# College Math Teaching

## August 5, 2011

### Blogging MathFest, 2011 (Lexington, KY)

Filed under: advanced mathematics, algebraic curves, conference, elementary number theory, number theory — collegemathteaching @ 1:50 am

I started the day by attending three large lectures:
Laura DeMarco, University of Illinois at Chicago who spoke about dynamical systems (that result from complex polynomials; for example if $f: C \rightarrow C$ is a function of the complex plane, one can talk about the orbit of a point $z \in C, z, f(z), f(f(z)) = f^{(2)}(z), f(f(f(z))) = f^{(3)}(z)....$. One can then talk about sets of points $w, w\in C$ and $sup|f^{(n)}| < \infty$ This is called the Filled Julia Set.

Ed Burger of Williams (a graduate school classmate of mine who made good) gave the second; he talked about Fibonacci numbers and their relation to irrational ratios (which can be obtained by continued fractions) and various theorems which say that natural numbers can be written uniquely as specified sums of such gadgets.

Lastly Manjul Bhargava of Princeton (who is already full professor though he is less than half my age; he was an Andrew Wiles student) gave a delightful lecture on algebraic curves.

What I noted: all three of these mathematicians are successful enough to be arrogant (especially the third). They could have blown us all away. Yes, they took the time and care to give presentations that actually taught us something.

Of the three, I was the most intrigued by the last one, so I’ll comment on the mathematics.

You’ve probably heard that a Pythagorean triple is a triple of integers $a, b, c$ such that $a^2 + b^2 = c^2$. For now, we’ll limit ourselves to primitive triples; that is, we’ll assume that $a, b, c$ have no common factor.

You might have heard that any Pythagorean triple is of the form: $a = m^2 - n^2, b = 2mn, c = m^2 + n^2$ for $m, n$ integers. It is true that $a, b, c$ being defined that way leads to a Pythagorean triple, but why do ALL Pythagorean triples come in this form?

One way to see this is to look at an algebraic curve; in this case, the curve corresponding to $x^2 + y^2 = 1$. Why? Start with $a^2 + b^2 = c^2$ and divide both sides by $c^2$ to obtain $((\frac{a}{c})^2 + (\frac{b}{c})^2 = 1$ One then notes that one is now reduced to looking to rational solutions to $x^2 + y^2 = 1$ (a rational solution to this can be put in the $((\frac{a}{c})^2 + (\frac{b}{c})^2 = 1$ form by getting a common denominator).

We now wish to find all rational points (both coordinates rational) on the circle; clearly $(-1,0)$ is one of them.
Easy claim: if $(u, v)$ is such a rational point, then the line from $(-1,0)$ to $(u, v)$ has rational slope.
Not quite as easy claim: if a line running through $(-1, 0)$ has rational slope $s < \infty$ then the line intersects the circle in a rational point.
Verification: such a line has equation $y = s(x+1)$ and intersects the circle in a point whose $x$ value satisfies $x^2 + s^2(x+1)^2-1 = 0$. This is a quadratic that has rational coefficients and root $x = -1$ hence the second root must also be rational. Let’s calculate the second root by doing division: $\frac{(s^2 + 1)x^2 +2s^2x + s^2-1}{x+1} = (s^2+1)x + s^2 - 1$. So the point of intersection has $x = \frac{1-s^2}{s^2 + 1}$ latex and $y = s(\frac{1-s^2}{s^2 + 1} + 1) = \frac{2s}{s^2 + 1}$. Both are rational.

Therefore, there is a one to one correspondence between rational slopes and rational points on the circle and all are of the form $(\frac{1-s^2}{s^2 +1}, \frac{2s}{s^2 + 1})$. Note: we obtain $(-1,0)$ by letting $s$ go to infinity; use L’Hopital’s rule on the first coordinate). So if we have any Pythagorean triple $(a,b,c)$ then $\frac{a}{c} = \frac{1-s^2}{s^2 + 1}, \frac{b}{c} = \frac{2s}{s^2 + 1}.$ But $s$ is rational hence we write $s = \frac{p}{q}$ where $p, q$ are relatively prime integers. Just a bit of easy algebra reveals $\frac{a}{c} = \frac{q^2 -p^2}{p^2 + q^2}, \frac{b}{c} = \frac{2pq}{p^2 + q^2}$ which gives us $a = q^2 - p^2, b =2pq, c = q^2 + p^2$ as required.

The point: the algebraic curve motivated the proof that all Pythagorean triples are of that form.

Note: we can extract even more: if $f(x,y) = 0$ latex is any quadratic rational curve (i. e., $f(x,y) = a_1 x^2 + a_2 x + a_3 + a_4 y^2 + a_5 y + a_6 xy$, all coefficients rational, and $(u, v)$ is any rational point and there is a line through $(u,v)$ of rational slope $s$ which intersects the curve in a second point (the quadratic nature forbids more than 2 points), the second point must also be rational. This follows by obtaining a quadratic in $x$ by substituting $y = s(x - u) + v$ and obtaining a quadratic with rational coefficients that has one rational root.

Of course, it might be the case that there is no rational point to choose for $(u, v)$. In fact, that is the case for $x^2 + y^2 = 3.$

Why? Suppose there is a rational point on this curve $x = \frac{p}{q}, y = \frac{a}{b}$ with both fractions in lowest terms. We obtain $(pb)^2 + (aq)^2 = 3(qb)^2$ Now let’s work Mod 4 (hint from the talk): note that in $mod 4, 2^2 = 0, 3^2 = 1$ therefore the sum of two squares can only be 0, 1 or 2. The right hand side is either 3 or 0; equality means that both sides are zero. This means that $pb, aq$ are both even and therefore $3(qb)^2$ is divisible by 4 therefore either $q$ is even or $b$ is even.
Suppose $b$ is odd. Then $q$ is even and because $pb$ is even, $p$ is even. This contradicts the fact that $p, q$ are relatively prime. If $q$ is odd, then because $aq$ is even, $a$ is even. This contradicts the fact that $a, b$ are relatively prime. So both $q, b$ are even which means that $p, a$ are odd. Write $q = 2^I m, b = 2^J n$ where $m, n$ are odd (possibly 1). Then $(p^2)(2^{2J})n^2 + (a^2)(2^{2I}) m^2 = 2^{2J + 2I}3 m^2n^2$. Now if $J = I$ we obtain $(pn)^2 + (am)^2$ on the left hand side (sum of two odd numbers squared) which must be 2 mod 4. The right hand side is still only 3 or 0; this is impossible. Now if, say, $J \ge I$ then we get $(pn)^2 2^{2(J-I)} + (am)^2 = 2^{2J} 3 (mn)^2$ which means that the odd number $(am)^2$ is the difference of two even numbers. That too is impossible.

Hence $x^2 + y^2 = 3$ contains no rational coordinates; that circle manages to miss that dense set.

The point of all of this is that algebraic curves can yield significant information about number theory.

Photos This is the German Enigma Coding machine (with plug board) at the NSA booth. This is another view of the Enigma Edward Burger; I had some bad timing here. Just prior to the first talk.