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.
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
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Posted by Jer
on 2012-09-25 16:17:25 |