Prove that $p$ divides $m$ and $3$

Question

Prove that if some prime of the form $p=4k-1$ for some $k \in \mathbb{Z}^+$ divides $m^2+9$, then $p$ divides $m$ and $3$.

We can't have $p$ divide $m$ and not $3$ or $p$ divide $3$ and not $m$. So we just have to consider the case where $p$ does not divide $m$ and $3$. How do we find a contradiction?

This question was asked yesterday... can't find it anymore though. — Servaes, Sep 13 '16 at 23:40
@dxiv How does that deal with the case that $p$ does not divide $m$ and $3$? — user19405892, Sep 13 '16 at 23:42
If $p>3$ then $m^2\equiv -9\pmod p$ implies that $-1$ is a square mod $p$. — lulu, Sep 13 '16 at 23:43
If you are unfamiliar with the theory of quadratic residues, note that $-1$ has order $2\pmod p$, clearly, so if $m$ were a square root of $-1$ it would have order $4\pmod p$. But the order of any element must divide $p-1$ hence $p-1=4r$ for some $r$, contradicting your assumption. — lulu, Sep 13 '16 at 23:57
Well, sure. Quadratic residues come up often in problems like these. Any time you have a quadratic congruence, you can get results of this form. — lulu, Sep 14 '16 at 00:21
@Servaes: I think this question has been already asked and then deleted by the same user, but I'm not sure... — Watson, Sep 14 '16 at 11:37

score 2 · Answer 1 · answered Sep 13 '16 at 23:49

Note that $p|3$ implies $p=3$ (as dxiv said). So, the problem is equivalent to the following statement:

Prove that if $p = 4k-1$ is a prime dividing $m^2+9$, then $p=3$.

Furthermore, we may notice that if $3|m^2+9$, $3|m$, so $p|m$. Thus, it suffices to prove:

The only prime $p\equiv 3\mod 4$ that can divide $m^2+9$ is $p=3$.

We will prove this using the notion of quadratic reciprocity.

Let $p$ be a prime $p \equiv 3\mod 4, p\neq 3$, and assume for the sake of contradiction that $p|m^2+9$ for some $m$, or equivalently that $m^2 \equiv -9 \mod p$. That would mean that

$$\left(\frac{-9}{p}\right) = 1$$

where $\left(\frac{a}{p}\right)$ is a Legendre Symbol, which is $1$ if $a$ is a quadratic residue mod $p$ and $-1$ otherwise. Using the property that

$$\left(\frac{a}{p}\right)\left(\frac{b}{p}\right) = \left(\frac{ab}{p}\right)$$

we get that

$$\left(\frac{-9}{p}\right) = \left(\frac{-1}{p}\right)\left(\frac{9}{p}\right)$$

Since $9$ is obviously a quadratic residue modulo any prime $p$, $\left(\frac{9}{p}\right) = 1$, so we reduce our condition to

$$\left(\frac{-1}{p}\right) = 1$$

This Legendre symbol is well known to be $1$ is $p\equiv 1\mod 4$ and $-1$ is $p\equiv 3\mod 4$, so we get a contradiction.

Note that this line of reasoning fails for $p=3$ since $\left(\frac{-9}{3}\right)$ is not $1$ or $-1$ by the way the Legendre symbol is defined. These manipulations do not work when the denominator of the Legendre symbol divides the numerator.

score 0 · Answer 2 · answered Sep 14 '16 at 00:06

0

Carl's proof is correct, but I think the Legendre symbol is over killing here.

If p <> 3, then m^2=-9 (mod p). There exists an integer a such that 3a=1 (mod p), thus (am)^2 = -1 (mod p)...

answered Sep 14 '16 at 00:06

S. Y

1,193

You're correct, but I think using the Legendre symbol requires less justification (for instance you would probably have to justify that $3^{-1}$ exists $\mod p$, and the only way I've seen the result that $m^2\equiv -1\mod p$ has solutions only when $p\equiv 1\mod 4$ (or $p=2$) is with the Legendre symbol. – Carl Schildkraut Sep 14 '16 at 00:19
There are other ways to show -1 is not a quadratic residue of a prime of type 4K-1. But anyway, I don't think my solution is materially different from Carl's. I tried to add it as a comment. Thanks. – S. Y Sep 14 '16 at 00:46

Prove that $p$ divides $m$ and $3$

2 Answers2