How do I prove that $x^2+8y^2$ does not properly represent 3? To me, it seems obvious that the binary quadratic form $x^2+8y^2$ does not properly represent 3. However, I have managed to prove that it does so I think I must be doing something stupid.
I have used the following:
Let f be a a binary quadratic form and n an integer. We say that f properly represents n if there exists [x,y]∈$\mathbb Z^2$ such that (x,y)=1 and f(x,y)=n. (x,y) is defined as the greatest common divisor of x and y.
Lemma 5.3 (iii) Some form of discriminant d properly represents n if and only if $u^2\equiv d\pmod {4n}$
Then here is my 'proof':
Let $f(x,y)=x^2+8y^2$. Then, by Lemma 5.3(iii), $f(x,y)$ properly represents n=3 if and only if there is a solution to $u^2\equiv d\pmod{4\cdot3}$
$d=b^2-4ac=0^2-4\cdot1\cdot8=-32$ so
$u^2\equiv -32\pmod{12}$  which gives us
$u^2\equiv 4\pmod{12}$
which clearly has the solution $u=2$ so $f(x,y)$ properly represents n=3.
I know that this is wrong and it's very likely wrong for a stupid reason but I can't figure out what that is so any help would be appreciated.

Lemma $5.3$ is meant the following way:
Theorem: Let $n > 0$ and $d$ be given integers. There exists a
binary quadratic form of discriminant $d$ that represents $n$ properly if and only if
the congruence $x^2\equiv d\bmod 4d$ has a solution.
So we only know that there exists a binary quadratic form $f(x,y)$ with discriminant $d=-32$ representing $n=3$ properly. Certainly it is not $x^2+8y^2$.
