Elementary results with p-adic numbers I'm giving a talk for the seminar of the PhD students of my math departement. I actually work on Berkovich spaces and arithmetic geometry but, of course, I cannot really talk about that to an audience that includes probabilists, computer scientists and so on.
I'd rather like to do an introduction to $p$-adic numbers and $p$-adic analysis. I think these kind of things come up to be really cool when you work on it even just for a short time, but I have the ambitious aim to show them something nice and elementary whose statement will be understood by everyone (of course the proof may also be really hard, but there I could give them just its general idea).
In other words the question is: if I had prepared something about Galois theory I would have finished with the application to resolubility of polynomial or compass and straightedge constructions; if it had been something about modular forms, it would have been for sure Fermat's last theorem; with 3-surfaces it would have been Poincaré conjecture and so on. What if it's about p-adic numbers or p-adic analysis?
I thought about results on valuations of roots of polynomials, but it seems to me already too complicated (par ailleurs since I'm introducing valuations at the beginning of the talk, it won't turn out to be an application to something they already knew).
 A: Chris Wuthrich has mentioned that the structure of the group $({\bf Z}/p^n{\bf Z})^\times$ can be easily determined by $p$-adic methods.  These local methods are really indispensable for determining the structure of the group $({\mathfrak o}/{\mathfrak p}^n)^\times$ in general, where ${\mathfrak o}$ is the ring of integers in a number field and ${\mathfrak p}\subset{\mathfrak o}$ is a prime ideal.  See Chapter 15 of Hasse's Number Theory.
As a related application, consider Wilson's theorem ($(p-1)!\equiv-1\pmod p$ for a prime number $p$).  More generally, Gauss (Disquisitiones, $\S$78) determined the product of all elements of $({\bf Z}/a{\bf Z})^\times$ for every $a>0$.  There are now two possibilities, $+1$ and $-1$, and Gauss proves that the product is $-1$ precisely when $a$ is $4$, or
$p^m$, or $2p^m$ for some odd prime $p$ and integer $m>0$.
For an ideal ${\mathfrak a}\subset{\mathfrak o}$ in the ring of integers of a number
field, what is the product of all elements in $({\mathfrak o}/{\mathfrak a})^\times$ ?  There are now four possibilities, and one can say which one occurs when.  This turns out to be a result of Laššák Miroslav (2000) but I had a lot of fun a few years ago giving a simple $p$-adic proof in JTNB  (or arXiv).
Addendum. Laššák's paper is available online now-a-days. 
A: 
In other words the question is: if I had prepared something about Galois theory I would have finished with the application to resolubility of polynomial or compass and straightedge constructions; if it had been something about modular forms, it would have been for sure Fermat's last theorem; with 3-surfaces it would have been Poincaré conjecture and so on. What if it's about p-adic numbers or p-adic analysis?

These results have very different levels of sophistication. I would propose the Weil conjectures as an application : http://en.wikipedia.org/wiki/Weil_conjectures#Statement_of_the_Weil_conjectures
The first statement (rationality of the Zeta function) was originally proved by Dwork using purely $p$-adic methods. It's a beautiful application of $p$-adic analysis and $p$-adic functional analysis. In addition, it should not be too hard to state the theorem, since it's about counting solutions of polynomials in finite fields.
Finally, you can also say that Kedlaya now has given a purely $p$-adic proof of the complete conjecture (previously proved by Deligne using other methods).
A: I'm a bit surprised that nobody mentions the interesting fact that $\log(-1)=0$ in $\mathbb{Q}_2$.
This result could be stated without involving $p$-adic numbers: the partial sums of the series $\sum_{i=1}^\infty 2^i/i$ have $2$-adic valuations tending to infinity.
Same for the dilogarithm $\log_2$: one has $\log_2(-1)=0$ in $\mathbb{Q}_2$.
A: The suggestion of Maurizio Monge about the Hensel lemma is a good one in my opinion ; as an application, I suggest to study which integers have square roots in the various completions of the rationals, and hence see they are not isomorphic.
Mahler's theorem about continuous functions would probably be pretty nice.
Amice's exposition might be a nice basis, too.
A: Here is a beautiful and essentially elementary result using the $p$-adics:  the Skolem-Mahler-Lech theorem.

Theorem.  (Skolem-Mahler-Lech)  Let $(a_i)$ be a sequence defined by an integer linear recurrence.  Then the set of $i$ such that $a_i=0$ is the union of a finite set with finitely many arithmetic progressions.

A quick proof may be found on Terry Tao's blog, here.  Essentially, the $p$-adic step of the proof works by defining a $p$-adic analytic function with infinitely many zeros, and then concluding that this function is identically zero--by the definition of this function, this gives some congruence information about the structure of the zero set of the linear recurrence, as desired.  The proof is quite elementary and beautiful, and I think accessible to people seeing the $p$-adics for the first time.
A: For a mind bending example, there are sequences of rationals that converge both p-adically and in the real sense to rational numbers, but not the same rational number.
A: Here are three more elementary results whose most natural proof involves $p$-adic ideas.  All three can be found in Cassels' book on Local Fields (edit mentioned by Laurent Berger in a comment).
The first one is Witt's proof of the theorem of Clausen and von Staudt in Chapter 1.  It requires nothing more than the definition of the $p$-adic valuations; the idea is that $a\in{\bf Q}$ is in ${\bf Z}$ if $a\in{\bf Z}_p$ for every prime $p$. 
The second one says that the order of every finite subgroup $G\subset{\mathrm GL}_n({\mathbf Q})$  divides
$$
\prod_l l^{\beta(l)}
$$
where $l$ runs over the primes and 
$$\beta(l)=\lfloor n/(l-1)\rfloor+\lfloor n/l(l-1)\rfloor+\lfloor n/l^2(l-1)\rfloor+\cdots
$$
for $l\neq2$ and $\beta(2)=n+2\lfloor n/2\rfloor+\lfloor n/2^2\rfloor+\lfloor n/2^3\rfloor+\cdots$.  See Theorem 2.1 in Chapter 4. 
The third one is a theorem of Selberg which says that every finitely generated subgroup $G\subset{\mathrm GL}_n(k)$, where $k$ is a field of characteristic $0$, contains a normal torsionfree subgroup of finite index.  See Theorem 4.1 in Chapter 5.
Note finally that the (Skolem)-Mahler-Lech theorem is Theorem 5.1 in the same chapter.
Addendum. Since I had already TeXed Witt's proof for my Notes, it is easy to reproduce it here :
Theorem (von Staudt--Clausen, 1840)
Let $k>0$ be an even integer, and let $l$ run through the primes.  Then the
number 
$$
(1)\quad\quad\quad W_k=B_k+\sum_{l-1|k}{1\over l}
$$
is always an integer.  For example, $\displaystyle
W_{12}=B_{12}+{1\over 2}+{1\over3}+{1\over5}+{1\over7}+{1\over 13}=1$.  
[The British analyst Hardy says in his Twelve lectures
  (p. 11) that this theorem was rediscovered by Ramanujan ``at a
  time of his life when he had hardly formed any definite concept of
  proof''.]
Proof (Witt) : The idea is to show that $W_k$ is a $p$-adic integer
for every prime $p$.  More precisely, we show that $B_k+p^{-1}$ (resp. $B_k$)
is a $p$-adic integer if $p-1|k$ (resp. if not).
For an integer $n>0$, let $S_k(n)=0^k+1^k+2^k+\cdots+(n-1)^k$.  Comparing the
coefficients on the two sides of
$$
1+e^T+e^{2T}+\cdots+e^{(n-1)T}={e^{nT}-1\over T}{T\over e^T-1},
$$
we get $\displaystyle S_k(n) =\sum_{m\in[0,k]}{k\choose m}{B_m\over
  k+1-m}n^{k+1-m}$.  To recover $B_k$ from the $S_k(n)$, it is tempting to
take the limit $\displaystyle\lim_{n\to0}S_k(n)/n$, which doesn't make sense
in the archimedean world.  If, however, we make $n$ run through the powers
$p^s$ of a fixed prime $p$, then, $p$-adically, $p^s\to0$ as $s\to+\infty$,
and
$$
(2)\quad\quad\quad\lim_{s\to+\infty}S_k(p^s)/p^s=B_k.
$$
Let us compare $S_k(p^{s+1})/p^{s+1}$ with $S_k(p^s)/p^s$.  Every
$j\in[0,p^{s+1}[$ can be uniquely written as $j=up^s+v$, where $u\in[0,p[$ and
$v\in[0,p^s[$.  Now,
$$
\eqalign{
S_k(p^{s+1})
&=\sum_{j\in[0,p^{s+1}[}j^k=\sum_{u\in[0,p[}\sum_{v\in[0,p^s[}(up^s+v)^k\cr
% &\equiv p\left(\sum_{v\in[0,p^s[}v^k\right)
%  +kp^s\left(\sum_{u\in[0,p[}u\sum_{v\in[0,p^s[}v^{k-1}\right)
% \pmod{p^{2s}}\cr  
&\equiv p\left(\sum_v v^k\right)
 +kp^s\left(\sum_u u\sum_v v^{k-1}\right)\pmod{p^{2s}}\cr 
}$$
by the binomial theorem.  As $\sum_{v}v^k=S_k(p^s)$ and
$2\sum_uu=p(p-1)\equiv0\pmod p$, we get
$$
S_k(p^{s+1})\equiv pS_k(p^s)\pmod{p^{s+1}},
$$
where, for $p=2$, the fact that $k$ is even has been used.  Dividing
throughout by $p^{s+1}$, this can be expressed by saying that
$$
{S_k(p^{s+1})\over p^{s+1}}-{S_k(p^s)\over p^s}\in{\mathbf Z}_{(p)}
$$
is a $p$-adic integer, and therefore
$$
{S_k(p^r)\over p^r}-{S_k(p^s)\over p^s}\in Z_{(p)}
$$
for any two integers $r>0$, $s>0$, since ${\mathbf Z}_{(p)}$ is a subring of $\mathbf Q$.
Fixing $s=1$ and letting $r\to+\infty$, we see that $B_k-S_k(p)/p\in{\mathbf Z}_{(p)}$,
in view of $(2)$.  We need a
Lemma.
$S_k(p)=\sum_{j\in[1,p[}j^k$ is $\equiv-1\pmod p$ if $p-1|k$ and 
$\equiv0\pmod p$ otherwise.
This is clear if $p-1|k$.  Suppose not, and let $g$ be a generator of
$({\mathbf Z}/p{\mathbf Z})^\times$.   We have $g^k-1\not\equiv0$, whereas
$$
(g^k-1)\left(\sum_{j\in[1,p[}j^k\right)
\equiv (g^k-1)\left(\sum_{t\in[0,p-1[}g^{tk}\right)
\equiv g^{(p-1)k}-1\equiv0.
$$
It follows that $B_k+p^{-1}\in{\mathbf Z}_{(p)}$ if $p-1|k$ and $B_k\in{\mathbf Z}_{(p)}$
otherwise.  In either case, the number $W_k$ (1), which can be written as
$$
W_k=\cases{
(B_k+p^{-1})+\sum_{l\neq p}l^{-1}&\hbox{if }p-1|k\cr
(B_k)+\sum_l l^{-1}&\hbox{otherwise},\cr
}$$ 
(where $l$ runs through the primes for which $l-1|k$) turns out to be a
$p$-adic integer for every prime $p$.  Hence $W_k\in{\mathbf Z}$, as claimed.  
A: If you don't mind doing something really elementary, you can prove that the ten-adic integers $\mathbb{Z}_{10}$ is not an integral domain and show it is actually $\mathbb{Z}_{2}\oplus\mathbb{Z}_{5}$. 
A: You could mention Fermat's last theorem: Kummer's proof (in the regular case) uses properties of Bernoulli numbers that are close to the existence of the p-adic zeta function, and Wiles's proof uses p-adic numbers in at least 3 ways.
