# How do I differentiate between $\sqrt{5}$ and $-\sqrt{5}$ in the $11$-adics - and should I even try to?

I've been playing around with the $$11$$-adic numbers lately, and in particular, the values of $$\sqrt{5}$$. I have two different approximations of this value:

• $$a_1=\ ...937785904$$A$$44_{11}$$
• $$a_2=\ ...1733251$$A$$6067_{11}$$

Of course, as how in the real numbers the equation $$x^2-5=0$$ has two solutions ($$\sqrt{5}$$ and $$-\sqrt{5}$$), the same equation has two solutions in the 11-adics, as shown above. In my mind, it makes sense that these two solutions match up with the two solutions in the reals, down the the fact that $$a_1+a_2=0$$.

However, if that is the case, which is negative, and which is positive?

Part of me feels like this is a silly question to ask. In the $$p$$-adics, the line between negatives and positives can be blurred. For instance, in the $$11$$-adics, $$...11111_{11}\equiv-\frac{1}{10}$$, whereas $$...11112_{11}\equiv\frac{9}{10}$$. There's not the "luxury", so to speak, of being able to tell between one and the other by checking if there's a negative sign or not.

But still, something nags at me that the two are different numbers where one is the negative of the other, and I don't know which one's which. I can't find any clever little trick to suss out which is negative and which is positive. Doing $$(1+a_n)^2$$ yields $$6+2a_n$$, as one would expect, but this is of little value for our purposes.

All this leads up to the question, is there even any real purpose in trying to figure this out? In the reals, the equation $$x^2-5=0$$ has two roots, $$b_1$$ and $$b_2$$ whose sum is $$0$$. This is the same in the $$11$$-adics. In the real numbers, one being negative and one being positive is of consequence, because those concepts mean something. We can say definitively that $$b_1 even though $$|b_1|=|b_2|$$. The same cannot always be said of the $$11$$-adics, as is seen here. So, is there a concrete answer, and in the grand scheme of things, does there need to be one?

• Jan 4 at 1:30
• There is no meaning to the question. There is a convention in $\mathbb R$ showing how to choose one of the two square roots. But there is no such convention in $\mathbb Q_{11}$. Similarly, in the extension of $\mathbb R$ by appending zeros for $X^2+1$, there is no way to distingulsh the two zeros. Jan 4 at 1:31
• The question is the same in the finite field $\Bbb{F}_{11}$ or the cyclic group of $10$-th roots of unity. Note that in $1+11\Bbb{Z}_{11}$ there is a canonical choice of square root (the one which is again in $1+11\Bbb{Z}_{11}$) Jan 4 at 1:47
• Some people do call $5^{(11+1)/4}\equiv 4\pmod{11}$ the principal branch of the square root. That would make your $a_1$ the principal branch. Be very suspicious when someone tells tells you that something is not done, or doesn't exist, in terms of terminology or definitions. Almost always false.
– plop
Jan 4 at 2:10
• @owl good take. Unfortunately this approach works (as written) only when $p$ is one less than a multiple of $4$. We're still stuck e.g. with $\sqrt{14}$ in $5$-adics. Jan 4 at 17:17

There is no reasonable notion of "positive" in the $$p$$-adic numbers, in the following precise sense. Say that a field $$F$$ is formally real if it admits a total order $$\le$$ making it an ordered field, meaning that $$a \ge b$$ iff $$a - b \ge 0$$ and that the set of positive elements $$\{ a \in F : a > 0 \}$$ contains $$1$$ and is closed under addition and multiplication (there are a few equivalent ways to state this definition).

Claim: $$\mathbb{Q}_p$$ is not formally real, for any prime $$p$$.

Proof. In any ordered field, either $$a \ge 0$$ or $$a \le 0$$ (trichotomy). If $$a \ge 0$$ then $$a^2 \ge 0$$, and if $$a \le 0$$ then $$-a \ge 0$$ so $$(-a)^2 = a^2 \ge 0$$. So squares are non-negative, and so a sum of squares must be non-negative. In particular, $$-1$$ is negative and so cannot be a sum of squares. (This is an equivalent way to define formally real fields.) But

$$(1 - 4p)^{\frac{1}{2}} = \sum_{n \ge 0} {\frac{1}{2} \choose k} (-4p)^k$$

converges $$p$$-adically for every prime $$p$$ (the factor of $$4$$ is needed only to handle the case $$p = 2$$), from which it follows that we can write

$$-1 = \sqrt{1 - 4p}^2 + (4p - 2) 1^2$$

as a sum of at most $$4p - 3$$ squares in $$\mathbb{Q}_p$$. (We can do much better than this; by applying Lagrange's four-square theorem we can replace $$4p - 3$$ with $$5$$, and I think applying it more carefully we should be able to replace $$5$$ with $$4$$. But we don't need to.)

But this means that $$\mathbb{Q}_p$$ is not formally real as desired. $$\Box$$

So, all we can really say is that one of the square roots is $$4 \bmod 11$$ and the other one is $$7 \bmod 11$$ and that's all. There is no reasonable sense in which one is "positive" or "larger" than the other. This is analogous to the situation with the two square roots of $$-1$$ in the complex numbers $$i$$ and $$-i$$, which also don't really come in any kind of preferred or distinguished order. (This is a little disguised if you define $$\mathbb{C}$$ as $$\mathbb{R}[i]/(i^2 + 1)$$ since this definition privileges $$i$$ over $$-i$$ explicitly, but there are alternatives that avoid this. You can, for example, define $$\mathbb{C}$$ as the algebra of matrices in $$M_2(\mathbb{R})$$ of the form $$\left[ \begin{array}{cc} a & b \\ -b & a \end{array} \right]$$.)

• $-1$ is already a sum of two squares in $\mathbb Q_p$ for $p \equiv 3$ mod $4$, and is itself a square in $\mathbb Q_p$ for $p \equiv 1$ mod $4$. Only in $\mathbb Q_2$ do we need four squares. Jan 6 at 1:49
• @Torsten: I couldn't find a nice argument for $p \equiv 3 \bmod 4$; what do you have in mind? I see that $-1 = 1 + (-2)$ works if $p \equiv 3 \bmod 8$. Jan 6 at 2:00
• See user Ted's beautiful answer to math.stackexchange.com/q/149520/96384. Jan 6 at 2:46
• Ah, that makes sense, thanks! Jan 6 at 3:44
• Accepting some theory, one can actually find solutions where the squares are integers: Since the squares in $\mathbb F_p^\times$ form a subgroup of index $2$, there is at least one $n \in \{1, \dots, p-2 \}$ such that $n$ is a square (in $\mathbb Z_p$) and $n+1$ is not. But then, if $-1$ is not a square itself, $-n-1$ is a square (modulo $p$, and then in $\mathbb Z_p$ via Hensel). And $-1 = n + (-n-1)$. Jan 6 at 17:34

It is a good exercise to check that the closure of the set of negative rational numbers in $$\mathbb{Q}_p$$ is all of $$\mathbb{Q}_p$$. It follows that there is no meaningful notion of "negative" in $$\mathbb{Q}_p$$.

This doesn't mean that we are hopeless and there is no way at all to tell one square root of $$5$$ from the other in $$\mathbb{Q}_{11}$$. For instance, one is 4 mod $$11$$ and the other is $$7$$ mod $$11$$. You could define a continuous "square root map" on $$(\mathbb{Z}_p^\times)^2$$ (the set of $$p$$-adic units that are squares) by, for example, insisting on always taking the square root with least positive residue. I don't know of practical implications of such a map, though, unlike in the real case.

If the radicand is $$\equiv1\bmod p$$ for odd $$p$$ (meaning the $$p$$-adic representaion ends in $$1$$)*, then we may plug in the Maclaurin series

$$\sqrt{1+x}=1+(x/2)-(x^2/8)+...$$

and will converge to a specific value, which will be one of the square roots, specifically having the form $$1+py$$ where $$|py|<1$$ in the $$p$$-adic norm. We could call such a root the "positive" square root in the sense that the multiplicative identity $$1$$ absolutely dominates the series and its sum. But there are two caveats:

1. Not all $$p$$-adic integers will lend themselves to a convergent Maclaurin series even if roots exist; as noted above the radicand has to have residue $$1\bmod p$$. We thus cannot universally define a "positive" root in this way. For instance, the Maclaurin series for $$\sqrt{1+4}$$, rendered into $$11$$-adics, will not converge to either $$11$$-adic square root of $$5$$.

2. Because of the difference between $$p$$-adic and real norms, what are ordinarily negative numbers can appear "positive" through their difference from the multiplicative identity having a reduced $$p$$-adic norm. Thus in your $$11$$-adic example, $$\sqrt{1+99}=\sqrt{100}$$ gives a convergent Maclaurin series -- whose sum is $$-10$$ instead of $$+10$$. If we use the modification denoted by * for the case $$p=2$$, the $$2$$-adic square root of $$1+8=9$$ from the Maclaurin series is $$-3$$ instead of $$+3$$.

*For $$p=2$$ we would need the radicand to be $$\equiv1\bmod 8$$, thus to end $$2$$-adically in $$...001$$. The sum of the series then converges to the root $$2$$-adically closest to the multiplicative identity, having the form $$1+4y$$.

You may also wonder whether it is possible to define a canonical square root on $$(\mathbb Q_p^\times)^2$$.

When $$p$$ is odd, there is a canonical choice of square root for elements in $$\mathbb Z_p$$ that are $$1$$ modulo $$p$$, because \begin{align}1 + p \mathbb Z_p & \to 1 + p \mathbb Z_p \\ x & \mapsto x^2\end{align} is an isomorphism. Now $$\mathbb Q_p^\times \cong \mathbb Z \times \mathbb F_p^{\times} \times (1 + p \mathbb Z_p)$$ canonically. So to define a canonical square root on all of $$(\mathbb Q_p^\times)^2$$ becomes therefore equivalent to defining a canonical square root on $$(\mathbb F_p^\times)^2$$.

When $$p \equiv 3 \pmod 4$$ there is a unique choice of square root that gives a group homomorphism $$\sqrt{}: (\mathbb F_p^\times)^2 \to \mathbb F_p^\times$$. Explicitly, \begin{align}\sqrt{} : (\mathbb F_p^\times)^2 &\to \mathbb F_p^\times \\ x & \mapsto x^{(p+1)/4} \,. \end{align} Alternatively, when $$g$$ is any primitive root mod $$p$$, send $$g^2$$ to $$-g$$ and extend multiplicatively to $$g^{2k}$$.

When $$p \equiv 1 \pmod 4$$, no such group homomorphism exists, or we would reach the classical high school student's paradox that $$1 = \sqrt{1^2} = \sqrt{(-1)^2} = \sqrt{-1}^2 = -1 \,.$$

I remember reading in Neal Koblitz's book that there are examples of rational polynomials where you can feed a rational approximation into Newton-Raphson and get convergence to a rational root, but the real and p-adic versions give different ones.