## Monday, February 26, 2018

### In praise of the Rational Roots Theorem

First some personal historical background. In my school district, you could do Algebra in middle school but unlike a standard class it only covered linear equations.  When I was in High School after doing  Geometry in 9th grade, I entered a 3 year accelerated math sequence that terminated with AP Calculus BC.  For the first year we did a semester going over quadratics leading up to the derivation of the quadratic formula and a semester of trigonometry.   So for all intents and purposes, I didn't learn anything from the standard Algebra II curriculum. Interestingly, this didn't have any particular consequences and as time went by I learned some of the topics when necessary and required for something else. I remember thinking in College, "I wish I had covered more Linear Algebra/Matrices" but never really "What's Descartes' rule of signs?"

However, over the last few years my affection for two particular tools from there has grown quite a bit: The Rational Roots Theorem and Polynomial division.  First, these are often under attack and dropped (just as in my own experience).  Its not unusual to see people wonder online: what are the real world applications of these or will they ever be used again?  In High School, I might have said you can always graph and use approximation techniques like Newton's Method when these come up. More significantly,  the existence of Wolfram Alpha has made generalized solutions to cubic and quartic equations easily accessible (if not derivable) From my perspective, they are two basic polynomial analysis techniques that offer a gateway to understand higher degree polynomials.  That understanding is valuable in itself but in addition they offer a fairly general technique for a lot of algebraic puzzles that I try out and I find it extremely satisfying to be able to analyze these with  just pencil and paper.

#### Example 1:

$$x^2 - 13 =\sqrt{x + 13}$$

This looks not to hard at first until you square both sides to get rid of the radical and realize its a quartic in disguise:

$x^4 - 26x^2 + 169 = x + 13$ => $x^4 - 26x^2 -x + 156 = 0$

A common strategy at this point is to look for clever factorizations.  But its often really hard to see where to start. In fact, I find these are often easier to derive backwards after you know the roots anyway.

So let's start with a quick graph of the  functions. This could be done by hand but I'll use geogebra here. The left hand side is a parabola with vertex at (0,-13) while the the right hand side is half of the rotated 90 degree version of the same parabola with a vertex at (-13,0).  If you're looking for factorization this symmetry is something that provides an avenue of attack. But for our purposes it also shows us there are only 2 real roots in the quartic and approximately where they lie.

Here's where the Rational Roots test comes in. Since 156  = $2^2\cdot 3 \cdot 13$ It says that if there is rational root its going to be either: $\pm 1, \pm 2, \pm 3, \pm 4\, \pm6, \pm 12, \pm 13, \pm26, \pm 39, \pm 52, \pm 78$ or $\pm156$  That's a bit daunting but looking at the graph or the behavior of the functions indicate we really only need to test smaller values and 4 is probably the most promising.

I just plugged that back into the original problem rather than doing the quartic and indeed -4 works out. (This is a bit of a cheat since not all the quartic solutions are also solutions to the original problem due to sign issue with the radical but if it does work then you're golden.)

At this point we know know x+4 is a factor of the original quartic and we can divide it out to get a simpler cubic equation.

Apply polynomial division $\frac{x^4 - 26x^2 - x + 156 }{x+4} = x^3 -4x^2 - 10x + 39$  to get the remaining cubic part of the equation.

Now once again we can apply the rational roots test but on the much smaller set {1,3,13,39}.  Its clear from the graph that none of these are going to be a solution to the original problem and that again the smaller ones are more likely. So starting at 1, I find that 3 works out  (27 - 36  - 30 + 39 = 0).   That mean x -3 is another factor.  Interestingly you can see why it doesn't work:  9 - 13 = -4 while the square root of 13  + 3 = 4.  So the inverse sign changes have interfered (but if the bottom of the sideways parabola were present that would be an intersection point).

Once again apply polynomial division  $\frac{x^3 -4x^2 - 10x + 39}{x - 3} = x^2 - x +13$   Having factored the quartic down to an approachable quadratic we can now  apply the quadratic formula to find two more solutions:  $\frac{1 \pm \sqrt{53}}{2}$.   Either by testing or looking at the graph we can see $\frac{1 - \sqrt{53}}{2}$ is the second solution while its converse again lies on the intersection of missing bottom half of the sideways parabola.

Extension for another time:  We have 3 and 4 wouldn't it be nice if 5 also showed up (and this is tantalizingly close to the generator function for pythagorean triples in the complex plane)?  Is there a general form to the intersections of this type i.e. a parabola and its rotated counterpart?

#### Example 2:

Find the integer solutions to: $x^3y^3 - 4xy^3 + y^2 + x^2 - 2y - 3 = 0$

This again looks fairly complex and of degree 6 on first glance. But lets try experimenting with values of x and see what falls out:  [I'm going to only consider the x >= 0 for simplicity here but somewhat similar logic applies for the negatives.]

if x = 0 this simplifies to:

$$y^2 - 2y - 3 = 0$$ which has 2 integer roots.

if x  = 1 this simplifies to:
$$-3y^3 + y^2 -2y - 2 = 0$$
We can applies the rational root test and check the possible integers $\pm1, \pm2$ with no hits.

if x = 2 this simplifies to:
$$y^2 -2y + 1 = 0$$ which has one integer solution.

Note the constant term flipped from negative to positive at this point and now something interesting happens

if x = 3 this simplifies to
$$15y^3 + y^2 -2y + 6 = 0$$
The rational roots test now is only going to give positive candidates and the higher degree terms start to dominate making it impossible for this to reach 0 with an integer. I.e. $15y^3 + y^2 > 2y$ for all integers > 1.

Continuing on the same trend continues
if x = 4 this simplifies to
$$48y^3 + y^2 -2y + 13 = 0$$  For the same reason this is even worse  $48y^3 + y^2 > 2y$ for all integers > 1.

So we can infer with this logic that no integer solutions exist above x = 2.

## Wednesday, February 21, 2018

### Aloha

I'm on mid winter break with the kids for the week in Oahu.

In the meantime check out my collected problems which are almost at 30 in total: http://mymathclub.blogspot.com/p/collected-problems-2.html

## Friday, February 16, 2018

### 2/13 Farey Sequences

Today was a special occasion for Math Club. Instead of just me or a vicarious video, we once again had a guest lecture from the UW Applied Math department.

This time, Professor Jayadev Athreya came out to the middle school to give a talk on Farey Sequences. That was fairly propitious, since I had meant to get to this subject during this session: http://mymathclub.blogspot.com/2017/11/1128-egyptian-fractions.html  but quickly realized I didn't have enough time to cover even Egyptian Fractions.  So there was a good thematic fit with some of the other things we've done.

My favorite moment of the day came early on when Jayadev had each of the kids talk about why they came to math club. (I usually do this on the first session too) There were a smattering of "I like competitive math" responses but then we reached a girl who roughly said "I don't know why I came originally but I like it so I keep coming." That's victory in my book!

What's also interesting here is a chance to more closely observe all the kids and another person's teaching style.  Jayadev's basic structure was fairly similar to what I might have done.

• Start with having the kids map out all the reduced form fractions where the denominator was less than or equal to 10 and then arrange them by size from smallest to largest. He then graphed this on a number line as a group.

• Closely investigate the numerators of the fractions (in suitable common denominator form) when comparing them to notice a trend: they always differed by one.

• Do a formula proof that the mediant is always 1 apart from its generators if they are 1 apart.

A lot of this was structured as group work at the tables with discussions after a few minutes where things were consolidated.  I always find these transitions a bit tricky to time so it was useful watching someone else.  I would also probably have done a few of these parts as group work on the whiteboard itself and then gallery walked for the discussions but there was good work done at everyone's seats.

Like last time, I also noticed some unexpected hesitancy with operating on fractions. It took a bit more time to draw out the kids and have them explain how to compare fractions with different denominators. Although create common denominators was mentioned as well as variety of numeracy instincts ("for unit fractions, the fraction become smaller as the denominator increases" or "you can also create common numerators to do informal comparisons").   Again, I felt like this was a useful practice/review for some.  Alternate hypothesis: the kids were more reluctant to volunteer at points which was more social rather than indicating any gaps.  If this is the case, I'd like to work on activities to bring out more questions. One idea I have toyed with in the past is, is selecting one student to be "the skeptic" during any demo and come up with at least one question about the logic. If I do go this way, I'll probably start with having them do this with something I discuss and depending on how it goes try it also during all whiteboard discussions.

Overall I was really pleased. We now have an invitation to visit the Applied Math Center on the UW campus. I have to investigate whether the logistics are workable.

## Thursday, February 8, 2018

### 2/6 Olympiad #3 and AMC 10

I almost cancelled this week's math club due to feeling ill the night before. But in the end I was well enough and the activities were straightforward so I went ahead with the meeting. We started with the candy I had forgotten to bring last week. My wife picked up some red vines for me, the reception of which I was curious to see.  They were all eaten by the end so there may be more licorice in the future.

Participation in the problem of the week was lighter that I would like but I had enough kids to still demo solutions. In particular with this problem: http://www.cemc.uwaterloo.ca/resources/potw/2017-18/English/POTWD-17-NA-16-S.pdf the key is to count the total number of pips in the set
of dominos.

The students demonstrated two different methods which was good. Most approaches end up with a triangular table since when you calculate the combinations you often end up with  n pips | m pips and m pips | n pips which are the same domino.  I'm trying to elicit more questions from the other kids. This time it worked well when I asked "Does anyone have any questions about X's diagram and how X did ...."  I also spent some time modelling asking questions about their strategies and why they had created the triangles to draw this out.

Once done we participated in the third MOEMS olympiad. My feeling is this was the most unbalanced of the set so far. The starter questions were all fairly easy and they gave a hint that unwound most of the complexity of one of them and then it ended with a really interesting
Diophantine fraction equation that was quite a bit more difficult to do.  One followup question I have for myself: is even in cases that simplify is it enough to consider just the factors of the denominator of a sum i.e. if   K1 / A + K2 / B = K3/K4 where all the cases are constant.

Finally I chose another AMC based question for the Problem of the Week: