# Sergei Yakovenko's blog: on Math and Teaching

## Tuesday, March 22, 2011

### Symmetry: new course for the high school math teachers

Filed under: lecture,Rothschild Course "Symmetry" — Sergei Yakovenko @ 2:53

## What is a symmetry?

There is no “rigorous definition”. Informally, the notion of symmetry appears wherever we have:

• a collection (set) of transformations (actions, operations, …) which can be applied to these objects and transform them again into objects of the same class;
• a collection (set) of objects (points, figures, variables, functions, rules of the game…).

Then usually, but not always, the result of the transformation of an object is a new, different object. However, there may be exceptions: some objects may stay the same after a nontrivial transformation. Then we say that such objects are symmetric and the transformations preserving the object are called its symmetries.

Sometimes the mere understanding of the symmetry can be a key to solving the problem.
Problem. For which values of the parameters $a,b\in\mathbb R$ the system of algebraic equations

$x+y+z=1$
$x^2+y^2+z^2=a$
$x^3+y^3+z^3=b$

Solution. Consider the transformation of cyclic change of variables, $x\mapsto y,\ y\mapsto z,\ z\mapsto x$. This transformation can be applied to each of the three equations and preserves all of them. Therefore if we have a solution $(x_0,y_0,z_0)$, then the triple $(y_0,z_0,x_0)$ is also a solution. If the system has only one solution, then it should stay the same, i.e., $x_0=y_0,\ y_0=z_0,\ y_0=z_0$. In other words, all three components should be equal to the same number $c\in R$. The first equation instantly implies that $c=\frac13$. Substituting this value to the remaining equations, we see that $3\times\frac19=\frac13=b$ and $3\times\frac1{27}=b=\frac19$.

Problem. Can the above system have exactly two solutions for some values of $a,b$?

Problem. Consider the regular polygon with $n$ vertices, inscribed in a circle. Denote by $O$ the center of the circle and $A_1,\dots,A_n$ the vertices (see the picture for $n=5$):

Prove that the vector sum $\overline{OA_1}+\overline{OA_2}+\cdots+\overline{OA_n}=0$.

Solution. Consider the rigid rotation of the plane around the origin $O$ by the angle $360^\circ/n$. Then the circle and the polygon will be preserved, only the labels of the vertices will change as follows: $A_1\mapsto A_2,\ A_2\mapsto A_3,\ \dots,\ A_n\mapsto A_1$. Thus all terms in the above  vector sum will undergo a cyclic permutation, which does not change the result. If we denote this sum by $S$, then the conclusion is that the rotation of $S$ by  $360^\circ/n$ is equal to $S$. Yet the only vector that does not change after a rotation (other than by $360^\circ$) is the zero vector.

Problem. Prove that for $a\in\mathbb N$ integer and $p$ prime, the difference $a^p-a$ is divisible by $p$. (Little Fermat theorem)

Solution. Consider necklaces (מחרוזת) of $p$ beads (חרוזים) each, made in $a$ distinct colors, and let’s count them. We can choose each bead by $a$ distinct ways, so the total number appears to be $a^p$.  However, this is not the right answer, since the necklaces which differ by rotation we counted several times:

A “completely asymmetric” necklace, without any regularity in the pattern, will be counted $p$ times, since we have exactly $p$ rotations. On the other hand, the monochromatic (single-color) necklaces will be counted only once: there exist $a$ of them.

Are there any intermediate symmetric necklaces? If after rotation by $q$ “steps” we obtain the initial necklace, then this would mean that the color pattern is periodic with period $q$. But the whole necklace should consist of an entire number of such “periods”, i.e., $p$ should be divisible by $q$. But by assumption, $p$ is prime! This means, that except for the monochromatic necklaces, there are no “partially symmetric” ones, only completely asymmetric (i.e., such, that any nontrivial rotation produces a different coloring scheme). Now we conclude that

(total count of coloring schemes) $a^p$ = (number of single-color schemes) $a$ + $p\times$(number of non-symmetric colorings, an integer number).

This proves that $a^p-a$ is divisible by $p$.

Problem. Prove the little Fermat theorem by induction in $a$. Hint: consider the binomial coefficient $\binom {p}{k}=\frac{p!}{k!(p-k)!}$ and show that for $p$ prime it is always divisible by $p$ except when $k=0$ or $k=p$.

1. Strictly speaking, your solution of the first problem is incomplete. The symmetry consideration shows only that the only possible values for a and b are the ones you obtained. But you still need to prove that for these values there is indeed only one solution of the equations in question.

Comment by Andrei Zelevinsky — Monday, March 28, 2011 @ 9:56

• Andrei, thanks. Indeed, we have to do that, but it seems to be more technically complicated. The only way I know would be to reduce the system to the standard symmetric polynomials. solve it and show that the corresponding cubic equation has a triple root. Any better ideas?

Comment by Sergei Yakovenko — Tuesday, March 29, 2011 @ 7:44

• Maybe I’ll think of something when I have a moment. Meanwhile you have another typo there: since you made the sum of your unknowns equal to 3, then a = b = 3, as well.

Comment by Andrei Zelevinsky — Tuesday, March 29, 2011 @ 1:06

• If we look for real solutions then the first equation defines a plane in R^3, and (1,1,1) is the closest to the origin point of this plane (the base of the perpendicular). So already the first two equations have only one (real) solution.

Comment by Andrei Zelevinsky — Tuesday, March 29, 2011 @ 1:20

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