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 counterexample.
You want b = yz^2 where y is not a square, but if y were a square, say y = a^2 then you'd have b= a^2*z^2 = (az)^2 which is a square not the multiple of a square as sought.
This proves that y cannot be a square, so there will be no counterexamples. Now to find if any solutions actually exist.
I found the first 3 values of b (5, 29, 169) and put them in OEIS which returned http://oeis.org/A001653
The sequence contains 195025 which is 7801*5^2

Posted by Jer
on 20120925 16:17:25 