# Hensel's lemma in $\mathbb Z_2$

Can you give me a concrete example for a quadratic form $$f(x,y)=ax^2+bxy+cy^2 \in \mathbb Z_2[x,y]$$ which has a primitive solution $(x^*,y^*) \in \mathbb Z_2 \times \mathbb Z_2$ (mod 4) with the property $$\frac{\partial f}{\partial x}(x^*,y^*) \not\equiv 0 \text{ (mod 4)},$$ but $f$ has no solution in $\mathbb Z_2 \$? So, Hensel's lemma fails in this case.

Note: Hensel's lemma requires a solution (mod 8) for lifting.

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Not sure that I understand your derivative condition. But would $f(x,y)=x^2-5y^2$ work? The squares of odd numbers are congruent to $1\pmod8$, so any $(x,y)$ pair of odd numbers is a solution $\pmod 4$, the partial derivatives are $\equiv2\pmod4$, and there are no solution with $x,y$ units in $\Bbb{Z}_2$, because for such $x,y$ we have $f(x,y)\equiv4\pmod8$?? –  Jyrki Lahtonen Nov 15 '13 at 14:34
Ok, but I don't unterstand why the solution $x,y$ in $\mathbb Z_2$ satisfies $f(x,y) \equiv 4 \text{ (mod 8)}$. Please can you explain this to me? –  Lisa Mainhard Nov 15 '13 at 21:57
It depends. If $x,y$ are both odd, then $x^2$ and $y^2$ are both $\equiv1\pmod8$, so $x^2-5y^2\equiv1-5\equiv4\pmod8$. If one of them is odd and the other is even, then $x^2-5y^2$ is also odd. I don't know what you mean by $x^*,y^*$ - I thought that you mean that both of them are 2-adic units. –  Jyrki Lahtonen Nov 16 '13 at 6:09