TITLE: How to prove that $53^{103}+ 103^{53}$ is divisible by 39?
QUESTION [9 upvotes]: 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.

REPLY [2 votes]: ${\rm mod}\ \ \ 3\!:\ n = 53^{\large j}\! + 103^{\large k}\equiv (\color{#0a0}{-1})^{\large j}\ \ +\,\ \ \color{#c00}1^{\large k} \equiv\, \color{#0a0}{-1}\color{#c00}{+1}\equiv 0,\ $ assuming $\,j\,$ is odd.
${\rm mod}\ 13\!:\ n = 53^{\large j}\! + 103^{\large k}\equiv \ \ \ \  \color{#c00}1^{\large j}\, +\,(\color{#0a0}{-1})^{\large k} \equiv\, \color{#c00}{{+}1}\color{#0a0}{-1}\equiv 0,\ $ assuming $\,k\,$ is odd.
So $\ 3,13\mid n\,\Rightarrow\, 39\mid n\  $ by $\ {\rm lcm}(3,13) = 3\cdot 13\ $ (or by CCRT) $\ $ QED
Remark $\ $ The above innate $\rm\color{#c00}{sym}\color{#0a0}{metry}$ above can be brought to the fore as in this answer, i.e.
$$\{a,b\} \equiv \{\color{#c00}{+c},\,\color{#0a0}{-c}\,\}\,\ {\rm  mod}\,\ m\,\&\,n\ \Rightarrow\ {\rm lcm}(m,n)\mid a\!+\!b\qquad\qquad$$
The congruence arithmetic above employs the basic Congruence Sum and Power Rules.