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.
This is basically the same as roy smith's excellent comment, but I'd like to put a slightly different spin on it.
A normal variety is a variety that has no undue gluing of subvarieties or tangent spaces.
Let me explain what I mean by gluing.
Given a variety $X$, a closed sub-scheme $Y \subseteq X$ and a finite (even surjective) map $Y \to Z$, you can glue $X$ and $Z$ along $Y$ (identifying points and tangent information). This is the pushout of the diagram $X \leftarrow Y \rightarrow Z$.
You might not always get a scheme (although you do in the affine case) but you always get an algebraic space. In the affine case, this just corresponds to the pullback in the category of rings.
Example 1: $X = \mathbb{A}^1$ glued to $Z = \bullet$ (one point) along $Y = \bullet, \bullet$ (two points) is a nodal curve.
Example 2: $X = \mathbb{A}^1$ glued to $Z = \bullet$ (one point) along $Y = \star = \text{Spec } k[x]/x^2$ a fuzzy point gives you a cuspidal curve.
Example 3: $X = \mathbb{A}^2$ glued to $Z = \mathbb{A}^1$ along one of the axes $Y = \mathbb{A}^1$ via the map $Y \to Z$ corresponding to $k[t^2] \subseteq k[t]$ gives you the pinch point / Whitney's umbrella = $\text{Spec } k[x^2, xy, y]$.
If I recall correctly, all non-normal varieties $W$ come about this way for some appropriate choice of normal $X$ (the normalization of $W$) and $Y$ and $Z$ (NOT UNIQUE). Roughly speaking, if you are given $W$ and want to construct $X, Y, Z$, do the following: Let $X$ be the normalization, let $Z$ be some sufficiently deep thickening of the non-normal locus of $X$ and let $Y$ be some appropriate pre-image scheme of $Z$ in $X$.
Edit: There is a proof available now HERE
Assuming this is true, you can see that all non-normal things are non-normal because they either have some points identified (as in 1 or 3) or some tangent space information killed / collapsed (as in example 2), or some combination of the two.
Best Answer
Indefinite binary quadratic forms, integer coefficients and discriminant not a square, possess an automorphism group; taking the Hessian matrix $H,$ an automorphism element is an integer matrix $P$ such that $P^T H P = H.$ The oriented part ($P$ has positive determinant) is infinite cyclic. For the form $A x^2 + B xy + C y^2,$ with discriminant $\Delta= B^2 - 4 AC$ positive but not a square, Hessian $$ H = \left( \begin{array}{cc} 2A & B \\ B & 2 C \end{array} \right), $$ all positive determinant $P$ are given by integer solutions to $\tau^2 - \Delta \sigma^2 = 4,$ with $$ P = \left( \begin{array}{cc} \frac{\tau - B \sigma}{2} & - C \sigma \\ A \sigma & \frac{\tau + B \sigma}{2} \end{array} \right) $$
For the case of $x^2 + Bxy+y^2$ with $B^2 > 4,$ the automorphisms take on the familiar Vieta appearance, and the negative determinant ones can be taken to be interchanging the variables. With a fixed target $T,$ any expression $x^2 + B xy + y^2 = T$ can be transported, by automorphisms, to a region satisfying desired inequalities. These desired solutions can be thought of as representative points in a group orbit. Siegel's description of counting solutions comes down to counting the representative solutions, that is the number of orbits, as the number of literal representations of a fixed target number is infinite.
Probably worth pointing out that finding a generator for the (oriented part) of the automorphism group requires solving $\tau^2 - \Delta \sigma^2 = 4,$ where $\Delta > 0$ is not a square. This would be a bit much. However, for the special case $x^2 + B xy + y^2,$ the discriminant is $\Delta = B^2 - 4,$ so that $B^2 - \Delta 1^2 = 4.$ There is also the single-variable "jumping" argument, done extremely well in Hurwitz 1907
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
LEMMA 1
Given integers $$ M \geq m > 0, $$ along with positive integers $x,y$ with $$ x^2 - Mxy + y^2 = m. $$ Then $m$ is a square.
PROOF.
First note that we cannot have integers $xy < 0$ with $ x^2 - Mxy + y^2 = m, $ since then $ x^2 - Mxy + y^2 \geq 1 + M + 1 = M + 2 > m.$ If we have a solution with $x > 0$ and $xy \leq 0,$ it follows that $y=0.$
This is the Vieta jumping part, with some extra care about inequalities. We begin with $$ y > x $$ and $$ y < Mx. $$ We have $$ x^2 - Mxy + y^2 > 0, $$ $$ x^2 > Mxy - y^2 = y(Mx - y) > x(Mx-y), $$ $$ x > Mx - y > 0. $$ That is, the jump $$ (x,y) \mapsto (Mx - y,x) $$ takes us from one ordered solution to another ordered solution while strictly decreasing $x+y.$ Within a finite number of such jumps we violate the conditions we were preserving; we reach a solution $(x,y)$ with $y \geq Mx,$ that is $x > 0$ but $Mx-y \leq 0.$ Since $(Mx - y,x) $ is another solution we know that $Mx-y = 0.$ Therefore $x^2 = m$ and $m$ is a square.
......
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
This one takes a little more work.
LEMMA 2
Given integers $$ m > 0, \; \; M \geq m+3, $$ there are no integers $x,y$ with $$ x^2 - Mxy + y^2 = -m. $$
The contrapositive of lemma 2 is that when there are solutions, $M \leq m+2.$ The bound is sharp, achieved at $x=y=1 \; .$ As a side note, if $M \leq 2,$ then $x^2 - M xy + y^2$ is positive or positive semi-definite. So, the contrapositive gives the other example at the wikipedia article, that if $xy$ divides $x^2 + y^2 + 1$ and $x,y >0,$ then actually $x^2 + y^2 + 1 = 3xy.$
Compare the contrapositive with Lemma 3 below...
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
This one took a whole bunch more work. I also needed some help from Gerry Myerson.
LEMMA 3
Given integers $$ m > 0, \; \; M > 0, $$ such that there are integers $x,y$ with $$ x^2 - Mxy + y^2 = -Mm, $$ then $$ M \leq (m+1)^2 + 1 \; . $$ The bound is sharp, achieved with $x=1, \; y=m+1 \; .$