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?
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 nonnegative 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 20150112 15:17:01 