# A question for students you dislike

At the recent CNTA conference, Professor Joe Silverman gave an explicit homomorphism of $\text{GL}_3(\mathbb{R})$ into $\text{GL}_6(\mathbb{R})$ which he jokingly called a great question to ask undergraduates to work out explicitly… if you don’t like them very much. In a similar vein, here is a question that one might ask undergraduates they don’t particularly like:

Let $\mathbf{a} = (\alpha, \beta, \gamma)$ be a triple of co-prime integers which are not all zero and such that $\alpha \gamma > 1$. Prove that the ternary quadratic form given by:

$K_{\mathbf{a}} (F) = \dfrac{1}{8\alpha^3} \bigg( 72 \beta^2 \gamma A^2 + 9 \alpha(\beta^2 + 4 \alpha \gamma) B^2 + 8 \alpha^3 C^2 - 18\beta (\beta^2 + 4 \alpha \gamma)AB$
$+ 12 \alpha (3 \beta^2 - 4 \alpha \gamma)AC - 24 \alpha^2 \beta BC \bigg)$

takes on integer values whenever the triplet $(A,B,C)$ lies in the lattice defined by the congruence conditions

$4 \beta \gamma x - (2 \beta^2 + \alpha \gamma) y + 2 \alpha \beta z \equiv 0 \pmod{\alpha^2}$

and

$\gamma(2 \beta^2 + \alpha \gamma)x - \beta(\beta^2 + \alpha \gamma)y + \alpha \beta^2 z \equiv 0 \pmod{\alpha^3}.$

I will reveal the solution in due time (and it does not involve explicit computation of congruences), but if come up with a solution let me know!

Last night, I had the honour of attending a panel discussion featuring eminent mathematician Manjul Bhargava. During the panel, the moderator, Professor Kumar Murty asked the very productive Fields Medalist to give some advice to graduate students in the audience who may be struggling with their research. Professor Bhargava’s response, paraphrased, is essentially the following:  always work on several problems at a time (at least three), of varying difficulty. There should be a problem which is quite difficult and if you make any progress on it it will be a major breakthrough; there should be one of moderate difficulty, and there should be an ‘easy’ problem that you know you can make progress on eventually. Further, never think too much about a problem at a time and instead rotate between the problems to change your mindset. Sometimes when you approach a problem with a fresh perspective you will gain some insight that would’ve been impossible if you stared at the same problem continuously, since you are subconsciously trying to apply the same techniques.

I thought Professor Bhargava’s advice was very helpful to the graduate students in the audience. It is something I started doing a few years ago, but it wasn’t something that I was aware of consciously. Hopefully heeding this advice will be helpful to your work.

# Pure Mathematics Analysis Exam – 2016 edition

Hi all,

I have included the solutions to the 2016 analysis exam here. Enjoy!

# Elizabeth Keen is not dead

More so than that, it appears that my observations watching the episode where ‘died’ were basically right on the money. Feels good to be right!

# On binary forms II

The following joint paper of myself and my advisor Professor C.L. Stewart has been released on the arxiv. In this follow-up post, I would like to describe in some detail how to establish an asymptotic formula for the number of integers in an interval which are representable by a fixed binary form $F$ with integer coefficients and non-zero discriminant.

There are essentially three ingredients which go into the proof, each established decades apart. The first essential piece of the puzzle was established by Kurt Mahler in the 1930’s. He showed that if we examine the number of integer points in the region $\{(x,y) \in \mathbb{R}^2 : |F(x,y)| \leq Z\}$, then the number of such points is closely approximated by the area of the region. Since the region is homogeneously expanding, the area itself is well-approximated by scaling the fundamental region’ given by $\{(x,y) \in \mathbb{R}^2 : |F(x,y)| \leq 1\}$. Indeed, let $A_F$ denote the area of this fundamental region and let $N_F(Z)$ denote the number of integer pairs $(x,y) \in \mathbb{Z}^2$ such that $|F(x,y)| \leq Z$. Then Mahler proved that

$\displaystyle N_F(Z) \sim A_F Z^{\frac{2}{d}}.$

More precisely, he proved a very good error term. He showed that when $d \geq 3$, we have

$\displaystyle N_F(Z) = A_F Z^{\frac{2}{d}} + O_F\left(Z^{\frac{1}{d-1}} \right).$

The question then becomes is there some way to remove the redundancies in Mahler’s theorem? For example, if $F$ has even degree, then $F(x,y) = F(-x,-y)$ for all $(x,y) \in \mathbb{R}^2$, so the pairs $(x,y), (-x,-y) \in \mathbb{Z}^2$ represent the same integer. Is it true that this is the only way that this can happen? Unfortunately, the answer is no. For example, consider the binary form $F_n = x^n + (x - y)(2x - y) \cdots (nx-y)$. Then clearly the points $(1,1), \cdots, (1,n)$ all represent 1, and this construction works for any positive integer $n$. Therefore, there does not appear to be a simple way to count the multiplicities of points representing the same integer in Mahler’s theorem.

While examples like the above exist, perhaps it is possible that this happen sufficiently rare as to be negligible. For instance, if only $O\left(Z^{2/d - \delta}\right)$ many points counted by $N_F(Z)$ are such that there exist many essentially different’ (precise definition to come) other points which represent the same integer, and even in the worst case there can be at most say $O\left(Z^{\frac{\delta}{2}}\right)$ many essentially different pairs, then we have shown that in total, the contribution from these bad points to $N_F(Z)$ is only $O\left(Z^{\frac{2}{d} - \frac{\delta}{2}}\right)$, which is fine.

We shall now make some definitions. We say that an integer $h$ is essentially represented by $F$ if there exist two integer pairs $(x_1, y_1), (x_2, y_2)$ for which $F(x_1, y_1) = F(x_2, y_2) = h$, then there exists an element

$\displaystyle T = \begin{pmatrix} t_1 & t_2 \\ t_3 & t_4 \end{pmatrix} \in \text{GL}_2(\mathbb{Q})$ such that

$\begin{pmatrix} x_1 \\ y_1 \end{pmatrix} = T \begin{pmatrix} x_2 \\ y_2 \end{pmatrix}$

and such that

$F_T(x,y) = F(t_1 x + t_2 y, t_3 x + t_4 y) = F(x,y)$

for all $(x,y) \in \mathbb{C}^2$. Otherwise, we say that $h$ is not essentially represented.

Now put $R_F(Z)$ to be the number of integers up to $Z$ which are representable by $F$, and let $R_F^{(1)}(Z)$ be the number of essentially represented integers and $R_F^{(2)}(Z)$ be the number of non-essentially represented integers. If we can show that $R_F(Z) \sim R_F^{(1)}(Z)$, then we are basically done. This amounts to showing that $R_F^{(2)}(Z)$ is small compared to $Z^{\frac{2}{d}}$.

Christopher Hooley proved this for both the ‘easy cubic case’ and the ‘hard cubic case’. However, it was D.R. Heath-Brown who showed that $R_F^{(2)}(Z)$ is always small compared to $R_F^{(1)}(Z)$. This paved the way to our eventual success at this problem.

It remains to account for the interaction between those $T \in \text{GL}_2(\mathbb{Q})$  which fix $F$ and $R_F^{(1)}(Z)$. These elements are called the rational automorphisms of $F$ and we denote them by $\text{Aut} F = \text{Aut}_\mathbb{Q} F$. The most novel contribution we made to this topic is that we accounted for the exact interaction between $\text{Aut} F$ and $R_F^{(1)}(Z)$ with the so-called ‘redundancy lemmas’. This will be discussed at a future time.

# Why Elizabeth Keen is (obviously) not dead

I just watched the latest episode of The Blacklist. MAJOR SPOILERS ARE AHEAD, THIS IS YOUR WARNING.

So in the latest episode, “Mr. Solomon – Conclusion”, there is a cataclysmic plot twist in which Elizabeth Keen, whose mysterious familial background is the driving force of the show, seemingly perished after a medical complication. The shock value of this development cannot be underestimated… however, I claim that she is (obviously) not dead. Here are the clues:

1) When Keen was assessed by the first doctor at the hospital, her diagnosis was fine. There are no indications at that time that she has suffered any serious trauma.

2) When Mr. Kaplan went to pick her up from the hospital, Keen was talking angrily about how she’s in this situation because of Reddington. Mr. Kaplan looked sympathetic. Most likely they planned what to do right after.

3) Notice that Keen started developing medical complications only after being treated at the nightclub. During this time Mr. Kaplan, Keen, and Nick had time to converse in private; Reddington (and therefore the audience) is not aware of what happened. After this conversation, all of a sudden Keen’s medical condition started deteriorating, and Nick asked for additional medical equipment to be brought. Who took care of this? Mr. Kaplan.

4) Mr. Kaplan very squarely blamed Red for Keen’s predicament. It is not at all surprising if she would secretly plan to fake Keen’s death just so that Reddington could not be in Keen’s life any longer.

5) After Keen has been pronounced dead, Mr. Kaplan, who earlier in the episode showed that she obviously cares for Keen, displayed no distress. Reddington almost fell over and Samar balled her eyes out, even Nick was extremely distraught. So why didn’t Mr. Kaplan react at all? Probably because she knows Liz is not really dead. Moreover, she again murmured something to Nick while Dembe was helping Red into the car. Clearly there is something astray here.

The obvious need to ‘kill’ Liz is because Megan Boone is pregnant for real and probably has to go give birth for real, and take some mat leave on top of that. This is a pragmatic solution to this very gnarly logistical issue. Do not be surprised if Keen pops up all of a sudden on episode 20 or earlier.

# A grade school question from China

My wife showed me a homework problem of a friend’s son in China that he could not solve, and his mother also could not solve. I thought it was quite an interesting problem, so I thought I would share.

I do not have an acceptable diagram (which admittedly would make the problem much easier to digest), so I will be as precise as possible in describing the problem. Suppose that you have a triangle $ABC$ with a right angle at vertex $B$. The length of the side $AC$ is equal to 14. Construct squares $AGFB$ (read counter-clockwise) on side $AB$ and $BEDC$ on side $BC$. Draw the line from vertex $A$ to the vertex $D$ of square $BCDE$, and let $H$ be the intersection of the line segment $AD$ and $BC$. Suppose further that the line segment $GH$  is parallel to $AC$. What is the area of the quadrilateral $BHDE$?

We let $x$ denote the side length of $AB$, $y$ denote the side length of $BC$, so that $x^2 + y^2 = 14^2 = 196$. Put $z$ for the length of $BH$. Since $AC$ and $GH$ are parallel, it follows that $z = y - x$. Therefore the desired area is given by $[AED] - [ABH]$ (the square brackets denote the area of the polygon with the given vertices), or

$\frac{y(x+y)}{2} - \frac{x(y-x)}{2} = \frac{x^2 + y^2}{2} = 98.$

This problem is nice because it requires adroit use of many different geometric facts, and once the proper principles are applied, the solution is beautiful and elegant. Hard to imagine an 11 year old being able to do this regularly though!

Edit: in the original post there was an error where I forgot to divide by 2 in the penultimate step. Silly mistake!