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The Question:

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


My Thoughts:

So from what I understand

\begin{align} \ & \phi:\Bbb Z/p\Bbb Z \rightarrow \Bbb Z/p\Bbb Z \quad\text{is a homomorphism} \\ \ \iff & \phi(xy) \equiv \phi(x)\phi(y) \pmod p \quad \forall \; x,y\\ \ \iff & (xy)^{13}\equiv x^{13}y^{13} \pmod p\quad \forall \; x,y \end{align}

But isn't $(xy)^{13}\equiv x^{13}y^{13} \pmod p\quad \forall \; x,y$ always true, regardless of $p$?

Or is there something I am not understanding?

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    $\begingroup$ What you're missing is you also need to show that $(x+y)^{13}=x^{13}+y^{13}$. $\endgroup$ – David C. Ullrich Apr 28 '18 at 16:27
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The notation "$xy$" is really $x\cdot y$ where $\cdot$ is the operation in $\mathbb{Z}/p\mathbb{Z}$, ie $\cdot=+$, where "+" is the sum $\mod p$.

Then, "$(xy)^{13}=x^{13}y^{13}$" means $(x+y)^{13}=x^{13}+y^{13}$

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That is always a homomorphism regard to multiplicity. But if you think to addition then you have to $p=13$ .

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  • $\begingroup$ What about p = 2? $\endgroup$ – Samuel Barkes May 5 '20 at 15:46

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