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How do you prove that $3n^2-1$ is never a perfect square

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closed as off-topic by Hayden, Michael Albanese, M Turgeon, William, RecklessReckoner Jul 15 at 2:19

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Hint: More generally, $3m-1$ is never a perfect square when $m$ is an integer. –  Thomas Andrews Jul 15 at 0:29

3 Answers 3

Hint: Look at the equation modulo $3$. Any integer $x$ can be only congruent to $0,1$ or $-1$ modulo $3$. What can you say about $x^2$?

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Another approach. First establish that perfect squares are either $0$ or $1$ modulo $4$:

$$(2k)^2 = 4k^2 \equiv 0\pmod 4$$

$$(2k+1)^2 = 4k^2 + 4k + 1 \equiv 1 \pmod 4$$

If $n$ is even, $n = 2m$ and,

$$3n^2 - 1 = 12m^2 - 1 \equiv -1 \equiv 3 \pmod 4$$

If $n$ is odd, $n = 2m+1$ and,

$$3n^2 - 1 = 12m^2 + 12m + 3 - 1 \equiv 2 \pmod 4$$

In neither case is the condition for a perfect square modulo $4$ met.

Hence $3n^2 - 1$ is never a perfect square.

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Let $3n^2-1=b^2, \text{ for a } b \in \mathbb{Z}$

$$3n^2-1 \equiv -1 \pmod 3 \equiv 2 \pmod 3$$

$$b=3k \text{ or } b=3k+1 \text{ or } b=3k+2$$

Then:

$$b^2=9k \equiv 0 \pmod 3 \text{ or } b^2=3n+1 \equiv 1 \pmod 3 \text{ or } b=3n+1 \equiv 1 \pmod 3$$

We see that it cannot be $b^2 \equiv 2 \pmod 3$,so the equality $3n^2-1=b^2$ cannot be true.

Therefore, $3n^2-1$ is never a perfect square.

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