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Old 2012-10-27, 18:18   #2
Batalov's Avatar
Mar 2008

22·3·7·109 Posts

For even values of n, this doesn't divide. For them, there's no denominator, but the logic below still applies.

For odd values of n: Suppose you have a composite n = k*m, with k>=3. Then,
(an+bn)/(a+b) = (akm+bkm)/(a+b) =
= (ak+bk)/(a+b)*(a(k-1)m - bm*a(k-2)m + b2m*a(k-3)m + ... + b(k-1)m)
where (ak+bk)/(a+b) is > 1. QED, composite.

For proper Mersenne numbers a+b = 2+(-1) = 1, which makes for an interesting case - there's no denominator.

When b=1, an+bn is a proper Fermat number, and for b>1 and odd (prime) n, (an+bn)/(a+b) has the property that you sought for.

Last fiddled with by Batalov on 2012-10-27 at 18:24
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