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Many triplets (Posted on 2010-05-10) Difficulty: 2 of 5
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    
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Solution Another set | Comment 4 of 6 |

Not sure if this is included in any of the previous:

For all non-negative integers i,
Let J=i^2 + 1

X=J^2
Y=sqrt(J^5 - J^4) = i * J^2
Z=J

X^2 + Y^2 = J^4 + J^5 - J^4 = J^5 = Z^5

The first few are:
4, 4, 2
25, 50, 5
100, 300, 10

Edited on May 11, 2010, 11:23 pm

Edited on May 11, 2010, 11:34 pm
  Posted by Larry on 2010-05-11 23:23:09

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