Is it true that $3$ is the only prime of the form $n^2-1$?

Question

One less than a perfect square is prime if and only if the prime is 3. Is this really, really true and do we have proof?

Seeing that a restatement of Legendre's conjecture can be made based on the answer to this question, why is it on hold? — Jeffrey Young, Sep 30 '14 at 16:08

score 13 · Accepted Answer · answered Sep 28 '14 at 08:25

13

Hint: $n^2-1=(n-1)(n+1)$

Should be obvious enough.

answered Sep 28 '14 at 08:25

fixedp

1,726

Are you absolutely certain of this? – Jeffrey Young Sep 28 '14 at 08:26
Expand the brackets on the right hand side if you have trouble believing this. – fixedp Sep 28 '14 at 08:27
n^2+1n-1n-1 Is this right? – Jeffrey Young Sep 28 '14 at 08:29
Yes, that is correct. – fixedp Sep 28 '14 at 08:29
n^2-1 Is this right? – Jeffrey Young Sep 28 '14 at 08:30
So n^2-1=n^2-1. – Jeffrey Young Sep 28 '14 at 08:32
Yep, as the equation suggests, $(n-1)(n+1)$ is another way of writing $n^2-1$. – fixedp Sep 28 '14 at 08:35
Is it because n squared minus one factors when n>2? – Jeffrey Young Sep 28 '14 at 08:40
2

This factorisation is true for any real number. The point is that $n^2-1$ has two factors $(n-1)$ and $(n+1)$, so it is prime iff either $(n-1)$ or $(n+1)$ is $1$. – fixedp Sep 28 '14 at 08:59
@fixedp the case $n+1=1$ implies $n=0$, which is not prime – Jakob W. Mar 13 '19 at 22:23

score 4 · Answer 2 · answered Sep 28 '14 at 08:26

4

Yes. In general, we may factor $$n^2 - 1 = (n + 1) (n - 1),$$ and both factors are larger that $1$ provided $n > 2$, or equivalently, that $n^2 - 1 > 3$.

answered Sep 28 '14 at 08:26

Travis Willse

99,363

Is it true that $3$ is the only prime of the form $n^2-1$?

2 Answers2

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