# 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 $n$.

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).

• $$n(n+1)+(an+b)(an+b+1)=4(cn+d)^2$$ $n,b,d -$ we set them as variables and unknown quantities and solve this Diophantine equation. For any number $a,c -$ there are always solutions. You can use the formula. math.stackexchange.com/questions/794510/… – individ Oct 11 at 4:41

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 $$x_{j+2} = 6 x_{j+1} - x_j$$
$$\begin{array}{cccc|c|c} a & c & \beta & d & 4cd-1& a \beta \\ \hline 1 & 1 & 3 & 1 & 3 & 3 \\ 7 & 5 & 17 & 6 & 119 & 119 \\ 41 & 29 & 99 & 35 & 4059 & 4059 \\ 239 & 169 & 577 & 204 & 137903 & 137903 \\ \end{array}$$
• @martycohen your equation amounted to $a^2 - 2 c^2 = -1,$ then $(2b+1)^2 - 8d^2 = 1,$ finally $a(2b+1)+ 1 = 4cd.$ The recurrence I show above is Cayley-Hamilton applied to both relevant matrices, [3,4//2,3] and [3,8//1,3] – Will Jagy Oct 8 at 19:33
• @martycohen sorry for delay. Probably messy to see in terms of separate linear recurrences. However, easy enough for the one-step mapping $$(a,c,\beta,d)\mapsto (3a+4c,2a+3c,3 \beta +8d, \beta + 3d).$$ And it is interesting. – Will Jagy Oct 10 at 2:13