Number Theory of Polynomials!

we have $x^{13} = x (mod 13) , x^5 = x mod 5$ because of FLT $5x^{13}+13x^5+9ax = 3x+4ax (mod 5) = (3+4a)x (mod 5)$ -> a = 3 mod 5 $5x^{13}+13x^5+9ax = 5x+9ax (mod 5) = (5+9a)x (mod 5)$ -> a = 11 mod 13

combining, a = 63 mod 65 , a = 63

It suffices to tackle the problem $\bmod 65$

When $x=1$ we want $5+13+9a\equiv 0 \bmod65 \iff 9a\equiv -18\bmod 65$ which is a linear modular equation, so it has unique solution $-2$ . Another way to obtain this is multiplying both sides by the inverse of $65$ which is $29$ to get $a\equiv-504\equiv -2\bmod 65$

So if there is a solution it is $-2$ . Lets see $5x^3+13x-18x\equiv 0 \bmod 65$ for every value of $x$ , so that $-2$ is indeed a solution.

It suffices to verify it $\bmod 13$ and $\bmod 5$ .

To see it $\bmod 5$ notice $5x^{13}+13x^{5}-18x\equiv 3x^{5}-3x\equiv 3x-3x\equiv 0 \bmod 5$ by fermat's little theorem.

To see it $\bmod 13$ notice $5x^{13}+13x^{5}-18x\equiv 5x^{13}-5x\equiv 5x-5\equiv 0 \bmod 13$ by fermat's little theorem.

This concludes the proof. Notice that Instead of using fermat we could check the $5$ possible values of $x\bmod 5$ and the $13$ possible values of $x\bmod 13$ . In any case we have the solution $\bmod 65$ is $-2$ , and the smallest positive integer in this congruence class is $63$ , the answer.

Satyajit Mohanty, Please check wether my solution is right or wrong.

If $f(x)$ is divisible by 65 then it should be divisible by $f(1)$ (given). Now the result would be $18+9a$ for $f(1)$ .

$65|18+9a \\ =65|9(2+a)$

Now 9, is not divisible by 65 therefore $2+a=65n$ for minimum we will take n=1 $2+a=65$ $\boxed{a=63}$