Local-global property for Diophantine equations over $\mathbb{Z}$ It is known that the Pell's equation
$$
x^{2} - 223 y^{2} = -3
$$
doesn't have any integer solution. This can be proved by computing fundamental units of $K=\mathbb{Q}(\sqrt{223})$. However, I tried to find a single prime $p$ s.t. the equation doesn't have a solution mod $p$, but it seems that there's no such prime for $p<1000$ (according to SAGE, if I write a correct program). Can we prove that the equation has a solution mod $p$ for al $p$?
 A: Sure.


*

*For $p \neq 2, 223$, let's look at the sets $S_1 = \{x^2 | x \in \mathbb{F}_p\}$ and $S_2 = \{223y^2 - 3 | x \in \mathbb{F}_p\}$. Each of them has size $\frac{p+1}{2}$ - e.g. for $S_1$, if $x = 0$ then $0^2 = 0$, otherwise quadratic residues/nonresidues split evenly - hence there are $\frac{p-1}{2}$ distinct values for $x^2$ if $x \neq 0$; this adds up to $\frac{p+1}{2}$.
Now $x^2 = 223y^2 - 3$ has solution in $\mathbb{F}_p$ if and only if $S_1$ and $S_2$ intersects. But if they don't intersect, then $S_1 \cup S_2$ has $p+1$ elements, which has more elements than $\mathbb{F}_p$, contradiction. Hence there must be a solution.

*For $p = 2$, $(x,y) = (1,0)$ is a solution.

*For $p = 223$, it's easy to check by quadratic reciprocity that $-3$ is a quadratic residue mod $223$. 
A: We might as well homogenize the quadratic equation under the form $x^2-223y^2+3z^2=0$ and solve $x^2=223y^2-3z^2$ in $\mathbf F_p$. Supposing  $p\neq 2,3$, the question becomes whether $3.223$ is a square in $\mathbf F_p$. Now make a wholehearted use of quadratic residues and the quadratic recipocity law : $(\frac {3.223}p)=(\frac 3p)(\frac {223}p)=(\frac p3)(\frac p{223})$, so $3.223$ is a square in $\mathbf F_p$ iff $(\frac p3)=(\frac p{223})=1$ or $(\frac p3)=(\frac p{223})=-1$. But the natural surjection $(\mathbf Z/223)^* \to (\mathbf Z/3)^*$ (these are cyclic groups) induced by the map $a \to a^{111}$ respects squares (resp. non squares), and we are done. 
It remains to treat the special cases $p=2,3$. The case $p=2$ admits obvious solutions. In the case $p=3$, Euler's criterion applies : $223^2\equiv 13^2 \equiv 1$ mod $3$ . 
