20170621, 10:16  #1 
May 2004
316_{10} Posts 
Modified Fermat's theorem
Modified Fermat's theorem: Let a belong to the ring of Gaussian integers. Then
a^(p^21) = = 1 (mod p). Here p is a prime number with shape 4m + 1 or 4m + 3. 
20170621, 17:21  #2  
Dec 2012
The Netherlands
2^{3}·3^{2}·5^{2} Posts 
Quote:
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 \(p1\) 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 \(p1\), which is a factor of \(p^21\), leading to the statement you gave. Last fiddled with by Nick on 20170622 at 07:33 Reason: Fixed typos 

20170623, 04:39  #3 
May 2004
2^{2}·79 Posts 
Modified Fermat's theorem
Yes, Nick; you are right  I forgot to add that a and p should be coprime. Thank you for a simple proof.

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