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)
$$ 5^{2n}+5^n+1=31t,~n\not=3k, $$ $n\in \mathbb N$

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 $n=1,2$
