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
Note that
$$\begin{split} 3^{2(k+1)+1} + 2^{(k+1)+2} &= 9\cdot 3^{2k+1} + 2\cdot 2^{k+2}\\ &= 7 \cdot 3^{2k+1} + 2 (3^{2k+1} + 2^{k+2}). \end{split}$$
$3^{2n+1}+2^{n+2}$ 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 $(7+2)^n=7a+2^n$.
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.
You're not allowed to use mod, but here's a proof using mod:
$$3^{2n+1}+2^{n+2}=3\cdot 9^{n}+4\cdot 2^n$$
$$\equiv 3\cdot 2^n+4\cdot 2^n\equiv 7\cdot 2^n\equiv 0\pmod{7}$$
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
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 $n\ge 1$
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 $3^3\equiv 6$......
