20170425, 03:07  #1 
May 2004
2^{2}×79 Posts 
Conjecture pertaining to Gaussian integers
Let a + ib be a Gaussian integer. Let p be a rational integer prime of shape 4m + 3. Then ((a + ib)^(p^21)  1) == 0 (mod p). This is subject to a + ib and p being coprime.

20170425, 08:56  #2 
Dec 2012
The Netherlands
1,453 Posts 
Yes, this follows from what we did on Gaussian integers in the discussion group (assuming I have understood you correctly).
Let \(R=\mathbb{Z}[i]/p\mathbb{Z}[i]\), the set of Gaussian integers modulo \(p\). Then \(R\) has \(p^2\) elements. As \(p\equiv 3\pmod{4}\), \(p\) remains prime in \(\mathbb{Z}[i]\) (see theorem 62) so \(R\) is a finite integral domain and hence a field, and therefore \(R^*\) has \(p^21\) elements. If \(a+bi\) and \(p\) are coprime then \(\overline{a+bi}\) is a unit in \(R\) so raising it to the power \(p^21\) gives \(\bar{1}\) by Lagrange's theorem (theorem 83). 
20170426, 05:44  #3  
May 2004
316_{10} Posts 
Conjecture pertaining to Gaussian integers
Quote:


20170426, 11:33  #4 
Dec 2012
The Netherlands
1,453 Posts 

20170427, 04:59  #5 
May 2004
2^{2}·79 Posts 
Conjecture pertaining to Gaussian integers
Thank you very much. Would be glad if you wouldlet me have your full name; my id: dkandadai@gmail.com. Incidentally, as founder of maths corner on fb let me invite you to join that group.

20170427, 08:44  #6 
"Brian"
Jul 2007
The Netherlands
3264_{10} Posts 
@devarajkandadai
Nick goes strictly by first name only, being the privacyenthusiast that he is. You won't catch him on Facebook. And he's also too polite to mention that this result would be a simple observation for undergraduate students, so mentioning him by name in that context is not really necessary. 
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