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A Pythagorean piglet (Posted on 2012-09-22) Difficulty: 3 of 5

Finding solutions to x2- b2=b2-1 is a bit too easy these days, so let b itself be the multiple of a square, say, yz2

For both y>1 and z>1, prove that y is not a square, or find a counter-example.

See The Solution Submitted by broll    
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re(2): Oinking solution. | Comment 3 of 4 |
(In reply to re: Oinking solution. by broll)

Ok you you are considering (az)^2 to be the multiple of a square where y=a^2, even though it is also a square itself.  Of course it is and that was my mistake.

  Posted by Jer on 2012-09-26 12:41:34

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