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.

Ah for the heady days when I still had not heard of the Pell equation!

Using trusty Dario Alpern supplies the recurrence relation

Xn+1 = P Xn + Q Yn
Yn+1 = R Xn + S Yn

P = 127
Q = 672
R = 24
S = 127

The first few terms for P are {1,127,32257,...} etc. WolframAlpha resolves this to P = ±1/2 (-(127-48 sqrt(7))^n-(127+48 sqrt(7))^n), and we need to take the negative value, multiply by 2 and add 2:

(127-48 sqrt(7))^n+(127+48 sqrt(7))^n+2 (I)

Next we look at the values of the square roots of this: {2,16,254,4048,64514 ...}, A090727 in Sloane with formula (8+sqrt(63))^n + (8-sqrt(63))^n.

Squaring this gives: (8-3 sqrt(7))^(2n)+(8+3 sqrt(7))^(2n)+2((8-3 sqrt(7))(8+3 sqrt(7)))^n.

More conveniently:

(8-3 sqrt(7))^(2n)+(8+3 sqrt(7))^(2n)+2 (II)

Now, (8-3 sqrt(7))^(2) = 127-48 sqrt(7); while (8+3 sqrt(7))^(2) =127+48 sqrt(7)   (III)

So at last we have the equality:

(127-48 sqrt(7))^n+(127+48 sqrt(7))^n+2  = (127-48 sqrt(7))^n + (127+48 sqrt(7))^n+2. (I),(III)

This number is always square since it is the square of (8+sqrt(63))^n + (8-sqrt(63))^n, see (II).

And the mystery is solved at last.

Edited on September 30, 2013, 2:14 am

Posted by broll on 2013-09-30 02:09:06