How do I prove that $3^n - 3$ is never a square number? 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.?
 A: 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.
A: You would need $3\mid3^{n-1}-1$, and this cannot occur if $n > 1$.
A: First, you can notice that $a\equiv 0\mod 3$. That said, you now know that $a=3k$, meaning you need to prove that there do not exist positive integers $k$ and $n$ such that
$$9k^2 + 3 = 3^n$$
Or, dividing by $3$:
$$3k^2 + 1 = 3^{n-1}$$
Can you do the final step?
EDIT:
You can also continue what you started:
$$a^2=3(3^{n-1} - 1)$$
From this equation, you can see that $a^2$ must be divisible by $3$, and because it is a perfect square, it must be divisible by $9$, which is impossible.
