[Math] Smallest positive integer a such that $5n^{13}+13n^5+a(9n)\equiv 0\pmod{65}$

Question

How would you find the smallest positive integer $a$ such that
$$5n^{13}+13n^5+a(9n)\equiv 0\pmod{65}$$
I simplified the polynomial in terms of $\bmod 5$ and $\bmod 13$ to get $3n+4na \equiv 0 \pmod{5}$ and $5n+9an \equiv 0 \pmod{13}$ but from here I'm stuck on how to proceed to solve for $a$.

Answer 1

Remember lil' Fermat can be written as $$\forall n,\:n^5\equiv n\mod5\qquad\text{ and }\qquad\forall n,\:n^{13}\equiv n\mod13.$$ So we have to solve the system of simultaneous congruences: $$\forall n,\begin{cases} 13n+9an\equiv 3n-an\equiv0\mod 5\\ 5n+9an\equiv 5n-4an\equiv0\mod 13 \end{cases}\iff \begin{cases} a\equiv \color{red}{3\mod 5},\\ 4a\equiv5\iff a\equiv -3\cdot5\equiv \color{blue}{-2\mod13}\end{cases}.$$ Now, a Bézout's relation between $5$ and $13$ is $\;2\cdot 13-5\cdot 5=1$, so solutions of the system of congruences is $$x\equiv\color{red}3\cdot2\cdot \color{blue}{13}-(\color{blue}{-2})5\cdot \color{red}{5} \equiv 63\mod65.$$ Thus the smallest positive integer $a$ satisfying the given congruence for all $n$ is $\color{green}{63}$.