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?
Start with n^2+(n+1)^2 = (m(m+1))^2+1
Turning all this into triangular numbers:
T(n)+2T(n+1)+T(n+2) = (2T(m))^2+1
T(n)+2T(n+1)+T(n+2) = (2T(m))^2+1
But then (T(m))^2 = T(n+1)
T(n)+2T(n+1)+T(n+2) = 4T(n+1)+1
T(n)+T(n+2) = 2T(n+1)+1
T(n)+T(n+2) = 2(T(m))^2+1
Following up on xdog's idea, the integer solutions {1,6,35,204,1189,..} apart from 1 and 6 will be found to differ from their corresponding T(n) {36,210,1225,7140,41616,...} by the values in the same series {1,6,36, 210,1225,...}, namely A096979 in Sloane, 'Sum of the areas of the first n+1 Pell triangles', evaluating to 1/32 (4+2 (1)^n+(32 sqrt(2))^n+(3+2 sqrt(2))^n), a series that starts with 2 zeroes for n=0,1, thereafter increasing monotonically.
So it seems that 1 and 36 are indeed the only triangular numbers that are squares of other triangular numbers.
Edited on January 12, 2015, 11:01 pm

Posted by broll
on 20150112 22:37:21 