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Old 2017-06-21, 10:16   #1
devarajkandadai
 
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Default 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.
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Old 2017-06-21, 17:21   #2
Nick
 
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Quote:
Originally Posted by devarajkandadai View Post
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:
http://www.mersenneforum.org/showthread.php?t=22223

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
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Old 2017-06-23, 04:39   #3
devarajkandadai
 
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Default 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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