Today is all about more issues regarding integers  no primes, no modulo, just integers  and we will continue the point of view of the integer lattice to frame our thoughts.
But first, I just have to be gratuitous; try evaluating this next cell, and seeing whether you can get a GCD which is not 3...
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Do you think this pattern continues? It's an interesting question$\ldots$
Also, I want to kill the suspense on the conductor.
A very neat solution! Proving all this about it is a little tedious with the tools we have at hand, though. Remember, for the one you did, you had to work with all the remainders with respect to four or three, and extending that in general is a mouthful. Soon we'll return to it for an easier proof, I promise!
Okay, now to integer lattice points/integeronly number theory. Let's review what was on the last homework assignment.
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What we are really asking is how many integer lattice points lie between the intercepts. One way to think about this is the distance between solutions. Let's assume that the equation is $ax+by=c$, and $gcd(a,b)=1$. Then from solution $(x_0,y_0)$ we get solution $(x_0+b,y_0a)$, and the distance between any two solutions is $$\sqrt{[(x_0+b)x_0]^2+[(y_0a)+y_0]^2}=\sqrt{a^2+b^2}\, .$$ Our strategy is now to see how many times that distance fits between the intercepts of the line  does that make sense? It doesn't give an exact answer, but should give a good ballpark estimate.
Those intercepts are $\frac{c}{a}$ and $\frac{c}{b}$, respectively (check it out above e.g. with $a=3$, $b=2$, $c=4$). Using the Pythagorean Theorem again, we see that the whole length available is $$\sqrt{\Big(\frac{c}{a}\Big)^2+\Big(\frac{c}{b}\Big)^2}=\frac{c}{ab}\sqrt{a^2+b^2}\, .$$ The ratio of this total length and the length between solutions is thus $\frac{c}{ab}$, which is a very nice pat answer.
There are two problems with it, though!
We can deal with each of these problems. We'll want to introduce the greatest integer function to do so (also called the floor function). Examples should suffice to define it: $$\lfloor 1.5\rfloor=1\; , \lfloor 1\rfloor=1\; ,\lfloor 1.99\rfloor=1\; ,\lfloor 0.99\rfloor=0\; ,\lfloor .01\rfloor=1\; .$$
There are a lot of other interesting questions that one can ask about pure integers, and equations they might satisfy. However, answering many of those questions will prove challenging without additional tools, so we will have to take a detour soon. But one such question is truly ancient, and worth exploring more today.
It is also quite geometric. We just used the Pythagorean Theorem above  but you'll note that we didn't really care whether the hypotenuse was an integer there. Well, when is it? More precisely, when are all three sides of a right triangle integers?
We call a triple of integers $x,y,z$ such that $x^2+y^2=z^2$ a Pythagorean triple, though there isn't necessarily evidence that Pythagoras thought this way about them. However, Euclid certainly did, and so will we. For that matter, we should also think of them as $x,y,z$ that fit on the quadratic curve $x^2+y^2=z^2$, given $z$ ahead of time.
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It seems quite random for which $z$ we get a Pythagorean triple even existing! (We'll return to that question later as well!) First, let's see what triples are possible.
So we can find all primitive Pythagorean triples by finding coprime integers $p$ and $q$ which have opposite parity, and then using this formula! And we get all Pythagorean triples by multiplying. Let's try to find some by hand now.
Of course, you could generate some as well...
3 squared plus 4 squared is 5 squared  True 15 squared plus 8 squared is 17 squared  True 35 squared plus 12 squared is 37 squared  True 63 squared plus 16 squared is 65 squared  True 5 squared plus 12 squared is 13 squared  True 21 squared plus 20 squared is 29 squared  True 45 squared plus 28 squared is 53 squared  True 77 squared plus 36 squared is 85 squared  True 7 squared plus 24 squared is 25 squared  True 55 squared plus 48 squared is 73 squared  True 9 squared plus 40 squared is 41 squared  True 33 squared plus 56 squared is 65 squared  True 65 squared plus 72 squared is 97 squared  True 11 squared plus 60 squared is 61 squared  True 39 squared plus 80 squared is 89 squared  True 13 squared plus 84 squared is 85 squared  True 15 squared plus 112 squared is 113 squared  True 17 squared plus 144 squared is 145 squared  True 3 squared plus 4 squared is 5 squared  True 15 squared plus 8 squared is 17 squared  True 35 squared plus 12 squared is 37 squared  True 63 squared plus 16 squared is 65 squared  True 5 squared plus 12 squared is 13 squared  True 21 squared plus 20 squared is 29 squared  True 45 squared plus 28 squared is 53 squared  True 77 squared plus 36 squared is 85 squared  True 7 squared plus 24 squared is 25 squared  True 55 squared plus 48 squared is 73 squared  True 9 squared plus 40 squared is 41 squared  True 33 squared plus 56 squared is 65 squared  True 65 squared plus 72 squared is 97 squared  True 11 squared plus 60 squared is 61 squared  True 39 squared plus 80 squared is 89 squared  True 13 squared plus 84 squared is 85 squared  True 15 squared plus 112 squared is 113 squared  True 17 squared plus 144 squared is 145 squared  True 
Historically, one of the big questions one could ask about such Pythagorean integer triangles was about its area. For primitive ones, the legs must have opposite parity (do you remember why?), so the areas will be integers. (For ones which are not primitive, the sides are multiples of sides with opposite parity, so they are certainly also going to have an integer area.)
So what integers work? You all know one with area 6, and it should be clear that ones with area 1 and 2 can't work (because the sides would be too small and because $2,1$ doesn't lead to a triple); can you find ones with other areas?
The primitive triple 3 4 5 gives a triangle of area 6 The primitive triple 15 8 17 gives a triangle of area 60 The primitive triple 35 12 37 gives a triangle of area 210 The primitive triple 63 16 65 gives a triangle of area 504 The primitive triple 5 12 13 gives a triangle of area 30 The primitive triple 21 20 29 gives a triangle of area 210 The primitive triple 45 28 53 gives a triangle of area 630 The primitive triple 77 36 85 gives a triangle of area 1386 The primitive triple 7 24 25 gives a triangle of area 84 The primitive triple 55 48 73 gives a triangle of area 1320 The primitive triple 9 40 41 gives a triangle of area 180 The primitive triple 33 56 65 gives a triangle of area 924 The primitive triple 65 72 97 gives a triangle of area 2340 The primitive triple 11 60 61 gives a triangle of area 330 The primitive triple 39 80 89 gives a triangle of area 1560 The primitive triple 13 84 85 gives a triangle of area 546 The primitive triple 15 112 113 gives a triangle of area 840 The primitive triple 17 144 145 gives a triangle of area 1224 The primitive triple 3 4 5 gives a triangle of area 6 The primitive triple 15 8 17 gives a triangle of area 60 The primitive triple 35 12 37 gives a triangle of area 210 The primitive triple 63 16 65 gives a triangle of area 504 The primitive triple 5 12 13 gives a triangle of area 30 The primitive triple 21 20 29 gives a triangle of area 210 The primitive triple 45 28 53 gives a triangle of area 630 The primitive triple 77 36 85 gives a triangle of area 1386 The primitive triple 7 24 25 gives a triangle of area 84 The primitive triple 55 48 73 gives a triangle of area 1320 The primitive triple 9 40 41 gives a triangle of area 180 The primitive triple 33 56 65 gives a triangle of area 924 The primitive triple 65 72 97 gives a triangle of area 2340 The primitive triple 11 60 61 gives a triangle of area 330 The primitive triple 39 80 89 gives a triangle of area 1560 The primitive triple 13 84 85 gives a triangle of area 546 The primitive triple 15 112 113 gives a triangle of area 840 The primitive triple 17 144 145 gives a triangle of area 1224 
It is worth asking why there are no odd numbers in the list so far. In fact, we can prove quite a bit about these things.
Remember, $x$ and $y$ can be written as $x=q^2p^2$ while $y=2pq$, for relatively prime opposite parity $q>p$. Then the area must be $$pq(q^2p^2)=pq(q+p)(qp)\, .$$ So can the area be odd?
To find out more about what areas are possible, we show briefly that all four of the factors above are relatively prime to each other in pairs, at least in a primitive triple.
So one could analyze a number to see if it is possible to write as a product of four relatively prime integers as a starting point. E.g. $30=2\cdot 3\cdot 5\cdot 1$ is the only way to write $30$ as a product of four such numbers (assuming no more than one of those is 1!), and since $q+p$ must be the biggest, we must set $q+p=5$. Quickly one can see that $q=3,p=2$ works with this, so there is such a triangle. What are the sides?
This turns out to be a very deep unsolved problem. This article gives some background on the congruent number problem, which asks the related question of which Pythagorean triangles with rational side lengths give integer areas. This page in particular is interesting from our present point of view.
But we can ask another question, which led Fermat to some of his initial investigations into this theory. Namely, when is the area of a Pythagorean triple triangle a perfect square?
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If you'll notice, we don't see to be getting a lot of these. In fact, none.
What would we need to do to investigate this? Well, remember  we already did a lot of work in showing that the area is $$pq(q^2p^2)=pq(q+p)(qp)\; ,$$ where all four of those quantities are relatively prime. If the area is also a perfect square, then (here's one of those unproved things again!) since they are coprime, they themselves are all perfect squares!
Now we will do something very clever. Something the Greeks used occasionally; something Fermat used for many of his proofs. We are going to take that (hypothetical) triangle, and produce a triangle with strictly smaller sides with the same properties  including integer sides and square area! That means we could apply the same argument to our new triangle, and then the next one... but the WellOrdering property won't allow that. So the original triangle was impossible to begin with.
So let's make that smaller triangle!
And there is a bonus to all this. In the proof, we really showed that there is no pair $p$ and $q$ of (coprime) squares such that $q^2p^2$ is also a perfect square $t^2$; that is what we started with, after all. So, if $p=u^2$ and $q=v^2$ (forget our previous use of $u$ and $v$), we have that $$v^4u^4=t^2$$ is impossible. In your homework, you will use this to prove the famous first case of "Fermat's last theorem": $$x^4+y^4=z^4$$ is not possible for any three positive integers $x,y,z$.
All that remains is your homework. Remember, do it, even though it's not being handed in Monday! It is your practice  and preparation for the next class period.
