Prove that $3a^2-1$ is never a perfect square when $a$ is an integer.
I'm not sure how to go about this proof or what form of an integer to use. I know an integer can be represented using
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$2k$, $2k+1$, or
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$3k$, $3k+1$, $3k+2$, or
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$4k$, $4k+1$, $4k+2$, $4k+3$…
but how do I know which form to use for this problem?
Best Answer
Hint $ $ If not, let $\,n\,$ be the least natural such that $\,\color{#c00}{n^2 =3k-1}\,$ for some $\,k\in \Bbb Z.\,$ Since $\,n\neq 1,2,3\,$ $\,n\!-\!3\,$ is a smaller such natural since $\,(n\!-\!3)^2\! = \color{#c00}{n^2}\!-\!6n\!+\!9 = \color{#c00}3(\color{#c00}k\!-\!2n\!+\!3)\color{#c00}{-1},\,$ contradiction.