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.
 A: $$53\equiv14\pmod{39},103\equiv-14\pmod{53}$$
$$53^{103}+103^{53}\equiv14^{103}+(-14)^{53}\pmod{39}\equiv14^{53}[(14)^{50}-1]$$
Now $\displaystyle14^1\equiv1\pmod{13},14^1\equiv-1\pmod3\implies14^2\equiv(-1)^2\equiv1$
$\displaystyle\implies14^{\text{lcm}(1,2)}\equiv1\pmod{3\cdot13}$
or directly by observation, $\displaystyle14^2=196\equiv1\pmod{195}\equiv1\pmod{39}$
$\implies14^{2n}\equiv1$ for non-negative integer $n$

Generalization :
We need $\displaystyle a\equiv1\pmod3,\equiv-1\pmod {13}$ and $\displaystyle b\equiv-1\pmod3,\equiv1\pmod {13}$ where $a,b$ are odd integers
So, $\displaystyle a=3A+1=13B-1$ for some integers $A,B$
$\iff\displaystyle3A=13B-2=13B-(2\cdot13-3\cdot8)\iff3(A-8)=13(B-2)$
$\displaystyle\implies\frac{3(A-8)}{13}=B-2$ which is an integer
$\displaystyle\implies13|3(A-8)\iff13|(A-8)$ as $(13,3)=1$
$\displaystyle\implies A=8+13C\implies a=3(8+13C)+1=39(C+1)-14\equiv-14\pmod{39}$
Similarly, we can derive $\displaystyle b\equiv14\pmod{39}$
But one condition, $a,b$ must be odd integers
We can further generalize this any two relative prime  odd integers
A: ${\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.
A: 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 $\displaystyle103\equiv1\pmod3\implies103^m\equiv1$
$\displaystyle\implies53^{103}+103^{53}\equiv-1+(1)\pmod3$
