$X(\mathbb{Z}/p\mathbb{Z})$ versus $\{X(\mathbb{Z})\pmod{p}\}$ Let $P_1$,...,$P_m$ be polynomials in $n$ variables with coefficients in $\mathbb{Z}$ and consider the set
$$X(\mathbb{Z})=\{(x_1,...,x_n)\in \mathbb{Z}^n \ |\ P_i(x_1,...,x_n)=0 ~ ,\ \forall i \in\{1,...,m\}\}.$$
In algebraic geometry, we are often interested in the number of elements in 
$$X(\mathbb{F}_{p})=\{(x_1,...,x_n)\in \mathbb{F}_p^n \ |\ P_i(x_1,...,x_n)=0 \pmod{p},\  \forall i \in\{1,...,m\}\}.$$
Are there known results about
$$\{X(\mathbb{Z})\pmod{p}\}:=\{(x_1,...,x_n) \pmod{p}\ |\ (x_1,...,x_m)\in X(\mathbb{Z})\}~ ?$$
Certainly, $|\{X(\mathbb{Z})\pmod{p}\}|\leq |X(\mathbb{F}_{p})|$. Can we say more? 
I'm interested in any notes/papers on the subject.
 A: I found the MSRI survey Strong approximation for algebraic groups by
Andrei Rapinchuk very well-written and informative (and it has a lengthy bibliography for further exploration).
To answer your question: the cubic hypersurface $X\subset \mathbb{A}^3$ defined by $3x^3 + 4y^3 + 5z^3 =0$ has only one $\mathbb{Z}$-point $(0,0,0)$ but has more than one $\mathbb{Z}/p\mathbb{Z}$-points for prime $p$.  
For example, when $p=2$ the locus is $\{(0,0,0),(0,1,0),(1,0,1),(1,1,1)\}$.
So your inequality is generally strict.
For a variety $X$ defined over $\mathbb{Z}$, say that $X$ has strong approximation if the reduction map $X(\mathbb{Z})\to X(\mathbb{Z}/m\mathbb{Z})$ is surjective for all $m\geq 1$.
Some necessary conditions for this property are that $X$ is absolutely irreducible, and also that the $\mathbb{Z}$-points are Zariski dense.  
An example where strong approximation does hold is $\mathrm{SL}_2$; and more generally simply-connected algebraic groups and homogeneous spaces whose $\mathbb{Z}$-locus is Zariski dense (see the above reference for a precise formulation).
