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Old 2017-10-19, 06:15   #1
devarajkandadai
 
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Default 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
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Old 2017-10-19, 09:07   #2
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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
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Old 2017-10-20, 10:51   #3
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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.
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Old 2017-10-20, 14:48   #4
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Quote:
Originally Posted by devarajkandadai View Post
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.
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Old 2017-10-27, 04:24   #5
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Quote:
Originally Posted by Nick View Post
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.
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Old 2017-10-27, 05:08   #6
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Quote:
Originally Posted by Dr Sardonicus View Post
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.
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Old 2017-10-27, 06:01   #7
devarajkandadai
 
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Quote:
Originally Posted by Nick View Post
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.
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Old 2017-10-27, 13:07   #8
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Quote:
Originally Posted by devarajkandadai View Post
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.
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Old 2017-10-28, 05:25   #9
devarajkandadai
 
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Quote:
Originally Posted by Dr Sardonicus View Post
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.
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Old 2017-10-28, 14:23   #10
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Quote:
Originally Posted by devarajkandadai View Post
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 View Post
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.
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Old 2017-10-28, 15:39   #11
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Quote:
Originally Posted by devarajkandadai View Post
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!
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