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 2017-06-21, 10:16 #1 devarajkandadai     May 2004 31610 Posts Modified Fermat's theorem Modified Fermat's theorem: Let a belong to the ring of Gaussian integers. Then a^(p^2-1) = = 1 (mod p). Here p is a prime number with shape 4m + 1 or 4m + 3.
2017-06-21, 17:21   #2
Nick

Dec 2012
The Netherlands

3·5·113 Posts

Quote:
 Originally Posted by devarajkandadai Modified Fermat's theorem: Let a belong to the ring of Gaussian integers. Then a^(p^2-1) = = 1 (mod p). Here p is a prime number with shape 4m + 1 or 4m + 3.
I assume you intend a and p to be coprime.

The case where p=4m+3 for some integer m is the one we already looked at in your earlier thread:

Suppose $$p=4m+1$$ for some integer $$m$$.
Then the prime factorization of $$p$$ in the Gaussian integers has the form $$p=q\bar{q}$$ for some Gaussian prime $$q$$ by Theorem 62 in our course,
where $$\bar{q}$$ is the complex conjugate of $$q$$, which is not an associate of $$q$$.
By the Chinese Remainder Theorem, we get
$\mathbb{Z}[i]/p\mathbb{Z}[i]\cong\mathbb{Z}[i]/q\mathbb{Z}[i]\times\mathbb{Z}[i]/\bar{q}\mathbb{Z}[i]$
so also
$\left(\mathbb{Z}[i]/p\mathbb{Z}[i]\right)^*\cong\left(\mathbb{Z}[i]/q\mathbb{Z}[i]\right)^*\times\left(\mathbb{Z}[i]/\bar{q}\mathbb{Z}[i]\right)^*$
Now $$N(q)N(\bar{q})=N(p)=p^2$$ so $$N(q)=N(\bar{q})=p$$ and therefore $$\left(\mathbb{Z}[i]/q\mathbb{Z}[i]\right)^*$$ has $$p-1$$ elements, and similarly for $$\left(\mathbb{Z}[i]/\bar{q}\mathbb{Z}[i]\right)^*$$.
It follows that the order of any element of $$\left(\mathbb{Z}[i]/p\mathbb{Z}[i]\right)^*$$ divides $$p-1$$, which is a factor of $$p^2-1$$, leading to the statement you gave.

Last fiddled with by Nick on 2017-06-22 at 07:33 Reason: Fixed typos

 2017-06-23, 04:39 #3 devarajkandadai     May 2004 22×79 Posts Modified Fermat's theorem Yes, Nick; you are right - I forgot to add that a and p should be co-prime. Thank you for a simple proof.

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