Trying to prove that the $p$-adic valuation of $a^2 + b^2$ is even for $p$-adic integers $a$ and $b$ when $p$ is a $3\bmod 4$ prime. The result for integer $a$ and $b$ is a lemma in elementary number theory, since an integer can be expressed as the sum of integer squares if and only if all $3\bmod 4$ primes which divide it do so to an even power. I suspect the proof that this extends to $a, b\in\mathbb{Z}_p$ has to do with the density of $\mathbb{Z}$ in $\mathbb{Z}_p$ and a property of the $p$-adic valuation, but I'm not quite sure where to go with it. Any advice?
Prove $v_p(a^2+b^2)$ is even for $a, b\in\mathbb{Z}_p$ and $p\equiv 3\pmod 4$.
number theoryp-adic-number-theoryvaluation-theory
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This is true for a non-trivial discrete (nonarchimedean) valuation $w$. Namely, we want to show $\mathcal{O} \subseteq \mathcal{O}_w$ or equivalently $w(x) \ge 0$ for all $x\in \mathcal{O}$, which I will write as $w(\mathcal O) \ge 0$. Now from the fact which you have, one deduces that for all $x \in \mathfrak{p}$ and all $l \in \mathbb{N}_{\geq 2}$ with $\gcd(l,p)=1$, $l$ divides $w(1+x)$. By discreteness of $w$ this enforces
$$w(1+\mathfrak{p}) =0.$$
Now the set $1+\mathfrak{p}$ makes up the most interesting part of $\mathcal{O}$. Actually, I assume you know, or can infer easily from what you know, that the multiplicative group of units $\mathcal{O}^\times$ has a direct decomposition
$$\mathcal{O}^\times \simeq \mu_F \times (1+\mathfrak{p}),$$
where $\mu_F$ is the group of roots of unity contained in $F$; and further the multiplicative monoid $\mathcal{O} \setminus\lbrace 0 \rbrace$ has a direct decomposition
$$\mathcal{O} \setminus\lbrace 0 \rbrace \simeq \pi^{\mathbb N_0} \times \mathcal{O}^\times$$
where $\pi$ is an element with $v(\pi)=1$. Now from the definition of a root of unity it is immediate that
$$w(\mu_F) = 0$$
as well, hence $w(\mathcal{O}^\times) =0$. So if, finally, we can show that
$$w(\pi) \ge 0,$$
then we are done. To do that, I would think it suffices e.g. that up to some element $u \in \mathcal{O}^\times$ (for which we already know that $w(u)=0$), some power of $\pi$ is $p \in \mathbb Z$, and $w(p) \ge 0$ because $w$ is a non-archimedean valuation. [Update: A much easier argument I saw in this answer by user reuns is: $w(\pi) < 0$ would, via the ultrametric maximum principle, contradict the already established $w(1+\pi)=0$.]
As for why I think this is not true without the discreteness assumption: The axiom of choice notoriously allows for field isomorphisms $\mathbb C_p \simeq \mathbb C_q$ for all primes $p,q$. Since your $F$ is contained in $\mathbb C_p$, via such an isomorphism we can pull back the canonical ("$q$-adic") non-archimedean valuation of $\mathbb C_q$ (which no longer has a discrete value group) to $\mathbb C_p$, and then restrict it to $F$, call this $w$. Now if the above statement still held true, it would follow that $\frac{1}{q} \in \mathcal{O} \subseteq \mathcal{O}_w$; however, that isomorphism, like any field isomorphism, fixes $\mathbb Q$, and of course in the $q$-adic valuation $v_q(\frac{1}{q})=w(\frac{1}{q}) <0$. So what the proof above shows is that such a restriction of a $q$-adic value to some local field inside $\mathbb C_p$, besides not being explicit, can never be a discrete valuation -- interesting!
One of my favourite classical results using $p$-adic methods in elementary number theory is the theorem of Skolem-Mahler-Lech:
This is a theorem about linear recurrence sequences, which are sequences of integers where each term is a fixed linear combination of $n$ previous ones. So fixing $n$ the sequence $s_i$ is defined by choosing the first $n$ terms $$s_0,\ldots, s_{n-1}\in \mathbf Z$$ and a relation for all $k$ $$s_{k + n} = \sum_{i=0}^{n-1} a_i s_{k+i}$$ for fixed $a_i$.
Some examples are the Fibonacci sequence ($n = 2$,$s_0 = 0, s_1 = 1$, $a_0=a_1= 1$), and simpler things like any eventually periodic sequence, or the sequence $s_k = k$ (here $n=2$, $s_0 = 0, s_1=1$, $a_0 = -1, a_1= 2$). We can make other such sequences easily by noting that that the sum of any two linear recurrence sequences is also a linear recurrence sequence.
An important fact about such sequences are that their generating functions $$f_s = \sum_{k= 0}^\infty s_k x^k$$ are always rational functions of the variable $x$ (one polynomial divided by another), where the numerator defines the initial terms $s_0, \ldots, s_{n-1}$ and the denominator defines the recurrence relation.
Of the examples I mentioned above, the fibonacci sequence grows (exponentially), eventually periodic sequences are bounded, and the sequence $s_k=k$ also grows, just less quickly than fibonacci.
One question that one might then ask is:
What is the set of $k$ for which $s_k = 0$?
from these examples (and others) we might conjecture that this set is periodic, except for finitely many exceptions (after all we can always change finitely many terms of any linear recurrence sequence to make a sequence with the same behaviour eventually but with zeroes wherever we want at the start).
How might one go about proving this? The first step of the proof is to use the rational generating function $f_s$ and write out its partial fraction decomposition over an algebraically closed field (like $\overline {\mathbf Q}$), this will be of the form
$$f_s = \sum_{i=1}^{\ell} \sum_{j = 1}^{r_i} \frac{\alpha_{ij}}{(x - \beta_{i})^j} $$
for some fixed roots $\beta_j$ of the original denominator of $f_s$.
Now using this decomposition we have $$f_s = \sum_{i=1}^{\ell} \sum_{j = 1}^{r_i} \alpha_{ij}{\left(\sum_{n=0}^\infty \beta_i^n x^n\right)^j} $$
what this gives is that $$s_n = \text{some polynomial expression involving terms }\beta^n $$
For example for the fibonacci sequence this recovers Binet's formula $$s_n = \frac{1}{\sqrt 5} \left(\frac{1+ \sqrt 5}2\right)^n-\frac{1}{\sqrt 5} \left(\frac{1- \sqrt 5}2\right)^n.$$ Or for the periodic sequence $0,1,0,1,0,1,\ldots$ this is $$ s_n = 1^ n - (-1)^n$$
So we've written $s_n$ as a sum of exponential-type functions in $n$ with different bases, that we want to describe the zeroes of this function for $n \in \mathbf N$.
Now the magic part: the function $e^x$ is an analytic function, and on a bounded domain analytic functions have only finitely many zeroes (unless they are zero everywhere). This would give us a lot of control over the zeroes of $s_n$ if the naturals were bounded. Which leads to the slightly strange question:
What if the natural numbers were bounded? And the functions $\beta^n$ were still analytic in some way?
Of course using the usual absolute value and metric on $\mathbf Q$ and $\mathbf C$ this is totally false.
But in the $p$-adic numbers this is true! The integers are all bounded ($p$-adically) by norm $\le 1$. So lets treat these functions as $p$-adic functions and control the zero sets in some way.
How does this prove the result? The functions $\beta^n$ are not $p$-adic analytic functions of $n$ on their own, but they are so on small enough $p$-adic disks though, what ends up happening is we get distinction between congruence classes of $n$ mod $p-1$ for some well-chosen $p$ such that in each congruence class there are only finitely many zeroes of $s_n$ or the function $s_n$ is identically zero on that congruence class. This gives us the theorem mentioned above, that the zeroes of $s_n$ are periodic, except for finitely many exceptions.
Best Answer
Suppose $v_p(a^2+b^2)=2k+1$ is odd, so $a^2+b^2\equiv 0\pmod{p^{2k+1}}$ but $a^2+b^2\not\equiv 0\pmod{p^{2k+2}}$. Now pick integers $x,y$ with $x\equiv a\pmod{p^{2k+2}}$ and $y\equiv b\pmod{p^{2k+2}}$. Now, we have $x^2+y^2\equiv 0\pmod{p^{2k+1}}$ and $x^2+y^2\not\equiv 0\pmod{p^{2k+2}}$, so $v_p(x^2+y^2)=2k+1$. This is a contradiction.