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.
flooble's webmaster puzzle