TITLE: How do I prove that $3^n - 3$ is never a square number?
QUESTION [10 upvotes]: Prove that there do not exist positive integers $a$ and $n$ such that $a^2 + 3 = 3^n$.
I have tried writing the equation in another way: $a^2 = 3^n - 3 = 3(3^{n-1} -1)$ but that doesn't get me anywhere. Should I consider different cases such as $a$ and $n$ being odd, $a$ and $n$ being even, etc.?

REPLY [4 votes]: First note that $n=1$ gives $a=0$, which, while square, is not positive as required. So $n$ must be greater than $1$.
Proceeding to search for a contradiction, assume that we can find $a^2 = 3^n-3 = 3(3^{n-1} -1)$, so we would need $3 \mid a^2$. Since $3$ is prime, this would mean that $3 \mid a$ and we can find integer $k$ such that $a=3k$.
However that gives $3k^2 = (3^{n-1}-1)$, requiring $3 \mid (3^{n-1}-1)$ which is not true - a contradiction which shows that there is no such $a$ and that $3^n-3$ cannot be the square of a positive number.