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 Many triplets (Posted on 2010-05-10)
Prove that the equation x^2+y^2=z^5 has an infinite number of positive integer solutions.

 No Solution Yet Submitted by Ady TZIDON No Rating

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 Some more infinities of solutions. | Comment 3 of 6 |
The original wording did not exclude zero, so this would include
(n^5)^2 + 0^2 = (n^2)^5

Others have posted variations where x=y
(2^(2n+2))^2 + (2^(2n+2))^2 = (2^(2n+1))^5

For any a,b,c that satisfies a^2 + b^2 = c
(ac^(5n+2))^2 + (bc^(5n+2))^2 = (c^(2n+1))^5

For any a,b,c that satisfy a^2 + b^2 = c^2
(ac^(5n+4))^2 + (bc^(5n+4))^2 = (c^(2n+2))^5

If a^2 + b^2 = c^3 then (ac)^2 + (bc)^2 = c^5
If a^2 + b^2 = c^8 then (ac)^2 + (bc)^2 = (c^2)^5
If a^2 + b^2 = c^(5n-2) then (ac)^2 + (bc)^2 = (c^n)^5

A way to extend any given solution of x^2 + y^2 = z^5 to infinitely more solutions:
(n^5 x)^2 + (n^5 y)^2 = (n^2 z)^5

I have not found a way of generating ALL possible solutions.

For example 1475^2 + 4282^2 = 29^5 does not fit any of the patterns found.

 Posted by Jer on 2010-05-10 15:36:11
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