I was trying to prove a counterexample to Hasse's principle. To do this, I was trying to prove that the equation $x^4-17=2y^2$ has no solutions on $\mathbb{Q}$ but has solutions in $\mathbb{Q}_p$ for every prime $p$.
I was able to prove that the equation has no solutions on $\mathbb{Q}$, but I was not able to do the second part. What I've done was to prove (using Riemman-Hurwitz) that $x^4-17=2y^2$ defines an elliptic curve and thus Hasse's theorem gives us that the curve is non empty over $\mathbb{F}_p$ (i.e. the equation has at least one solution over $\mathbb{F}_p$). The cases $p=2,17$ are special, but these are easily checked.
From here I know that I have to use Hensel's lemma to lift the solution to $\mathbb{Q}_p$. This is the Hensel's lemma I was trying to use (maybe it is not the correct form of Hensel's lemma):
Let $(K,|\cdot|)$ be a complete valuation field, $f\in R[X]$ and $a_0\in R$ such that $|f(a_0)|<|f'(a_0)|^2$, where $R=\lbrace x\in K\mid|x|\leq 1\rbrace$. Then, there exists a unique $a\in R$ such that $f(a)=0$ and $a-a_0\in\mathcal{M}$, where $\mathcal{M}=\lbrace x\in K\mid|x|< 1\rbrace$.
I'm not sure how to use this, since the theorem involves a polynomial in $x$, but I'm working with a polynomial in $x$ and $y$. Also, I'm not sure how to relate roots in $\mathbb{F}_p$ and roots in $\mathbb{Q}_p$. I know that the residual field of $\mathbb{Q}_p$ is $\mathbb{F}_p$ (i.e. $R/\mathcal{M}\cong\mathbb{F}_p$) but I'm not sure how this relates with Hensel's lemma.
Any hints or help will be thanked.
Let me just fill in the details of KCd's comment for completeness. First assume $p\neq 2,17$, and let $(a+p\mathbb{Z},b+p\mathbb{Z})$ be a root of our polynomial in $\mathbb{F}_p$. Then at least one of $a$ and $b$ is non-zero mod $p$; otherwise $17$ would be zero mod $p$, a contradiction. So we have two cases.
First suppose $a\notin p\mathbb{Z}$. Consider the polynomial $g(x)=x^4-(17+2b^2)$. Then $g(a)\equiv 0$ mod $p$ and $g'(a)=4a\not\equiv 0$ mod $p$, so by Hensel's lemma there is a root $a'$ of $g$ in $\mathbb{Q}_p$. Now $(a',b)$ is a $\mathbb{Q}_p$-rational point of our curve.
Now suppose $b\notin p\mathbb{Z}$. Then consider the polynomial $h(y)=2y^2+(17-a^4)$. Then $h(b)\equiv 0$ mod $p$ and $h'(b)=4b\not\equiv 0$ mod $p$, so by Hensel's lemma there is a root $b'$ of $g$ in $\mathbb{Q}_p$. Now $(a,b')$ is a $\mathbb{Q}_p$-rational point of our curve. Either way, we are done.
For the case $p=17$, you can use almost an identical argument as above; we just need to find an non-zero $\mathbb{F}_{17}$-rational point of the polynomial to work. (Why does this suffice?) For example, take $(1+17\mathbb{Z},3+17\mathbb{Z})$. Finally, for the case $p=2$, it would suffice to show that $17$ is a fourth power in $\mathbb{Q}_2$, and by Hensel's lemma it suffices for this to show that $17$ is a fourth power mod $2^5$. (Why?) But $3^4=81$ is congruent to $17$ mod $2^5$, as needed.