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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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re: Another set | Comment 5 of 6 |
(In reply to Another set by Larry)

Your infinite set is just a small portion of one of mine.

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


I consider these examples of one of mine:

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

a=1, b=1, c=2, n=0
a=1, b=2, c=5, n=0
a=1, b=3, c=10, n=0

In your equation J=i^2+1, J=c, i=b, 1=a

My n is your implied 0 (2n+1=1)

  Posted by Jer on 2010-05-12 16:53:19

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