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 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
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Posted by xdog
on 2015-01-12 15:17:01 |
No solution, maybe just an interesting byway
