Prove some divisibility results by induction Please hint me, I have two questions:
Prove by induction that:
1)
$$ {13}^n+7^n+19^n=39k,\,\, n\in\mathbb O$$
in which $\mathbb O$ is  the set of odd natural numbers.  
2)
$$ 5^{2n}+5^n+1=31t,~n\not=3k, $$ $n\in \mathbb N$
 A: With congruences instead of induction:
(1) You need to show that $13^n+7^n+19^n$ is divisible by both $3$ and $13$. This follows from
$$13^n+7^n+19^n\equiv 1^n+1^n+1^n=3\equiv 0 \ ({\rm mod\ } 3),$$ 
$$13^n+7^n+19^n\equiv 0^n+7^n+(-7)^n\ ({\rm mod\ }13),$$ where the last expression is $=0$ since $n$ is odd.
(2) Here, I'd like to rewrite $$5^{2n}+5^n+1=(5^n)^0+(5^n)^1+(5^n)^2={(5^n)^3-1\over 5^n-1}={5^{3n}-1\over 5^n-1},$$
using the formula for geometric series (assuming $n\neq 0$). Note that 
$$5^{3n}-1=125^n-1\equiv 1^n-1=0\ ({\rm mod\ }31).$$ 
If $n$ is not divisible by $3$ then $n=3k+1$ or $n=3k+2$, and we have 
$$5^{3k+1}-1=5\cdot 125^k-1\equiv 5\cdot 1^k-1=4\ ({\rm mod\ }31),$$ 
$$5^{3k+2}-1=25\cdot 125^k-1\equiv 25\cdot 1^k-1=24\ ({\rm mod\ }31)$$
So $31$ divides the product $(5^{2n}+5^n+1)(5^n-1)=5^{3n}-1$, but not the factor $5^n-1$. Since $31$ is a prime number, it has to divide the other factor $5^{2n}+5^n+1$.
A: For one, let $\displaystyle f(n): 13^n+7^n+19^n=39k$ holds true for $n=m$ where $m$ is odd positive integer
$\implies\displaystyle13^m+7^m+19^m=39k_1$
$\implies\displaystyle13^m+7^m+19^m=13k_2\implies 7^m+19^m=13k_3\ \ \ \ (1)$
Now for $n=m+2,$
$\displaystyle13^{m+2}+7^{m+2}+19^{m+2}\equiv7^m\cdot49+19^m\cdot361\pmod{13}\equiv-3(7^m+19^m)\pmod{13}\equiv0$ [using $(1)$]
$\displaystyle13^{m+2}+7^{m+2}+19^{m+2}-(13^m+7^m+19^m)=13^m(13^2-1)+7^m(7^2-1)+19^m(19^2-1)\equiv0\pmod3$
$\displaystyle\implies 13^{m+2}+7^{m+2}+19^{m+2}\equiv13^m+7^m+19^m\pmod3\equiv0$
$\displaystyle\implies 13^{m+2}+7^{m+2}+19^{m+2}$ is divisible by lcm$(13,3)=39$ 
if $f(n)$  holds for $n=m$
Show that $f(n)$ holds if $n=1$

For two, let $g(n):5^{2n}+5^n+1=31t$ holds true for $n=m$
Now, $5^{2m}+5^m+1=(5^2)^m+5^m+1\equiv(-6)^m+5^m+1\pmod{31}$ 
$\displaystyle\implies (-6)^m+5^m+1$ is divisible by $31$
Now for $n=m+3,$ 
$\displaystyle (-6)^{m+3}+5^{m+3}+1-\{(-6)^m+5^m+1\}=(-6)^m\{(-6)^3-1\}+5^m(5^3-1)=(-6)^m(-217)+5^m(124)\equiv0\pmod{31}$
$\displaystyle\implies  (-6)^{m+3}+5^{m+3}+1\equiv(-6)^m+5^m+1\pmod{31}\equiv0$
Now show that $g(n)$ holds for $n=1,2$
