Solving $x^2 - 11y^2 = 3$ using congruences I'm looking to find solutions to $x^2 - 11y^2 = 3$ using congruences. 
The question specifically asks "Can this equation be solved by congruences (mod 3)? If so, what is the solution? (mod 4) ? (mod 11) ?"
I know using (mod 4) I can show that there aren't any solutions:
$x^2 - 11y^2 \equiv 3 (mod 4)$
$x^2 - 11y^2 \equiv -1 (mod 4)$
This is unsolvable because of the fact that 11 is a prime, and it's congruent to 3 mod 4, and thus would follow that there are no solutions. Thus I know that the whole equation is unsolvable.
I'm curious because the back of the book said that this result could also be seen (mod 3), but I can't get the same result:
$x^2 - 11y^2 \equiv 3 (mod 3)$
$x^2 - 11y^2 \equiv 0 (mod 3)$
Thus x and y can both be congruent to 0 mod 3 to solve it.  Can someone help me understand why this equation is provably unsolvable mod 3?
 A: $$-11 \equiv -11+16 \equiv 5 \equiv 1 \pmod 4$$
So,the congruence becomes:
$$x^2+y^2 \equiv 3 \pmod 4$$
$$x \equiv 0 \pmod 4 \Rightarrow x^2 \equiv 0 \pmod 4$$ 
$$x \equiv 1 \pmod 4 \Rightarrow x^2 \equiv 1 \pmod 4$$ 
$$x \equiv 2 \pmod 4 \Rightarrow x^2 \equiv 0\pmod 4$$ 
$$x \equiv 3 \pmod 4 \Rightarrow x^2 \equiv 1 \pmod 4$$ 
$$$$
$$y \equiv 0 \pmod 4 \Rightarrow y^2 \equiv 0 \pmod 4$$ 
$$y \equiv 1 \pmod 4 \Rightarrow y^2 \equiv 1 \pmod 4$$ 
$$y \equiv 2 \pmod 4 \Rightarrow y^2 \equiv 0\pmod 4$$ 
$$y \equiv 3 \pmod 4 \Rightarrow y^2 \equiv 1 \pmod 4$$
We can see that it cannot be $x^2+y^2 \equiv 3 \pmod 4$
EDIT:
$$-11 \equiv 1 \pmod 3$$
So,the congruence becomes:
$$x^2+y^2 \equiv 0 \pmod 3$$
$$x \equiv 0 \pmod 3 \Rightarrow x^2 \equiv 0 \pmod 3$$ 
$$x \equiv 1 \pmod 3 \Rightarrow x^2 \equiv 1 \pmod 3$$ 
$$x \equiv 2 \pmod 3 \Rightarrow x^2 \equiv 1 \pmod 3$$ 
$$$$
$$y \equiv 0 \pmod 3 \Rightarrow y^2 \equiv 0 \pmod 3$$ 
$$y \equiv 1 \pmod 3 \Rightarrow y^2 \equiv 1 \pmod 3$$ 
$$y \equiv 2 \pmod 3 \Rightarrow y^2 \equiv 1 \pmod 3$$ 
We can see that $x^2+y^2 \equiv 0 \pmod 3$,only if $x \equiv 0 \pmod 3 \text{ AND } y \equiv 0 \pmod 3$
A: The point is not that it is unsolvable mod $3$, but that every solution modulo $3$ must satisfy $x\equiv y\equiv0\pmod3$.
Indeed, as $-11\equiv1\pmod3$ we have $x^2+y^2\equiv0\pmod3$. Some case checking learns that squares are only congruent to $0$ or $1$ $\pmod3$ (and this is worth being remembered!), so $x^2+y^2\equiv0$ implies $x^2\equiv y^2\equiv0$.
Now here comes the contradiction: if $3\mid x,y$, then $9\mid x^2,y^2$ and hence $9\mid x^2-11y^2=3$.
