Is it $a^{48m+1}+b^{48n+1} \equiv 0 \pmod{39} \Leftrightarrow a+b \equiv 0 \pmod{39}$. For all $a,b \in \mathbb{Z}$ and for all $m,n \in \mathbb{N}\setminus \left\lbrace0\right\rbrace$,
is $a^{48m+1}+b^{48n+1} \equiv 0 \pmod{39} \iff a+b \equiv 0 \pmod{39}$?
I think the answer is yes, but I can't prove it. Is there somebody who can help me? Thank you.
 A: Hint: Use Euler theorem $$a^{\varphi(n)}\equiv 1\pmod n$$
if $\gcd(a,n )= 1$. Notice that $$\varphi (39) = 2\cdot 12 =24$$

Edit: 12. 29. 2019


*

*a) If $(a,39) =1$ then clearly also $(b,39)=1$ so we have $a^{24} \equiv 1$ and $b^{24} \equiv 1$ so  $a^{48m}a \equiv a$ and $b^{48n}b \equiv b$ and thus: $$a^{48m+1}+b^{48n+1} \equiv a+b$$ and the equivalence follows. 

*b) If $(a,39) \ne 1$ then $3\mid a$ or $13 \mid a$ or $39\mid a$.  


*

*If $39\mid a$ then also $39\mid b$ and claim follows

*If $3\mid a$ and $13\nmid a$ then if $3\mid a^{48m+1}+b^{48n+1} $ we have also $3\mid b$ so $3\mid a+b$ and vice versa. And for $13$ you can repeat a process in case a). 

*If $13\mid a$ and $3\nmid a$ then repeat a process from previous case
A: Hint
Using http://mathworld.wolfram.com/FermatsLittleTheorem.html
For $3\mid x(x^2-1),$
and $13|x(x^{12}-1)$
$[3,13]$ will divide $x(x^{[2,12]}-1)$ and hence $x(x^{12m}-1)$ for any integer $m$
A: Note that $39=3\times13$, and $3$ and $13 $ are prime.
Using Fermat's little theorem, $a^3\equiv a\bmod 3$, 
and by induction $a^{2k+1}\equiv a\bmod 3$, so $a^{48m+1}\equiv a\bmod 3$.  
Likewise, $a^{13}\equiv a\bmod 13,$  so $a^{12n+1}\equiv a\bmod 13$, so $a^{48m+1}\equiv a\bmod 13$.
Therefore, $a^{48 m+1}\equiv a\bmod 39$.  Can you take it from here?  
