Suppose $n$ divides $3^n + 4^n$. Show that $7$ divides $n$. Suppose $n \geq 2$ and $n$ is a divisor of $3^n + 4^n$. Prove that $7$ is a divisor of $n$.
My work so far:
I had a hypothesis that if $n| 3^n + 4^n$, then $n = 7^k$ for some $k\in\mathbb{N}$. But this is not necessarily so. Take $n = 7⋅379$, where $3^7 + 4^7 = 7^2⋅379$. Then, $3^7+4^7$ divides $3^n + 4^n$, and since $n$ divides $3^7+4^7$, we must have $n|3^n+4^n$.

Assume $\enspace n\mid 3^n+4^n$,  for some $n\in\mathbb{N}$, $n\geq 2$.
Hence, $$3^n+4^n=nm, \quad for \enspace m\in\mathbb{N}.$$ 
Since $3^n+4^n$ is odd $\enspace (odd\cdot odd=odd, \enspace even\cdot even=even, \enspace  odd+even=odd)$, and since $\enspace n \mid 3^n+4^n$, we know that $n$ is also odd. Thus,  we can express $3^n+4^n$ as
$$3^n+4^n=(3+4)\bigg(3^{n-1}-3^{n-2}4+3^{n-3}4^2-\ldots +3^24^{n-3}-34^{n-2}+4^{n-1}\bigg)$$
$$3^n+4^n=7k \qquad$$ 
whereby $\enspace k=\big(3^{n-1}-3^{n-2}4+3^{n-3}4^2-\ldots +3^24^{n-3}-34^{n-2}+4^{n-1}\big).$
Since $\enspace 7 \mid 7k$, it follows that $\enspace 7 \mid 3^n+4^n$.
Thus, so far we have,
$$3^n+4^n=nm=7k$$
and hence, 
$$n=7\bigg(\frac{k}{m} \bigg)$$
If $\enspace m \mid k$, then great, we are done! 
So, let's assume the contrary that $\enspace m \nmid k$. Since $\enspace m \nmid k$, there exists $q\in\mathbb{N}, q>0$ such that $m=7q$ and $n=\frac{k}{q}$, whereby $q\mid k$. $\enspace 7\nmid n$, since $\frac{n}{7}=\frac{k}{m}$. Hence, $gcd(n,7)=1$. We have already shown that $7\mid 7k$ and $n\mid 7k$. Thus,
$$gcd(n,7)=1, \enspace 7\mid 7k, \enspace n\mid 7k \enspace \Longrightarrow \enspace 7n\mid 7k \enspace \Longrightarrow \enspace n\mid k \enspace \Longrightarrow \enspace \frac{k}{q}\mid k$$
Since $m\nmid k$, $k=mb+r$ for some $b,r\in \mathbb{Z}, \enspace 0\leq r < m$. For some $s\in\mathbb{N}$,
$$\frac{k}{q}\mid k \enspace \Longrightarrow \enspace \frac{mb+r}{q}\mid k \enspace \Longrightarrow \enspace \big(7b+\frac{r}{q}\big)\mid k \enspace \Longrightarrow \enspace k=s\big(7b+\frac{r}{q}\big)$$
$$k=7sb+\frac{rs}{q}=mb+r \enspace \Longrightarrow \enspace b(7s-m)=r\big(1-\frac{s}{q}\big) \enspace \Longrightarrow \enspace \frac{7bq}{r}=\frac{q-s}{s-q} \enspace \Longrightarrow \enspace \frac{7bq}{r}=-1 \enspace \Longrightarrow \enspace 7bq=-r \enspace \Longrightarrow \enspace r=-mb \enspace \Longrightarrow \enspace k=mb-mb=0$$
But we already know that $k>0$, hence a contradiction. Thus, our assumption is false, and $m\mid k$.
Hence, since $n=7\big(\frac{k}{m}\big)$ and $m \mid k$, therefore $7\mid n$.
