The LHS is a product of two squares, one even the other odd.
The RHS can be split into odd and even parts in two different
ways: 2b*(b + 1)orb*2(b + 1), so that four pairs of
equations are possible:

(1)a^{2} = 2b&(a + 1)^{2} = b + 1 givinga = 0 or -4,

(2)a^{2} = b&(a + 1)^{2} = 2(b + 1)givinga = 1,

(3)a^{2} = b + 1&(a + 1)^{2} = 2bgivinga = 3 or -1,

(4)a^{2} = 2(b + 1)&(a + 1)^{2} = bgivinga = -2.

a and b must be positive integers, so the only two
possibilities are those given in the question:

a = 1, b = 1:T_{1}^{2} = T_{1}anda = 3, b = 8:T_{3}^{2} = T_{8}