Finally, some abstract algebra pays off!
[sp]First of all, note that $65|(5x^{13}+13x^5 + 9ax) \iff 5|(5x^{13}+13x^5 + 9ax)$ and $13|(5x^{13}+13x^5 + 9ax)$.
This suggests looking at the given polynomial mod 5 and mod 13.
Mod 5 we have:
$3x^5 + 4ax = 0$ (since we want 5 to divide the expression on the left).
$x(3x^4 + 4a) = 0$
Since $x$ can be some integer co-prime to 5, we will assume $x \neq 0$ (mod 5). Thus:
$3x^4 + 4a = 3 + 4a = 0$, that is:
$4a = 2$ (mod 5) so that:
$a = 3$ (mod 5).
Mod 13 we have:
$5x^{13} + 9ax = 0$ (since we want the expression on the left to be divisible by 13), thus:
$x(5x^{12} + 9a) = 0$. Again, we may assume $x \neq 0$ (mod 13), so we have:
$5x ^{12} + 9a = 5 + 9a = 0$ thus:
$9a = 8$ (mod 13), so that:
$a = 11$ (mod 13).
Since 5 and 13 are prime, we can apply the Chinese Remainder theorem to obtain:
$a = (3)(13)[13^{-1}]_5 + 11(5)[5^{-1}]_{13} = (3)(13)(2) + (11)(5)(8) = 78 + 440 = 518 = 65*7 + 63 = 63$ (mod 65).
Depending on if you meant:
the smallest POSITIVE value for $a$, we obtain: $a = 63$
the smallest absolute value for $a$, we obtain $a = -2$
there is no "smallest" such $a$ given the usual ordering of the real numbers, since the set of solutions contains infinitely many negative integers.[/sp]
