Toying around with triangular numbers, it seems that there are
already 4 squares below 1000. At least I get this one:
?- between(1,1000,N), T is N*(N+1)//2, sqrtrem(T,_,0).
N = 1,
T = 1 ;
N = 8,
T = 36 ;
N = 49,
T = 1225 ;
N = 288,
T = 41616 ;
No
Are there infinitely many square triangular numbers?
To reproduce work done many times before.
If $T_n = m^2$, since $T_n = \dfrac{n(n+1)}{2}$, this becomes $n(n+1) = 2m^2$.
Completing the square, $n^2+n+\frac14 =2m^2+\frac14 $ or $(n+\frac12)^2 =2m^2+\frac14 $.
Clearing fractions, this is $(2n+1)^2 =8m^2+1 $ or $(2n+1)^2-8m^2 =1 $.
This is a case of a Pell equation $x^2-dy^2 = 1$, and the identity
$\begin{array}\\ (x^2-dy^2)(u^2-dv^2) &=x^2u^2-d(x^2v^2+y^2u^2)+d^2y^2v^2\\ &=x^2u^2\pm 2dxuyv+d^2y^2v^2-d(x^2v^2\pm 2xuyv+y^2u^2)\\ &=(xu\pm dvy)^2-d(xv\pm yu)^2\\ \end{array} $
shows that if there is one solution to $x^2-dy^2 = 1$ then there are an infinite number. If $(x_0, y_0)$ satisfies $x_0^2-dy_0^2 = 1$, the recurrence is $x_{n+1} = x_0x_n+dy_0y_n, y_{n+1}=x_0y_n+y_0x_n $.
If $d=8$, since $3^2-8\cdot 1^2 = 1$, the recurrence is $x_{n+1} = 3x_n+8y_n, y_{n+1}=3y_n+x_n $.
Starting with $(x_0, y_0) =(3, 1)$, and remembering that the $n$ in $T_n$ satisfies $2n+1 = x$, this gives $ (x_1, y_1) =(3\cdot 3+8\cdot 1, 3\cdot 1+3) =(17, 6) \quad (\dfrac{8\cdot 9}{2} = 6^2)\\ (x_2, y_2) =(3\cdot 17+8\cdot 6, 3\cdot 6+17) =(99, 25) \quad (\dfrac{49\cdot 50}{2} = 25^2)\\ $