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 Find additional pairs (Posted on 2016-08-08)
The pair of positive integers (4,6) satisfies the equation

(2x+1)*x=y^2

Find at least three additional pairs, both x and y below 10^6.

 See The Solution Submitted by Ady TZIDON No Rating

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 Pell Solution | Comment 4 of 5 |
Rearrange the equation to form (4x+1)^2 = 8y^2 + 1.  Let z=4x+1.  Then z^2 - 8y^2 = 1 is a Pell equation.  Its first nontrivial solution is (z,y)=(3,1).

Then the recursions z(n+1) = 3*z(n)+8*y(n) and y(n+1) = 3*y(n)+z(n) generate further solutions (z,y) = (3,1), (17,6), (99,35), (577,204), (3363,1189), (19601,6930), (114243,40391), (665857,235416), (3880899,1372105), (22619537,7997214), etc.

Only the even numbered pairs in the sequence have z = 1 mod 4.  Then the first five solutions to the original equation are (x,y) = (4,6), (144,204), (4900,6930), (166464,235416), (5654884,7997214)

 Posted by Brian Smith on 2016-08-09 00:11:46
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