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 Proof positive: squares of integers (Posted on 2007-04-07)
If p and q are positive integers that satisfy 3p²+p=4q²+q, prove that p-q, 3p+3q+1 and 4p+4q+1 are squares of integers.

 See The Solution Submitted by K Sengupta No Rating

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 Two of three proved | Comment 2 of 3 |

Take the given identity 3p²+p=4q²+q and multiply both sides by 16 and add 1:
48p^2+16p+1 = 64q^2+16q+1

Expressing each side in terms of polynomial squares gives:
(8p+1)^2 - 16p^2 = (8q+1)^2

Applying a common parameterization for the Pythagorean theorem gives:
8p+1 = x^2+y^2
2xy = 8p^2
8q+1 = x^2-y^2

From there take the sum and difference of 8p+1 and 8q-1:
(8p+1) + (8q+1) = (x^2+y^2) + (x^2-y^2)
2(4p+4q+1) = 2x^2
4p+4q+1 = x^2

(8p+1) - (8q+1) = (x^2+y^2) - (x^2-y^2)
8(p-q) = 2y^2
4(p-q) = y^2

This is sufficent to prove that if 3p²+p=4q²+q then p-q and 4p+4q+1 are both squares

 Posted by Brian Smith on 2007-04-09 11:34:14

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