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 Post subject: Divisibility
PostPosted: 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: Divisibility
PostPosted: 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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