Dividing...

ckck

Given that the $gcd(a,b)=1$ , find all ordered pairs $(m,n)$ such that $a^{m}+b^{m}$ divides into $a^{n}+b^{n}$ .

What I have so far

**Posted:** Mon Oct 26, 2009 4:57 pm

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SCII!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
김 동 현
BAM!

bbgun34

You need to show that $n$ or $m$ is odd, and the factorization does not work if there is an even exponent. Also, the factorization would only work for $a^n - b^n$ , for evens. (it's fine if the power is odd).

One obvious solution is $(m,n) = \boxed{(k,k)}$ , for some $k$ .

Are there any restrictions about the actual numbers $a,b,m,n$ themselves?

**Posted:** Mon Oct 26, 2009 6:56 pm

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Bomb

4+6+4x6=34; $2^5=(2^2)!+2^3$
$\prod \sum \square \ge \square \sum \prod$

ckck

The only other restrictions that I forgot to mention were that $a$ , and $b$ are positive integers.

**Posted:** Mon Oct 26, 2009 8:19 pm

_________________
SCII!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
김 동 현
BAM!

Potla

I have proceeded with this problem as follows:
Since $a^m+b^m|a^n+b^n$ hence $m\leq n\implies m+k=n$
Plugging this value of $n$ into the given relation we have $a^m+b^m | a^m \cdot a^k +b^m\cdot b^k$ .
So now we let $a^m \cdot a^k +b^m\cdot b^k=la^m+lb^m\implies a^m(a^k-l)+b^m(b^k-l)=0$
$\implies a^m(a^k-l)=b^m(l-b^k)$
Now since $\gcd (a,b)=1$ ; hence this equation is only possible when $a^k-l=l-b^k=0$
$\implies a^k=b^k\implies \boxed{k=0}$ .
Hence $m=n$ is forced; $\implies$ all possible pairs of $m,n$ are
$\boxed{(m,n)=(t,t)}$ where $t$ is any positive integer.