Hasse-Minkowski principle fails I am working on a counterexample to Hasse-Minkowski principle: 
$$F(X)=(X^2-2)(X^2-17)(X^2-34)=0$$
And the way the solution goes is we first consider the equation$\pmod p$,
 for prime $p\space s.t. p\space \neq  2\space, p\neq17$. And then for $p=2\space and\space p=17$. The part I am struggling with is the very first one. How can I properly apply Hensle's Lemma to show that we can lift the solution? Do we consider the whole equation or its parts? Do I need to find $x$ s.t. $F(x)\equiv 0\pmod p$ and $F(X)\not\equiv 0\pmod p$ and then apply Hensel's Lemma or do I consider $X^2-2\equiv 0\pmod p$,$X^2-17\equiv 0\pmod p$, $X^2-34\equiv 0\pmod p$ separately. Because it seems like if $X^2-2\equiv 0\pmod p$ then we do not really care about whether $X^2-17\equiv 0\pmod p$ or $X^2-34\equiv 0\pmod p$
Thank you! 
 A: There is a very detailed solution for this counterexample, and others, in the thesis on The Local-Global Principle in Number Theory by A. Schink, in chapter $7$. 
Case $1$. $p\neq 2,17$. If $2$ or $17$ is a quadratic residue, then $a^2=2$ or $a^2=17$ modulo $p$ is solvable in $\mathbb{Z}/p\mathbb{Z}$. Hence there exists an $a\in \mathbb{Z}/p\mathbb{Z}$ such that $f(a)\equiv 0\bmod p$. The derivative $2a$ is not congruent $0$ modulo $p$, since $p\neq 2$ and $p\nmid a$, so solutions lift by Hensel's Lemma. The other cases are solved similarly.
A: Here is a direct answer to your question.  You are trying to prove that $(X^2-2)(X^2-17)(X^2-34) = 0$ has a $p$-adic solution.  Since the $p$-adic integers are an integral domain, the product of those three factors is $0$ iff any one of them is $0$.  Therefore it is equivalent to consider each of the equations $X^2-2$, etc separately.
For your purposes you don't even need to use this fact: it's obvious that if, say, $X^2-2 = 0$ then so is the product, so it is certainly sufficient to show that one of the separate equations has a root.  Thus any thing you do to lift a solution to one of the three equations will give a solution to the original.
One can also see this from the congruence perspective: if $a(x) b(x) c(x) \equiv 0 \pmod{p^k}$ for some large exponent $k$ then one of $a(x), b(x), c(x)$ must be divisible by $p^{\lceil k/3 \rceil}$.  So if we can take $k$ arbitrarily large then the same must be true for at least one of $a(x), b(x), c(x)$.  The direction you're working in is even simpler: provided $a(x) \equiv 0 \pmod{p^k}$ then of course $a(x)b(x)c(x) \equiv 0 \pmod{p^k}$ as well.  It is totally ok to apply Hensel's lemma to $a(x)$ to get a root for $a(x) b(x) c(x)$.
