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 Register FAQ Search Today's Posts Mark Forums Read  2017-10-19, 06:15 #1 devarajkandadai   May 2004 4748 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 2017-10-19 at 06:16   2017-10-19, 09:07 #2 devarajkandadai   May 2004 22·79 Posts That should read a^(x+ k*Eulerphi(norm(m)) +c = = 0 (mod m). Last fiddled with by devarajkandadai on 2017-10-19 at 09:08   2017-10-20, 10:51 #3 Nick   Dec 2012 The Netherlands 6E516 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.   2017-10-20, 14:48   #4
Dr Sardonicus

Feb 2017
Nowhere

175B16 Posts Quote:
 Originally Posted by devarajkandadai That should read a^(x+ k*Eulerphi(norm(m)) +c = = 0 (mod m).
Assuming that is meant as

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.   2017-10-27, 04:24   #5

May 2004

22×79 Posts Quote:
 Originally Posted by Nick 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.
Yes-this can be tested if you have pari.   2017-10-27, 05:08   #6

May 2004

22×79 Posts Quote:
 Originally Posted by Dr Sardonicus Assuming that is meant as 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.
Eulerphi(4) is obviously not equal to 8. btw did you attend the AMS conference at Antwerp in May 1996? I was there.   2017-10-27, 06:01   #7

May 2004

22×79 Posts Quote:
 Originally Posted by Nick 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.
Yes- this can be easily tested if you have pari. btw did you attend the AMS-BENELUX conference at Antwerp in May 1996? I was there.   2017-10-27, 13:07   #8
Dr Sardonicus

Feb 2017
Nowhere

3·1,993 Posts Quote:
 Originally Posted by devarajkandadai Eulerphi(4) is obviously not equal to 8.
The formula you gave was

Quote:
 a^(x+ k*Eulerphi(norm(m))) + c == 0 (mod m)
I used m = 4. Taking norm(m) from Q(i) to Q per your formula, we have

norm(4) = 16.

And, Eulerphi(16) = 8.   2017-10-28, 05:25   #9

May 2004

22×79 Posts Quote:
 Originally Posted by Dr Sardonicus Assuming that is meant as 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.
My conjecture states that c has got to be a Gaussian Integer i.e. b cannot be 0.   2017-10-28, 14:23   #10
Dr Sardonicus

Feb 2017
Nowhere

3·1,993 Posts Quote:
 Originally Posted by devarajkandadai My conjecture states that c has got to be a Gaussian Integer i.e. b cannot be 0.
Let's see:

Quote:
 Originally Posted by devarajkandadai 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.
Since there isn't any "b" in your original formula, I can only guess at what you're claiming you said. My best guess is, you're claiming you said that c is not a rational integer. No, you never said that. You said c is a Gaussian integer. And, last I checked, the rational integers were a subset of the Gaussian integers.

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.   2017-10-28, 15:39   #11
Nick

Dec 2012
The Netherlands

5×353 Posts Quote:
 Originally Posted by devarajkandadai btw did you attend the AMS-BENELUX conference at Antwerp in May 1996? I was there.
No, I wasn't there, but I hope you enjoyed your visit to this part of the world!   Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post Cruelty Proth Prime Search 158 2020-07-31 22:23 carpetpool carpetpool 24 2017-10-29 23:47 devarajkandadai Number Theory Discussion Group 5 2017-04-27 08:44 Nick Number Theory Discussion Group 8 2016-12-07 01:16 Sam Kennedy Programming 3 2012-12-16 08:38

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