A fundamental tool used in the study of Diophantine equations... Notations:  $K$ is a perfect field, $\overline K$ an algebraic closure and  $V\subseteq \mathbb P^n(\overline K)$ is a projective variety on $\overline K$. If $V$ is defined over $K$, in symbols $V/K$, then I indicate with $V(K)$ the set of $K$-rational points of $V$, namely $V(K):=\mathbb P^n(K)\cap V$.
I'm sorry if this notational introduction is too quick, but this is a quite standard stuff.

This question arises from the very general remark you can find in the book "Tha arithmetic of elliptic curves - Silverman" at page 8:

Now, first of all it is not clear what is the formal statement of the proposition, let's try:

Let $f_1,\ldots,f_m$ be homogeneous polynomials with coeffs in
  $\mathbb Q$. If for every prime $p$ there exists $r\in\mathbb N$ such that they don't have a common (nontrivial) solution in 
  $\mathbb Z/p^r\mathbb Z$ then there isn't a (nontrivial) common
  solution in $\mathbb Q$.

Is it correct? Moreover, I'd like to see a proof of this important statement, could you please suggest any references or give any hints?
Finally, I don't understand the part "A more succinct way to phrase...", in particular I can't see where $p$-adic numbers are involved. 
 A: I think he's saying $f_1,\dots,f_m$ have a solution in $\Bbb Q$ implies they have a solution in $\Bbb Z/p^r\Bbb Z$ for all $p$ and all $r$.  Thus if you can find one $p$ and $r$ where there is no solution then there cannot be a solution in $\Bbb Q$.
You have overstated it as having to find no solution in $p^r$ for every $p$, for some $r$.  That's a much stronger statement.  There can be solutions for some $p$ yet not all $p$.
Furthermore note that it's not enough to have a solution in $\Bbb Z/p^r\Bbb Z$ for all $p^r$ to have a solution in $\Bbb Q$.  But to have one in $\Bbb Q$ you must have one in all $\Bbb Z/p^r\Bbb Z$.  Therefore to not have one in $\Bbb Q$ it suffices to show you do not have one in $\Bbb Z/p^r\Bbb Z$ for at least one $p$ and one $r$.
A: 1) What you write in yellow is true but is not what Silverman writes in black.
His criterion is much easier to use: you only have to check what happens for a single prime $p$.
2) Silverman's key observation is: 

If a point $P\in V(\mathbb Q)$ has projective coordinates $(q_0:q_1:\dots:q_n)$ [with the $q_i\in \mathbb Q$ not all zero], then the same point $P$ also has coordinates $(\lambda q_0:\lambda q_1:\dots:\lambda q_n)$, where $0\neq \lambda\in \mathbb Q$ is arbitrary.
  Hence we may assume, by suitably choosing $\lambda$, that $P=(z_0:z_1:\dots:z_n)$ where the $z_i$'s are integers with no common prime divisor.  

3) For the same reason one may also assume that the  polynomials $f_j\in \mathbb Q[X_0,X_1,\dots,X_n]$ defining $V$ also have coefficients in $\mathbb Z$.
Thus the  point $P\in V(\mathbb Q)$ has homogeneous coordinates satisfying $f_j(z_0,z_1,\dots,z_n)=0$ .  
4) We may now reduce modulo an arbitrary prime $p$ and obtain a solution $\overline P=(\overline z_0:\overline z_1:\dots:\overline z_n)\in V(\mathbb F_p)$.
Note carefully that our demand that the $z_i$'s have no prime factor ensures that $\overline P$ is well defined i.e. that $\overline z_i\neq \overline 0$  for at least one of its homogeneous coordinates $z_i$.  
5) Hence woe is us if $V(\mathbb F_p)=\emptyset$  for a single $p$: this implies (by contraposition) that $V(\mathbb Q)=\emptyset$ too, and this is the heart of Silverman's remark which motivated your question.
6) But we have only described a necessary condition  for having $V(\mathbb Q)\neq\emptyset$.
It can very well happen that for all primes $p$ one has  $V(\mathbb F_p)\neq\emptyset$ , but that nevertheless  $V(\mathbb Q)=\emptyset$.
This is the case for Selmer's notorious (smooth, irreducible) projective curve $V\subset \mathbb P^2$ given by the equation $$3X_0^3+4X_1^3+5X_2^3=0$$.  
