caluclate :

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- Mar 9th 2010, 11:07 PM #1

- Mar 10th 2010, 07:11 AM #2

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- Mar 10th 2010, 11:38 AM #3

- Mar 10th 2010, 01:16 PM #4

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- Mar 10th 2010, 01:20 PM #5

- Mar 10th 2010, 02:35 PM #6

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Ok, I see now.

So your conjecture is that gcd(5^m + 7^m,5^n + 7^n) is sometimes 2 and sometimes 5^gcd(m,n) + 7^gcd(m,n) but never any other value, and you don't yet have a choice function to tell which one it is. That's an improvement on what I had to say. I said essentially that it's 2 or some multiple of 2.

For what it's worth, I can confirm your conjecture for all m,n less than 1000.

- Mar 21st 2010, 11:30 AM #7

- Mar 21st 2010, 09:38 PM #8

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I was waiting for the original poster to say something.

It depends on the factorization

$\displaystyle a^n+b^n=(a^r+b^r)(a^{n-r}-a^{n-2r}b^r+a^{n-3r}b^{2r}-...+b^{n-r})$

which works if n/r is odd. So $\displaystyle a^n+b^n\text{ and }a^m+b^m$ have a common factor $\displaystyle a^r+b^r$ if r is a common factor of n and m, and n/r and m/r are both odd.

So in the prime factorizations of n, m, and r, the power of 2 is the same. In fact, we should have:

$\displaystyle gcd(a^n+b^n,a^m+b^m)=a^{gcd(n,m)}+b^{gcd(n,m)}$ if the prime factorizations of n and m contain the same power of 2.

I'm still foggy on the converse, though. Couldn't there be a common factor that's not of the form $\displaystyle a^r+b^r$?

- Hollywood