TITLE: Analogue of Fermat’s Theorem
QUESTION [0 upvotes]: The question is “Establish an analogue of Fermat’s Little Theorem for the ring $\mathbb{Z} [\sqrt{3}]$. I know you begin by letting $\alpha \in \mathbb{Z} [\sqrt{3}]$ so there exists integers $a$ and $b$ s.t $\alpha ^{p}=(a+b\sqrt3)^{p}$ where p is prime. Then I was told you need to use a binomial expansion and Fermat’s Little Theorem to get $\alpha ^{p} =a+b(\sqrt3)^{p}$ mod $p$ but I am unsure how you do this. Then I was told you can use Euler’s criterion and the definition of the Legendre symbol for $(\frac{3}{p})$ to establish an analogue but again i am unsure how you do this.

REPLY [1 votes]: As you said, if $\alpha \in \mathbb{Z}[\sqrt3],$ there exists integers $a$ and $b$ such that $\alpha = a+b\sqrt3$.
Now, we have to show $\alpha^p = a+b(\sqrt3)^p \pmod p$.
Raise $p$ power to the both sides of $\alpha = a+b\sqrt3$ to get:
$\alpha^p = (a+b\sqrt3)^p $.
$=a^p+ p a^{p-1}\sqrt3b+\frac{p(p-1) a^{p-2}}{2}3b^2 + \cdots + p a\sqrt3^{p-1}b^{p-1}+ b^p\sqrt3^p$
Using Wilson theorem we have $a^p =a \pmod p$ and $b^p =b \pmod p.$
Also, since we are reducing mod $p$, all the terms except the first and last will become zero.
Therefore $\alpha^p = a+0+b(\sqrt3)^p = a+b(\sqrt3)^p.$
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