# $P(X) = X^6 - 11X^4 + 36X^2 - 36$ has a root in $\mathbb{Q}_p$ for every $p$

Problem: Prove that $P(X) = X^6 - 11X^4 + 36X^2 - 36$ has a root in $\mathbb{R}$, has no roots in $\mathbb{Q}$, but has a root in $\mathbb{Q}_p$ for every $p$.

What I have done: I think this is actually false. We can find this factorization: $P(X) = (X^2 - 2)(X^2 - 3)(X^2 - 6)$. So we deduce that there are 6 different roots in $\mathbb{R}$, and there are no roots in $\mathbb{Q}$. For $p \not= 2,3$ combining Hensel's Lemma and multiplicativity of Legendre Symbol we can say that there is a root in $\mathbb{Q}_p$.

But for $p = 3$ we can't find $\sqrt{3}$ and $\sqrt{6}$ in $\mathbb{Q}_3$ beacuse they should have absolute value $|\sqrt{3}|_3 = |\sqrt{6}|_3 = 3^{-1/2}$ which is not possible cause, as sets, $|\mathbb{Q}_p|_p = |\mathbb{Q}|_p$. A root of $(X^2 - 2)$ in $\mathbb{Q}_3$ should be in $\mathbb{Z}_3$ because $\mathbb{Q}_3$ is the field of fractions of $\mathbb{Z}_3$ which is DVR and hence integrally closed (is that true?). But we can't solve $a_0^2 \equiv 2 \pmod 3$.

$\mathbb{Q}_2$ is almost the same cause we must find a root for $(X^2 - 3)$ but $3 \not \equiv 1 \pmod 8$.

Did I make any mistakes or is this problem just wrong?

Here is a valid counterexample for the Hasse principle of this sort. Take $$f(x)=(x^2-2)(x^2-17)(x^2-34)=0.$$ It has a real solution, but no rational one; and it has a solution in all completions of $\mathbb{Q}$. For $p\neq 2,17$ this goes as before with Hensel's lemma; and for $p=2$ we have now $17\equiv 1 \bmod 8$, so that $17$ is a $2$-adic square. For $p=17$ we have $6^2\equiv 2\bmod 17$, so that $f(6)\equiv 0\bmod 17$, but $f'(6)=12\not\equiv 0\bmod 17$, as required. So $f$ does not have a rational solution even though it has a solution in all $p$-adic fields, for $p$ prime and $p=\infty$.
In your example, for $p=3$ we would need an $n$ with $f(n)\equiv 0\bmod 3$, but $f'(n)\not\equiv 0\bmod 3$.