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Sums of sixth powers I (Posted on 2013-10-13) Difficulty: 3 of 5

In the context of integer right angled triangles, it is well-known that every prime factor of the hypotenuse must either be of the form 4k+1 or be a common factor of all three sides.

Analogously, start with a sum of sixth powers as a multiple of the sum of their squares: x^6 + y^6 = k(x^2 + y^2) .

If k is prime, the prime concerned is called a Sextan Prime, see A002647 in Sloane.

Naturally, the solution need not be prime; e.g {x,y} = {4,2} gives the solution k=208, which is 2^4*13.

Claim: every solution is of the form a^2-3b^4, and every prime factor of such solution is either of the form u^2-3v^2, or occurs to a power which is a multiple of 4.

Prove it, or find a counter-example.

No Solution Yet Submitted by broll    
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