20171019, 06:15  #1 
May 2004
2^{2}×79 Posts 
Gaussian integers use of norms
Let f(x) = a^x + c = m where a and x belong to N; c is a Gaussian integer.Then a^(x+k*norm(m)) + c = = 0 (mod m). Here k belongs to N.
Last fiddled with by devarajkandadai on 20171019 at 06:16 
20171019, 09:07  #2 
May 2004
2^{2}×79 Posts 
That should read a^(x+ k*Eulerphi(norm(m)) +c = = 0 (mod m).
Last fiddled with by devarajkandadai on 20171019 at 09:08 
20171020, 10:51  #3 
Dec 2012
The Netherlands
3345_{8} Posts 
There is a missing bracket in the 2nd post.
Are you claiming that this is true for all natural numbers a,x,k and all Gaussian integers c? There appear to be obvious counterexamples  but perhaps I have not understood you correctly. 
20171020, 14:48  #4  
Feb 2017
Nowhere
31×197 Posts 
Quote:
a^(x+ k*Eulerphi(norm(m))) + c == 0 (mod m) then substituting a = 2, x = 1, c = 2, m = 4 gives 2^(1 +8*k) + 2 == 0 (mod 4) which only holds for k = 0. Exercise: Supply an additional hypothesis, under which your statement becomes correct. 

20171027, 04:24  #5 
May 2004
2^{2}·79 Posts 
Yesthis can be tested if you have pari.

20171027, 05:08  #6  
May 2004
2^{2}×79 Posts 
Quote:


20171027, 06:01  #7 
May 2004
2^{2}×79 Posts 
Yes this can be easily tested if you have pari. btw did you attend the AMSBENELUX conference at Antwerp in May 1996? I was there.

20171027, 13:07  #8 
Feb 2017
Nowhere
31·197 Posts 

20171028, 05:25  #9  
May 2004
2^{2}×79 Posts 
Quote:


20171028, 14:23  #10  
Feb 2017
Nowhere
31·197 Posts 
Quote:
Quote:
No matter. Your attempt to obviate my counterexample by imposing an ad hoc, post hoc condition, is rendered nugatory by the following, just as easily constructed example. Taking a = 10, x = 1, c = 1 + 2*I, m = 11 + 2*I, norm(m) = 125 we obtain 10^(1 + 125*k) + 1 + 2*I == 0 mod (11 + 2*I) The only integer k for which this holds is k = 0. Now, please go wipe the egg off your face, and consider the exercise I proposed. 

20171028, 15:39  #11 
Dec 2012
The Netherlands
5×353 Posts 

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