Find all $n$ such that $3^{2n+1}+2^{n+2}$ is divisible by $7$ [duplicate]

Find all $n$ such that

$3^{2n+1}+2^{n+2}$ is divisible by $7$

Prove that your answer is correct

So I am not allowed to use mods, as is a calculus question, I have tried by induction but can't get to prove that it works for $k+1$, by multiplying the equation by powers of $2$ and $3$.

Thank you for your help

Solution 1:

Note that

$$\begin{split} 3^{2(k+1)+1} + 2^{(k+1)+2} &= 9\cdot 3^{2k+1} + 2\cdot 2^{k+2}\\ &= 7 \cdot 3^{2k+1} + 2 (3^{2k+1} + 2^{k+2}). \end{split}$$

Solution 2:

Let $A_n=3^{2n+1}+2^{n+2}$ then you will find that $A_{n+1}=11A_n-18A_{n-1}$

Rationale: If $u_n=A\alpha^n+B\beta^n$ it is easy to check that $u_{n+1}=(\alpha+\beta)u_n-\alpha\beta u_{n-1}$.

Set $\alpha = 3^2=9, \beta=2$.

You need two consecutive values to ensure the persistence of the factor $7$ (you could use $n=-1$ even though the value involves fractions), which makes it less attractive in some ways than the induction arguments with a single base case. However, this can also used to construct further examples of persistence, and is quick if you are doing multiple questions of the same type.

Solution 3:

$3^{2n+1}+2^{n+2}$ is divisible by $7$ for all $n$:

$ 3^{2n+1}+2^{n+2}=\\ =3\cdot 9^n+4\cdot 2^n\\ =3\cdot (7+2)^n+(7-3)\cdot 2^n\\ =3(7a+2^n)+7\cdot2^n-3\cdot 2^n\\ =7(3a+2^n) $

where I have used the binomial theorem for getting $(7+2)^n=7a+2^n$.