Finding solutions to x^{2}- b^{2}=b^{2}-1 is a bit too easy these days, so let b itself be the multiple of a square, say, yz^{2}

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.

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