# Number theory proofs regarding perfect squares [closed]

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

-

## closed as off-topic by Hayden, Michael Albanese, M Turgeon, William, RecklessReckonerJul 15 '14 at 2:19

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Hayden, Michael Albanese, M Turgeon, William, RecklessReckoner
If this question can be reworded to fit the rules in the help center, please edit the question.

Hint: More generally, $3m-1$ is never a perfect square when $m$ is an integer. – Thomas Andrews Jul 15 '14 at 0:29

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 $x^2$?

-

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.

-

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.

-