Let a,b,c,d, be positive integers, with a and d different, and b and c different.
If a^4-b^5=c^5-d^4, then twice the sum of the 5th powers is a sum of two squares.
Prove it, or find a counter-example.
My generalization is similar. I don't have a proof it's complete either.
a=x^5, b=y^4, c=x^4, d=y^5
so twice the sum of the 5th powers is x^20 + y^20 which is the sum of the two squares (x^10 + y^10) and (x^10 - y^10).
I noodled around with the equivalent sum of 4th powers and considered that number lay between consecutive squares but nothing came of it.
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Posted by xdog
on 2013-05-01 19:24:27 |