Gaussian integers- use of norms

[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.

[devarajkandadai]

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

[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.

[Dr Sardonicus]

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.

[devarajkandadai]

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.

[devarajkandadai]

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.

[devarajkandadai]

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.

[Dr Sardonicus]

Originally Posted by devarajkandadai

Eulerphi(4) is obviously not equal to 8.

The formula you gave was

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.

[devarajkandadai]

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.

[Dr Sardonicus]

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:

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.

[Nick]

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!