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 Post subject: DivisibilityPosted: Fri Jan 15, 2016 10:37 pm

Joined: Mon Nov 09, 2015 4:32 pm
Posts: 7
Prove that there are no positive integers $a,b,n > 1$ such that: $(a^{n}-b^{n})\mid (a^{n}+b^{n})$

P.S. I don't have a solution.

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 Post subject: Re: DivisibilityPosted: Fri Jan 15, 2016 10:39 pm

Joined: Mon Nov 09, 2015 11:52 am
Posts: 76
Location: Limassol/Pyla Cyprus
Suppose that $(a^n-b^n)|(a^n+b^n)$. We may assume that $a > b$. Then there is an integer $k > 1$ such that $a^n+b^n = k(a^n-b^n)$ which gives $(k+1)b^n = (k-1)a^n$.

Suppose $p>2$ is a prime such that $p|(k+1)$. Since $(k+1,k-1) = (k+1,2) \leqslant 2$ then $p \nmid (k-1)$. Looking at the powers of $p$ in the unique prime factorisations of $(k+1)b^n$ and $(k-1)a^n$ we deduce that the highest power of $p$ which divides $k+1$ is a multiple of $n$.

If $2|(k+1)$ then $2|(k-1)$ as well. Then one of $k-1$ and $k+1$ is a multiple of $4$ and the other a multiple of $2$ but not $4$. Thus the highest power of $2$ which divides $k+1$ is congruent to $1 \bmod n$.

Similar results apply for the divisors of $k-1$. Therefore there are positive integers $c$ and $d$ with $c > d$ such that either $k+1 = 2c^n$ and $k-1 = 2d^n$ or $k+1 = c^n$ and $k-1 = d^n$ . So either $c^n - d^n = 1$ or $c^n - d^n = 2$. This is impossible as $c^n - d^n \geqslant (d+1)^n - d^n \geqslant nd^{n-1} + 1 \geqslant 3.$

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