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.
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 $\overline{f}$.