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.
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.
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Posted by Jer
on 2013-05-01 15:00:18 |
Some (all?)
