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.
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)
Pell Solution
