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 T times T creates another T?? (Posted on 2015-01-11)
The squares of triangular numbers 1 and 6 are triangular numbers 1 and 36.

T1^2 = 1 * 1 = 1 = T1
T3^2 = 6 * 6 = 36 = T8

Are there additional triangular numbers whose squares form a triangular number?

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 No solution, maybe just an interesting byway | Comment 6 of 8 |
a^2(a + 1)^2 = 2b(b+1)

Considered as a quadratic in 'b', this equation will have discriminant equal to a square.

After eliminating a common factor of 4 we have 1 + 2a^2(a + 1)^2 = k^2 which is a form of the Pell equation x^2 - 2y^2 = 1.

This has an infinite number of non-negative solutions beginning with (1,0), (3,2), (17,12), (99,70), (577,408).  Online you can find parametric solutions and methods for solving by evaluating convergents of sqrt(2), or you can use a neat trick to find the next solution:

x_next = 3*x_last + 4*y_last
y_next = 2*x_last + 3*y_last

Edited on January 12, 2015, 3:18 pm
 Posted by xdog on 2015-01-12 15:17:01

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