elementary number theory - How to prove that $53^{103}+ 103^{53}$ is divisible by 39?

This is a problem in my number theory textbook. It is based on modular arithmetic but im not getting how to start off to prove this. Please give me some hints on how to solve it.

Answer

As $39=13\cdot3$

For non-negative integers $m,n$

$\displaystyle53\equiv1\pmod{13}\implies53^n\equiv1$ and $\displaystyle103\equiv-1\pmod{13}\implies103^{53}\equiv(-1)^{53}$

$\displaystyle\implies53^{103}+103^{53}\equiv1+(-1)\pmod{13}$

and $\displaystyle53\equiv-1\pmod3\implies53^{103}\equiv(-1)^{103}$ and $\displaystyle103\equiv1\pmod3\implies103^m\equiv1$

$\displaystyle\implies53^{103}+103^{53}\equiv-1+(1)\pmod3$