If equation has integer solution it has solution for every prime p. How to prove that if the equation in the form:
$a_0 x_0^2 + a_1 x_1^2 + \dots + a_nx_n^2 = 0$
where $a_0, a_1, \dots , a_n \in \mathbb{Z}$, has an integer solution, then it has solution in $\mathbb{Z}_p$. Ultimately I would like to prove that if there is no solution for some prime $p$, there is no integer solution. Or at least I would like to know if there is some simple proof.
I've read about Hasse principle, but it and its proof are somewhat hard to grasp.
 A: Don't worry, this is the easy direction of the Hasse principle (the other direction is much harder) -- you don't have to understand the Hasse principle to understand this.
The point is that "addition and multiplication are well-defined mod $n$.'' So if $f(x_1, \ldots, x_n)$ is a polynomial with integer coefficeints, and $\alpha_1, \ldots, \alpha_n$ are integers, you can either evaluate $f(a)$, then reduce it mod $n$, or you can reduce the $\alpha_i$'s and $f$ mod $n$ (meaning, reduce all the coefficients of $f$), and then evaluate $\overline{f}(\overline{\alpha_1}, \ldots, \overline{\alpha_n})$. 
To say the same thing in a different language that you may not have seen before, the reduction map
$$
\phi: \mathbb{Z} \to \mathbb{Z}/n\mathbb{Z}
$$
is a ring homomorphism.
Therefore, suppose you have a polynomial $f(x_1, \ldots, x_n)$ with coefficients in $\mathbb{Z}$. If it has a solution $(\alpha_1, \ldots, \alpha_n)$, then $(\overline{\alpha_1}, \ldots, \overline{\alpha_n})$ is a solution to the reduced polynomial $\overline{f}$.
