A trick I have seen several times: If you want to show that some rational number is an integer (i. e., a divisibility), show that it is an algebraic integer. Technically, it is then an application of commutative algebra (the integral closedness of $\mathbb Z$, together with the properties of integral closure such as: the sum of two algebraic integers is an algebraic integer again), but since you define algebraic number theory as the theory of algebraic numbers, you may be interested in this kind of applications.
Example: Let $p$ be a prime such that $p\neq 2$. Prove that the $p$-th Fibonacci number $F_p$ satisfies $F_p\equiv 5^{\left(p-1\right)/2}\mod p$.
Proof: We can do the $p=5$ case by hand, so let us assume that $p\neq 5$ for now. Then, $p$ is coprime to $5$ in $\mathbb Z$. Let $a=\frac{1+\sqrt5}{2}$ and $b=\frac{1-\sqrt5}{2}$. The Binet formula yields $F_p=\displaystyle\frac{a^p-b^p}{\sqrt5}$. Now, $a^p-b^p\equiv\left(a-b\right)^p\mod p\mathbb Z\left[a,b\right]$ (by the idiot's binomial formula, since $p$ is an odd prime). Note that $p$ is coprime to $5$ in the ring $p\mathbb Z\left[a,b\right]$ (since $p$ is coprime to $5$ in the ring $\mathbb Z$, and thus there exist integers $a$ and $b$ such that $pa+5b=1$). Now,
$\displaystyle F_p=\frac{a^p-b^p}{\sqrt5}\equiv\frac{\left(a-b\right)^p}{\sqrt5}$ (since $a^p-b^p\equiv\left(a-b\right)^p\mod p\mathbb Z\left[a,b\right]$ and since we can divide congruences modulo $p\mathbb Z\left[a,b\right]$ by $\sqrt5$, because $p$ is coprime to $5$ in $p\mathbb Z\left[a,b\right]$)
$\displaystyle =\frac{\left(\sqrt5\right)^p}{\sqrt5}$ (since $a-b=\sqrt5$)
$=5^{\left(p-1\right)/2}\mod p\mathbb Z\left[a,b\right]$.
In other words, the number $F_p-5^{\left(p-1\right)/2}$ is divisible by $p$ in the ring $\mathbb Z\left[a,b\right]$. Hence, $\frac{F_p-5^{\left(p-1\right)/2}}{p}$ is an algebraic integer. But it is also a rational number. Thus, it is an integer, so that $p\mid F_p-5^{\left(p-1\right)/2}$ and thus $F_p\equiv 5^{\left(p-1\right)/2}\mod p$, qed.
The Gödel Speedup Theorem provides some explanation why real numbers (and variants) are useful in proving statements in number theory.
Real numbers, complex numbers, and $p$-adic numbers are second-order objects over the natural numbers. Thus a proof of a number theoretic fact using such analytical devices is formally a proof of that fact in second-order arithmetic. The Gödel Speedup Theorem shows that there is a definite advantage to using second-order arithmetic to prove elementary number theoretic facts.
Gödel Speedup Theorem. Let $h$ be any computable function. There is an infinite family $\mathcal{H}$ of first-order (indeed $\Pi^0_2$) statements such that if $\phi \in \mathcal{H}$, then $\phi$ is provable in first-order arithmetic and if $k$ is the length of the shortest proof of $\phi$ in second-order arithmetic, then the shortest proof of $\phi$ in first-order arithmetic has length at least $h(k)$.
Since computable functions can grow very fast, this shows that there are true number theoretic facts that one can prove using second-order methods (e.g. complex analysis, $p$-adic numbers, etc.) but any first-order (a.k.a. elementary) proof is unfathomably long. Admittedly, the statements produced by Gödel to verify the theorem are very unnatural from a number theoretic point of view. However, it is a general fact that second-order proofs can be much much shorter and easier to understand than first-order proofs.
Addendum. This excellent post by Emil Jeřábek demonstrates another speedup theorem, which is in many ways more striking. The method of going from a first-order $T$ to a second-order $T^+$ is conservative, meaning that $T^+$ cannot prove more first-order theorems than $T$. However, the mere act of allowing sets to replace formulas and introducing the possibility of quantifying over such sets introduces speedups faster than any exponential tower. Introducing $\mathbb{R}$, $\mathbb{C}$, $\mathbb{Q}_p$ and so forth has a similar effect where one can package complicated ideas into conceptually simpler ones (e.g. replacing $\forall\exists$ statements by the higher-level idea of continuity) can lead to monumentally shorter proofs!
Best Answer
The Euler factors in the $L$-series of an elliptic curve at non-singular primes can be defined as integrals over the $p$-adic points of $E$. When one does the analogous integral over $E(\mathbb{Q}_p)$ for singular primes, then one gets the number of components, which is $\#E(\mathbb{Q}_p)/E_0(\mathbb{Q}_p)$, multiplied by the integral over the identity component $E_0(\mathbb{Q}_p)$. The integral over the identity component gives the Euler factor at the singular prime, i.e., one of $1$, $(1-p^{-s})^{-1}$, or $(1+p^{-s})^{-1}$. So there's this extra factor given by the number of components. Of course, the number of components is the Tamagawa number in this setting. This is all explained in detail in Tate's article in Antwerp IV (Springer Lecture Notes in Mathematics 476), which is where I learned about it. I expect this will be easier to read than the Bloch-Kato article.
BTW, in the original formulation of Birch and Swinnerton-Dyer, they didn't know where the extra factors came from, so they just said that they were small integer "fudge factors". I believe it was Tate who indicated why they should be Tamagawa numbers.