts:

OK, so now I understand that I am supposed to find all primes $p$ such that

$$(x+y)^{13} \equiv x^{13}+y^{13} \pmod p \quad \forall \; x,y \in \Bbb Z/p\Bbb Z$$

I expanded out the binomial to get

\begin{align} \ & \iff \sum_{k=0}^{13}C_k^{13}x^ky^{13-k} \equiv x^{13}+y^{13}\pmod p \quad \forall \; x,y \in \Bbb Z/p\Bbb Z \\ \ & \iff 13x^{12}y+78x^{11}y^2+\cdots+78x^2y^{11}+13xy^{12} \equiv 0 \pmod p \quad \forall \; x,y \in \Bbb Z/p\Bbb Z \end{align}

Clearly, $p=13$ is a solution. But $p=2$ also works, just by inspection.

How can I prove that there is no other solution?

Answer 1

a necessary condition is that $(1+1)^{13} \equiv 1^{13} + 1^{13} \pmod{p}$ which is saying $p \mid 2^{13}-2 = 2\cdot 3^2 \cdot 5 \cdot 7 \cdot 13$. You can verify which primes work (i think all do).