Prove that $x^2 \not \equiv 3$ (mod 9)

Question

I want to prove that $x^2 \not \equiv 3$ (mod 9). I understand why this is true, but I don't know how to prove it. Where would I start?

Answer 1

If $x^ 2\equiv 3 \bmod 9$ then $x^ 2$ is a multiple of $3$, by euclids theorem $x$ is a multiple of $3$ and then $x^ 2$ is a multiple of $9$. Contradiction.

Answer 2

In mod nine, there are only 9 possibilities for classes of numbers i.e $[0],[1],...,[8]$. Just plug these in to show $[x]^2 \not \equiv 3 \ \textrm{(mod)}\ 9$.

Answer 3

In case you want to read up more:

The most transparent way to see such things is to work in $\mathbb{Z}_9$, and you are asking whether or not $3$ is a quadratic residue mod $9$. Understanding quadratic residues is a very interesting and central subject. See:

See: http://mathworld.wolfram.com/QuadraticResidue.html for a first definition.

So, one way to prove your result would be to use the Jacobi symbol:

To get a flavor:

$( \frac{3}{9} ) =( \frac{3}{3}) (\frac{3}{3}) = 0 \cdot 0 = 0 \not= 1$. Thus, $3$ is not a quadratic residue mod $9$.

Answer 4

If $$x^2=3+9k=3(1+3k)$$ Being $3$ a prime number, $3^2$ must divide $x^2$ to be a perfect square. How $1+3k$ is not a multiple of $3$ we have a contradiction.