# How to check if $2$ is a square $\mod 3$?

I don't think I can use the Legendre or Jacobi symbol here because $2$ is an even prime. I'm not sure I've learned methods to deal with $2$ even though I know how to use quadratic reciprocity, it only works with odd numbers I think.

• $0^2\equiv0\pmod3$, $1^2\equiv1\pmod3$, $2^2\equiv1\pmod3$ Commented Apr 2, 2015 at 16:33
• Maybe you're looking for a method for deciding whether $2$ is a square mod $p$, where $p$ is an arbitrary odd prime, because the $p=3$ case is just trivial. Commented Apr 2, 2015 at 16:35
• I see a pattern that every square is either $0$ or $1$ $\mod 3$, but I don't know how to prove that every square number is either $0$ or $1$ $\mod 3$. Commented Apr 2, 2015 at 16:40
• That's quite unusual.Attempting to understand quadratic reciprocity without mastering basic modular arithmetic or congruences is an extreme example of trying to run before one learns how to walk! Commented Apr 2, 2015 at 18:46
• In any case, I strongly recommend that you master such before tackling more advanced topics. There are many answers here that will aid learning: simply type "mod" in the site search box. Commented Apr 2, 2015 at 18:57

$0^2 = 0 \equiv 0 \pmod3$

$1^2 = 1 \equiv 1 \pmod3$

$2^2 = 4 \equiv 1 \pmod3$

So, no.

• I don't see how this proves that $2$ is not a square ($\mod 3$). What is the justification for not having to check every square number $\mod 3$? Commented Apr 2, 2015 at 16:37
• @EYES, what other numbers exist mod 3? 0,1, and 2 are the only numbers here. Commented Apr 2, 2015 at 16:38
• @ᴇʏᴇs $(3n+k)^2 = 3(3n^2+2k)+k^2 \equiv k^2 \pmod3$ so you do not have to look at any more Commented Apr 2, 2015 at 16:41
• @ᴇʏᴇs $(3n + k)^2$ represents the square of a number $k$ more than a multiple of $3$. Modulo $3$ this square is equivalent to $k^2$, as I have shown in my comment, and $k^2$ is equivalent to $0$ or $1$ as I have shown in my answer. Commented Apr 2, 2015 at 16:46
• If you wanted to say $k$ was $3$ like $3n + 3$, for example, then I could just call that number $3(n+1) + 0$ and so Henry's argument still applies. If you wanted to tell me that $k$ was $43$, as in $3n + 43$, then I could just write $3(n+14) + 1$. The only possible remainders when dividing by $3$ are $0,1,2$. The fact is that every single integer is of the form $3n$, $3n + 1$, or $3n + 2$. There are no other possibilities. Commented Apr 2, 2015 at 17:05

In a more general context $2$ is a square mod $p$ where $p$ is an odd prime if and only if :

$$\begin{pmatrix}2\\p\end{pmatrix}=(-1)^{\frac{p^2-1}{8}}\text{ is } 1$$

In your case, because $3^2-1=9-1=8$ the answer is no.

• $2$ is a square mod $p$ when $p=2$ (ducks) Commented Apr 2, 2015 at 16:39
• I wanted my answer to be complete and I forgot to mention this case... I am confused, thank you... Commented Apr 2, 2015 at 16:40