Solve $x^2 + 8 = 3^n$, where $x\in\mathbb N$.

Question

First of all, $n$ is even (find that out by trying to see what kind of remainders you can possibly get when dividing everything by 4). So $n=2m$, $m\in N$.

$ x^2 + 8 = 3^{2m} \Rightarrow x^2 - 1 = 3^{2m} - 9 \Rightarrow (x+1)(x-1) = (3^m + 3)(3^m - 3) \Rightarrow x$ is odd.

That's all I've tried so far. I'd like to get some help. Thanks.

Answer 1

Now that you know $n$ is even, write $8=(3^{n/2}-x)(3^{n/2}+x)$. What are the possible factors of $8$? Which cases do we have, then?

Answer 2

As $n=2m$, we have $x^2+9=(3^m)^2$, i.e. $8=(3^m+x)(3^m-x)$. Thus $x=\frac{a-b}2$ where $ab=8$ and $a\ge b$, i.e. $x\in\{\frac{8-1}2,\frac{4-2}2\}$ and finally $x=1$.