Show that for any natural number $n$, there exists a natural number $m$ for which: $$4^{2n+1} + 3^{n+2} = 13m$$
I don't know where to start. I tried to use Mathematical Induction, denoting the top statement by $\rm P(n)$ and prove that $\rm P(0)$ is true, but I got stuck. Can anybody help?
Because $$4^{2n+1}+3^{n+2}=4\cdot16^n+9\cdot3^n=4(16^n-3^n)+13\cdot3^n.$$
The statement $P(n)$ is:
There exists $m\in\mathbb N$ such that $$4^{2n+1}+3^{n+2}=13m$$
Step $1$: Proving that $P(0)$ is true should be easy, you just have to calculate what $$4^{2\cdot 0+1}+3^{0+2}$$ is equal to.
Step $2$: Assume that $P(n)$ is true, and write
$$\begin{align}4^{2(n+1)+1}+3^{(n+1)+2} &= 4^{2n+1+2} + 3^{n+2 + 1} \\&= 16\cdot 4^{2n+1} + 3\cdot 3^{n+2}\\&=13\cdot 4^{2n+1} + 3\cdot 4^{2n+1} + 3\cdot 3^{n+2}\end{align}$$
Can you continue from here?
If $P(n)$ holds, then$$\begin{align}4^{2(n+1)+1}+3^{n+1+2}&=4^{2n+3}+3^{n+3}\\&=4^2\times4^{2n+1}+3\times3^{n+2}\\&=13\times4^{2n+1}+3\times\bigl(4^{2n+1}+3^{n+2}\bigr).\end{align}$$Since $13\mid4^{2n+1}+3^{n+2}$, $13$ divides this number.
The sequence $u_n=a\alpha^n+b\beta^n$ satisfies the recurrence $$u_{n+1}=(\alpha+\beta)u_n-\alpha\beta u_{n-1}$$
So your expression (take $\alpha=16, \beta=3$) satisfies $u_{n+1}=19u_n-48 u_{n-1}$
Now $u_0=13, u_1=91$ and because you can then get all the other $u_n$ by successively adding multiples of these (using the recurrence) divisibility by $13$ persists (formally you can use induction).
The existence of the recurrence helps to explain the persistence of divisibility in all these kinds of problems.
\begin{eqnarray*} 4^{2n+1} + 3^{n+2} &\equiv _{13}& 4\cdot 16^n +3^{n+2}\\ &\equiv _{13}& 4\cdot 3^n + 9\cdot3^n\\ &\equiv _{13}& 13\cdot 3^n\\ &\equiv _{13}& 0 \end{eqnarray*} so $13\mid 4^{2n+1} + 3^{n+2}$.
