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 Fifthing the quart (Posted on 2013-04-28)

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.

 See The Solution Submitted by broll No Rating

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A number n is a sum of two squares if and only if all prime factors of n of the form 4m+3 have even exponent in the prime factorization of n.
This implies if the sum of the 5th powers is the sum of two squares then twice the sum is too.

Searching each of a,b,c,d up to 50 I only found one solution to the equation: 32^4 - 1^5 = 16^5 - 1^4
which I immediately generalized to
a=x^5, b=1, c=x^4, d=1
so the sum of the 5th powers is x^20 + 1 which is the sum of two squares (x^10 and 1.)

There may be other solutions when b and d aren't both 1.

 Posted by Jer on 2013-05-01 15:00:18

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