If $T_n$ is the $n$-th triangular number, there are an infinite number of $a, b, c, d$ such that $T_n+T_{an+b} =(cn+d)^2 $ for all $n$.

Question

If $T_n$ is the $n$-th triangular number, show that there are an infinite number of positive integers $a, b, c, d$ such that $T_n+T_{an+b} =(cn+d)^2 $ for all $n$.

This is inspired by an article in the current Mathematics Magazine (October 2019).

My calculations show that the first two solutions are (a, b, c, d) =(1, 1, 1, 1) and (7, 8, 5, 6).

Answer 1

cleaner to use $\beta = 2 b + 1.$ Order as $(a,c,\beta,d) $ so that each separate variable obeys the same recurrence, with a dummy variable $$ x_{j+2} = 6 x_{j+1} - x_j $$

$$ \begin{array}{cccc|c|c} a & c & \beta & d & 4cd-1& a \beta \\ \hline 1 & 1 & 3 & 1 & 3 & 3 \\ 7 & 5 & 17 & 6 & 119 & 119 \\ 41 & 29 & 99 & 35 & 4059 & 4059 \\ 239 & 169 & 577 & 204 & 137903 & 137903 \\ \end{array} $$