Find all $n$ such that $3^{2n+1}+2^{n+2}$ is divisible by $7$ 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 
 A: Note that 
$$\begin{split}
3^{2(k+1)+1} + 2^{(k+1)+2} &= 9\cdot 3^{2k+1} + 2\cdot 2^{k+2}\\
&= 7 \cdot 3^{2k+1} + 2 (3^{2k+1} + 2^{k+2}). 
\end{split}$$
A: $3^{2n+1}+2^{n+2}$ 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 $(7+2)^n=7a+2^n$.
A: 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.
A: You're not allowed to use mod, but here's a proof using mod:
$$3^{2n+1}+2^{n+2}=3\cdot 9^{n}+4\cdot 2^n$$
$$\equiv 3\cdot 2^n+4\cdot 2^n\equiv 7\cdot 2^n\equiv 0\pmod{7}$$
A: Just another way: 
Inductive hypothesis: $3^{2n+1} = 7k - 2^{n+2}$
Inductive step:
$$3^{2n+3} + 2^{n+3} = 3^2 * 3^{2n+1} + 2 * 2^{n+2} $$
$=3^2(7k - 2^{n+2})+2*2^{n+2}$
$=7k(3^2) - 3^2*2^{n+2} + 2* 2^{n+2}$
$=7k(3^2) + 2^{n+2}(-3^2+2)$
$=7k(3^2) -7*(2^{n+2})$
Both terms are divisible by 7 
A: *

*Setting $n=1$, we get $$3^{2\cdot 1+1}+2^{1+2}=35$$ above number $35$ is divisible by $7$ hence it holds for $n=1$

*Assume the number $3^{2n+1}+2^{n+2}$ is divisible by $7$ for $n=k$ then $$3^{2k+1}+2^{k+2}=7\lambda \tag 1$$

*Setting $n=k+1$, we get $$3^{2k+2+1}+2^{k+1+2}$$
$$=9\cdot 3^{2k+1}+2\cdot 2^{k+1}$$
$$=9\cdot 3^{2k+1}+9\cdot 2^{k+1}-7\cdot 2^{k+1}$$
$$=9(3^{2k+1}+2^{k+1})-7\cdot 2^{k+1}$$
setting the value from (1), 
$$=9(7\lambda)-7\cdot 2^{k+1}$$
$$=7(9\lambda-2^{k+1})$$ 
since, $(9\lambda-2^{k+1})$ is some integer hence the number $7(9\lambda-2^{k+1})$ is divisible by $7$ 
hence the number $(3^{2n+1}+2^{n+1})$ is divisible by $7$ for all integers $n\ge 1$
A: Little Hint: 
Note that from little Fermat's theorem that $2^6\equiv 1$ and $3^6\equiv 1$. Then the remainders of $3$ $\pmod 7$ are $$3,2,6,4,5$$ while the remainders of $2$ $\pmod 7$ are $$2,4,1$$ If $n=1$, $2^3\equiv 1$ and $3^3\equiv 6$......
