Congruence Equation $3n^3+12n^2+13n+2\equiv0,\pmod{2\times3\times5}$ 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 $t_1^2+t_2^2+...+t_n^2$.

Work first modulo $2$. We can rewrite the congruence as $n^3+n\equiv 0\pmod{2}$. Note that this always holds.
Now work modulo $3$. The congruence can be rewritten as $n+2\equiv 0\pmod{3}$. This holds precisely if $n\equiv 1\pmod{3}$.
Finally, work modulo $5$. The congruence can be rewritten as $3n^3-3n^2+3n-3\equiv 0\pmod{5}$, and then as $(n-1)(n^2+1)\equiv 0\pmod{5}$. This has the solutions $n\equiv 1\pmod{5}$, $n\equiv 2\pmod{5}$ and $n\equiv 3\pmod{5}$.
So the conditions come down to $n\equiv 1\pmod{3}$ and $n\equiv 1$, $2$, or $3$ modulo $5$.
The solutions are therefore $n\equiv 1$, $n\equiv 7$ and $n\equiv 13\pmod{15}$. 
