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.

That should read a^(x+ k*Eulerphi(norm(m)) +c = = 0 (mod m).

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.

[QUOTE=devarajkandadai;470095]That should read a^(x+ k*Eulerphi(norm(m)) +c = = 0 (mod m).[/QUOTE]Assuming that is meant as

a^(x+ k*Eulerphi(norm(m))[b])[/b] + 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.

[u]Exercise:[/u] Supply an additional hypothesis, under which your statement becomes correct.

[QUOTE=Nick;470141]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.[/QUOTE]

Yes-this can be tested if you have pari.

[QUOTE=Dr Sardonicus;470145]Assuming that is meant as

a^(x+ k*Eulerphi(norm(m))[b])[/b] + 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.

[u]Exercise:[/u] Supply an additional hypothesis, under which your statement becomes correct.[/QUOTE]
Eulerphi(4) is obviously not equal to 8. btw did you attend the AMS conference at Antwerp in May 1996? I was there.

[QUOTE=Nick;470141]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.[/QUOTE]

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.

[QUOTE=devarajkandadai;470436]Eulerphi(4) is obviously not equal to 8.[/QUOTE]

The formula you gave was

[quote]a^(x+ k*Eulerphi(norm(m))) + c == 0 (mod m)[/quote]

I used m = 4. Taking norm(m) from Q(i) to Q per [i][b]your[/b][/i] formula, we have

norm(4) = 16.

And, Eulerphi(16) = 8.

[QUOTE=Dr Sardonicus;470145]Assuming that is meant as

a^(x+ k*Eulerphi(norm(m))[b])[/b] + 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.

[u]Exercise:[/u] Supply an additional hypothesis, under which your statement becomes correct.[/QUOTE]
My conjecture states that c has got to be a Gaussian Integer i.e. b cannot be 0.

[QUOTE=devarajkandadai;470489]My conjecture states that c has got to be a Gaussian Integer i.e. b cannot be 0.[/QUOTE]

Let's see:

[QUOTE=devarajkandadai;470090]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.[/QUOTE]

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 [i]not[/i] 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, [i]post hoc[/i] 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.

[QUOTE=devarajkandadai;470438]btw did you attend the AMS-BENELUX conference at Antwerp in May 1996? I was there.[/QUOTE]
No, I wasn't there, but I hope you enjoyed your visit to this part of the world!