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.

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