Our plan for today involves the following:
First, I will briefly comment on one thing I forgot last time!
What does quadratic reciprocity do for us?
It helps us solve Mordell equations! Or at least the Legendre symbol does; the easiest cases use $\left(\frac{2}{p}\right)$ and multiplicativity. But more advanced ones no doubt would need to compute more complicated square roots. Here are two examples, without proof; there are many others.
The Proof!
Recall the statement of quadratic reciprocity. For odd primes $p$ and $q$, we have that $$\left(\frac{p}{q}\right)\left(\frac{q}{p}\right)=(1)^{\frac{p1}{2}\frac{q1}{2}}$$ That is to say, the Legendre symbols are the same unless $p$ and $q$ are both of the form $4k+3$.
Proof:
We will use the criterion of Eisenstein's we've used throughout. Recall that $p$ and $q$ are odd primes in the context of this proof. Now let $$R=\sum_{\text{even }e,\; 0<e<p}\left\lfloor\frac{qe}{p}\right\rfloor\; $$ be the exponent in question, so that $$\left(\frac{q}{p}\right)=(1)^R\; .$$
The key will be geometrically interpreting $\left\lfloor\frac{qe}{p}\right\rfloor$ as being the biggest integer less than $\frac{qe}{p}$. The following features are present in the next graphic, which should clarify how we'll think of it geometrically.
It should be clear that each blue stack has the same height as $\left\lfloor\frac{qe}{p}\right\rfloor$ for some even $e$.
Click to the left again to hide and once more to show the dynamic interactive window 
What we will now do is try to convince ourselves that the number of blue points has the same parity as the total number of (positive) points in and on the green box under the dotted line. If we can do that, the following steps finish the proof of quadratic reciprocity.
This would definitely complete the theorem!
So we must show that the number of blue points (points under the line with even $x$coordinate) is the same as the number of positive points in the green box under the line. With Eisenstein, we call this second number $\mu$. In the next graphic, there is a lot going on, but we can clarify each of the pieces.
Click to the left again to hide and once more to show the dynamic interactive window 
Let's look at the two sets of green dots.
In other words, they are symmetric images  simply rotated around the point $$\left(\frac{p}{2},\frac{q}{2}\right)\; .$$ So in order to say that $\mu$ has the same parity as $R$ (which is our goal to finish the proof), we just have to show that either set of green points has the same parity as that of the set of blue points outside the green box.
Which means the parity of the points inside the triangle $(\mu)$ is the same as that of the blue points ($R$), which is what we wanted to prove!
Q.E.D.
Interlude of the student presentation.
Okay, I want to end with just a slight preview of what the final quarter of the course will be about  namely, functions and number theory. We have already run into a few functions, most notably $\phi(n)$.
However, I want to talk about one we've encountered a little more recently, the number of ways to write $n$ as a sum of squares, or $r(n)$. For instance, $r(25)=12$. Why? Because you can write it using the pairs $$(\pm 3,\pm 4),\; \; (\pm 4,\pm 3),\;\; (\pm 5,0)\text{ and }(0,\pm 5)\; .$$ That's a little different; here, we count ALL solutions, positive or negative, and in any particular order possible.
First off:
The function $r$ has a formula. Let's write primes of the form $4k+1$ as $p$, and primes of the form $4k+3$ as $q$. Then:
$$\text{If }n=2^dp_1^{e_1}\cdots p_k^{e_k}q_1^{f_1}\cdots q_\ell^{f_\ell},\text{ then }r(n)=\begin{cases}0 & \text{ if any }f_j\text{ is odd}\\4\prod_{i=1}^k(e_i+1) & \text{ otherwise}\end{cases}$$
Notice that the empty product (no primes of the form $4k+1$) is 1, just like a sum over zero elements is zero.
It turns out that every single proof of this is not very elementary. They all either go into some detail regarding factorization of Gaussian integers, or do some lengthy divisibility and congruence analysis. So we will skip them.
Secondly:
Just like with $\phi(n)$, there are some relations with multiplying. Now that we have a formula, that is nearly immediate. For instance, $$r(3)r(5)=r(15)$$ because both sides are zero! But also $$r(25)r(13)=12\cdot 4=48=r(325)$$
But it doesn't always work. For instance, $$r(25)r(4)=12\cdot 4=48\neq 12=r(100)\; .$$ Naturally, $\phi$ wasn't always multiplicative in this sense; however, here the inputs are relatively prime and it still doesn't work. Is there any way we can say when the outputs would multiply?
Finally:
We can use $r(n)$ to talk about limits with functions. Yes, limits in number theory!
The following graphic has as its basic content the circle with radius $\sqrt{n}$ and blue lattice points representing all pairs $(x,y)$ such that $x^2+y^2\leq n$. There is a little box of area one around each such lattice point; as you might expect, the boxes roughly cover the circle, but certainly not exactly.
Click to the left again to hide and once more to show the dynamic interactive window 
What does this have to do with $r(n)$?
Again, the total number of representations of any integer less than or equal to $n$ can be thought of as the area of unit boxes around each lattice point giving the representation, and this number would be $$\sum_{k=0}^{n}r(k)\, .$$ So the area of the boxes can give us information about $r$.
As we noted, they neither cover nor are covered by the circle in question. However:
Using these two pieces of information, in terms of area covered, $$\pi \left(\sqrt{n}\frac{1}{\sqrt{2}}\right)^2\leq \sum_{k=0}^{n}r(k)\leq \pi \left(\sqrt{n}+\frac{1}{\sqrt{2}}\right)^2\, , $$ which simplifes to $$\pi \frac{n\sqrt{2n}+1/2}{n}\leq \frac{1}{n}\sum_{k=0}^{n}r(k)\leq \pi \frac{n+\sqrt{2n}+1/2}{n}\, ,$$ which yields $$\pi \left(1\frac{\sqrt{2}}{n}+1/2n\right)\leq \frac{1}{n}\sum_{k=0}^{n}r(k)\leq \pi \left(1+\frac{\sqrt{2}}{n}+1/2n\right)\, .$$
We can interpret this as saying the average number of representations of a positive integer as a sum of squares is $\pi$.
WHAT?!
But true.
We will discuss these and many related ideas in the last few weeks.
Homework:

Appendix:
One can also prove quadratic reciprocity using the Gauss Lemma of the appendices. Jones and Jones essentially give the proof on pages 133135; it is similar in style, though less elegant, than Eisenstein's proof. The key thing to remember is that we can change the congruence condition on one variable, $x$, into a latticepoint condition on two variables, $x$ and $y$, and then use techniques of the past for that!
The following interactive graphic shows how this works.
Click to the left again to hide and once more to show the dynamic interactive window 
