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Hensel's lemma for odd primes, as I understand, allows you to determine solutions to various homogeneous quadratic equations modulo $p^n$ by solving an appropriate equation mod $p$ and lifting up.

However, for $p=2$ we run into problems, since the derivative of a quadratic form is zero mod 2. Though we can't reduce equations mod $2^n$ to equations modulo $2$, can we do something close? Like go down to $2^2, 2^3$ or something like this? Or is the situation hopeless?

Edit: Noting that the problem I posed arises with quadratic forms in particular.

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I think you are talking about the fact that odd squares are 1 modulo 8. So, in quadratic forms books, they refer to eight squareclasses in $\mathbb Z_2,$ notably $1,3,5,7,2,6,10,14.$ The comment is often made that, for (homogeneous) cubic forms, the uncomfortable prime would be 3 instead. – Will Jagy Nov 18 '12 at 3:14
Yes, I am thinking about quadratic forms in particular. Maybe I should have been clear. Because when you take the derivative of a quadratic (homogenous) polynomial mod 2, you get 0 of course. Would you expand your comment into an answer, Will? I would greatly appreciate it. – user21725 Nov 18 '12 at 3:18
up vote 0 down vote accepted

Maybe Andre has answered your question. I can say that your questions seem a bit too open-ended for precise answers.

Meanwhile, I can recommend $p$-adic Numbers by Fernando Q. Gouvea. I have the second edition.

With that in mind: first, by Ernst Selmer, 1957, we find that the cubic (homogeneous) form $$ 3 x^3 + 4 y^3 + 5 z^3 $$ has nontrivial zeroes in every $\mathbb Q_p$ including the reals as $\mathbb Q_\infty,$ but has no rational root except $(0,0,0).$ That's actually not in this book. By root I mean $ 3 x^3 + 4 y^3 + 5 z^3 = 0.$

Second, given a quadratic form (homogeneous degree 2) that has non-trivial roots in every $\mathbb Q_p,$ then it has a non-trivial rational root. This is the celebrated Hasse-Minkowski Theorem, Gouvea page 79, Theorem 3.5.2.

Third, given a form $f$ and a representation question $f(x_1, \cdots, x_r) = n,$ what you do is create $$ g(x_1, \cdots, x_r, x_{r+1}) = f(x_1, \cdots, x_r) - n x_{r+1}^2. $$ Then you ask whether $g$ has nontrivial roots. This leads quickly into unfamiliar territory and the word isotropic. See J. W. S. Cassels, Rational Quadratic Forms, up to page 63. Cassels lists the 2-adic squareclasses carefully on page 43, in the context of the Hilbert norm residue symbol.

Finally, page 73 in Gouvea, Problem 116: Show that if $b \in \mathbb Z_2,$ and $b \equiv 1 \pmod {8 \mathbb Z_2}$ (so that in particular $b$ is a 2-adic unit), then $b$ is a square in $ \mathbb Z_2.$ Conversely, show that any 2-adic unit which is a square is congruent to $1$ modulo $8.$ Conclude that the group $\mathbb Q_2^\times/ \left(Q_2^\times \right)^2$ has order $8,$ and is generated by the classes of $-1,5,$ and $2,$ so that a complete set of coset representatives is $\{ 1,-1,5,-5,2,-2,10,-10 \}.$

Well, that's a start.

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Edit: This is an answer to an earlier version of the question.

In the usual statement of Hensel's Lemma, there is no requirement that the prime $p$ be odd. Naturally, there arise problems with quadratic congruences modulo a power of $2$. But the analysis of $x^2\equiv a\pmod{2^n}$ is quite smooth for $n\ge 3$.

There are similar problems with congruences $x^3\equiv a \pmod{3^n}$.

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Thank you. My question was posed poorly, as usual. – user21725 Nov 18 '12 at 3:20

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