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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)

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Some Thoughts Some ideas and some guesswork | Comment 2 of 5 |

WolframAlpha gives integer p = +-1/2 (-(127-48 sqrt(7))^n-(127+48 sqrt(7))^n):

Hence 2p+2=((127-48 sqrt(7))^n+(127+48 sqrt(7))^n)+2 needs to be square, for the conjecture to hold.

The first term is a+b+a-b+2 which is 256, or 16^2. The second is a^2+2304*b^2+a^2+2304*b*2+2 = 64516, or 254^2 etc.

With a little tinkering it will be apparent that the square roots of these values correspond to integer values of some x such that x^n=(8+3*7^0.5)^n+(8-3*7^0.5)^n:

16,254,4048,64514,1028176,16386306,261152656, 4162056194, 66331746448, 1057145886974 etc.<o:p></o:p>

Thus we are asked to show that the square of the integer (x^n) is a perfect square; which is true.

  Posted by broll on 2010-05-02 12:13:31
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