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- Thread starter taitu
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- Thread starter
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- Jan 29, 2012

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So $x^2+ y= z^2$ for x, y, and z integers. That is the same as $x^2- z^2= (x- z)(x+ z)= y$. Look at the ways to factor y: y= mn and the x- z= m, x+ z= n. Adding those two equations, 2x= m+ n, x= (m+ n)/2. Subtracting, 2z= n- m, z= (n- m)/2.

added much later: I've noticed that I have a sign error: from $x^2+ y= z^2$, $y= z^2- x^2$, not $x^2- x^2$. So y= (z- x)(z+ x). Taking y= mn, z- x= m, z+ x= n so that 2z= n+m, z= (n+m)/2, 2x= n- m so x= (n-m)/2, just the opposite of what I had before.

added much later: I've noticed that I have a sign error: from $x^2+ y= z^2$, $y= z^2- x^2$, not $x^2- x^2$. So y= (z- x)(z+ x). Taking y= mn, z- x= m, z+ x= n so that 2z= n+m, z= (n+m)/2, 2x= n- m so x= (n-m)/2, just the opposite of what I had before.

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