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2*P + 2 = Perfect Square? (Posted on 2010-05-01) Difficulty: 4 of 5
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.

No Solution Yet Submitted by K Sengupta    
Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: a computerized start but no proof and too small a sample size | Comment 3 of 5 |
(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

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