TITLE: Prove that there are no integers $a\ge 2\:$ and $n\ge 1$ such that $a^2+3=3^n$
QUESTION [0 upvotes]: Prove that there are no integers $a\ge 2\:$ and $n\ge 1$ such that $a^2+3=3^n$. I can not find which method I use (like proof by contradiction, contraposition, proof by cases, etc.) Can you help me about how can I start?

REPLY [2 votes]: $a^2 = 3(3^{n-1} - 1)\implies 3 \mid a^2\implies 3 \mid a\implies a = 3m\implies 9m^2= 3(3^{n-1}-1)\implies 3m^2=3^{n-1}-1$. This means $n=1, a = 0$. But $a \ge 2$. so $n > 1$, but then $1 = 3^{n-1} - 3m^2 = 3(3^{n-2} - m^2)\implies 3 \mid 1$ . Contradiction. So there is no solution!
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