Let $t_n$ denote the $n$th triangular number. For what values of $n$ does $t_n$ divide $t_1^2+t_2^2+ \cdots +t_n^2$.  The hint says that because $t_1^2+t_2^2+ \cdots +t_n^2 = t_n(3n^3 + 12n^2 + 13n + 2)/30$, it suffices to determine those $n$ satisfying $3n^3+12n^2+13n+2 \equiv 0$ (mod 2*3*5). However, I don't see how to get $t_1^2+t_2^2+ \cdots +t_n^2 = t_n(3n^3 + 12n^2 + 13n + 2)/30$ this equiation in the first place. Also, I'm trying to solve the congruence $3n^3+12n^2+13n+2 \equiv 0$ for each mod 2, 3, 5 respectively but I have been unable so far to find the $n$ that satisfies these congruences. I would greatly appreciate any help.
How to solve the following congruence equation? $$3n^3+12n^2+13n+2\equiv0,\pmod{2\times3\times5}$$ If $t_n$ be the $n$th triangular number, then  $$t_1^2+t_2^2+...+t_n^2=\frac{t_n(3n^3+12n^2+13n+2)}{30}$$ so if we solve this congruence equation we find the values of $n$ that $t_n$ divides $t_1^2+t_2^2+...+t_n^2$.
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