Show that $15\mid 21n^5 + 10n^3 + 14n$ for all integers $n$. I'm not sure if it's correct, but what I have so far is;
$$21n^5 + 10n^3 + 14n ≡ (1 + 0 - 1) ≡ 0 \mod 5$$ but I'm having trouble solving it in $\bmod 3$.
I have:
$$21n^5 + 10n^3 + 14n ≡ (0 + (?) + 2),$$
I'm not sure how to solve the $10n^3$ part of the congruence in $\mod 3$.
Also, if I have any mistakes please let me know.
 A: You have correctly interpreted this as a desired congruence identity:
$$
21n^5  + 10n^3+14n\equiv0\pmod{15}\,.
$$
And you seem to have recognized that it’s enough to verify it for the relatively prime factors $3$ and $5$, that is, it’s enough to verify the congruence modulo $3$ and modulo $5$. But, modulo $3$, it’s $n^3-n\equiv0\pmod3$, and modulo $5$, it’s $n^5-n\equiv0\pmod5$. Both of these are true for all $n$, as I hope you know.
A: Since $n^3 \equiv n \pmod{3}$, we have
$$
21n^5 + 10n^3 + 14n \equiv (10+14)n \equiv 24n \equiv 0 \pmod{3}.
$$
Similarly, since $n^5 \equiv n \pmod{5}$ we have
$$
21n^5 + 10n^3 + 14n \equiv (21+14)n \equiv 35n \equiv 0 \pmod{5}.
$$
A: Hint $\ {\rm mod}\ 5\!:\ f(n)\equiv n^5-n\equiv 0,\,$ and $\,{\rm mod}\ 3\!:\ f(n)\equiv n^3-n\equiv 0\ $ by Fermat.
Or $ $ note $\ f(n) = 21(n^5-n) + 10(n^3-n) + 45n\,$ and $15$ divides all summands.
A: Subtract the given expression from $30n^5+15n$, which gives
$$
S=9n^5-10n+n=(n-1)n(n+1)(9n^2-1).
$$
It suffices to show that $S$ is divisible by $15$. Because $9n^2-1=35+9(n-2)(n+2)$, we have
$$
S=\color{blue}{(n-1)n(n+1)35}+\color{red}{(n-2)(n-1)n(n+1)(n+2)}.
$$
The $\color{blue}{\text{blue}}$ part is divisible by $15$ because it is the product of three consecutive integers with $35$, and the $\color{red}{\text{red}}$ part is divisible by $15$ because it is the product of five consecutive integers.
