Each of P and Q is a nonnegative integer satisfying the equation: P² = 28*Q² + 1

Is 2*P + 2 always a perfect square?

If so, prove it. Otherwise, provide a counterexample.

(In reply to a computerized start but no proof and too small a sample size by Charlie)

I noticed that Q and P that you found both have recursive structure.

Q(n) = 254Q(n-1) - Q(n-2)

P(n) = 254P(n-1) - P(n-2)

The next solution of this form is
Q=39327734, P=2091028097, 2P+2=4162056196=64514^2

This could lead to an inductive proof that
P(n)^2 = 28Q(n)^2 + 1
and 2P(n) +2 is a perfect square
but I couldn't get it to come out.

This would only prove that numbers of this form work.
Other solutions beyond these could still satisfy P^2 = 28Q^2 +1

Posted by Jer on 2010-05-03 02:03:04