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Old 2017-02-09, 04:08   #1
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Nov 2016

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Post Solutions to a^2-ab+b^2 = 3^n

The basic algebra rules for factoring sums and differences of powers are here. For a^n+-b^n when n is prime, (a^n+-b^n)/(a+-b) = 0 or 1 (mod n). If (a^n+-b^n)/(a+-b) = 0 (mod n), let n^k be the highest power of n dividing (a^n+-b^n)/(a+-b). Then (a^n+-b^n)/((a+-b)*n^k) = 1 (mod n). Would it ever be the case that (a, b) > 1, k > 1, (a^n+-b^n)/((a+-b)*n^k) = 1?

For n = 3, we are looking to find

Solutions to a^2+ab+b^2 = 3^n (I messed up in the question title btw, this is what I meant to ask.)

The only known solutions with integers are

1^2+1*1+1^2 = 3^1
2^2-2*1+1^2 = 3^1

It is probably the case that no more solutions with (a, b) > 0 exists other than these two. If you know of one, or have a proof that no solutions to a^2+ab+b^2 = 3^n exist except (1, 1) and (-2, 1), please post arguments here.

Last fiddled with by carpetpool on 2017-02-09 at 04:47
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