# Arithmetic & Geometry

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After a month of the unexpected break from mathematics, I will resume the regular weekly blog posts. It’s a kind of relaunch of this blog, and I  will begin with the discussion of an arithmetic problem with a geometric solution.  This is problem 103 from  The USSR Olympiad Problem Book:

Prove that
$\displaystyle{t_1+t_2+\ldots + t_n = \left\lfloor\frac{n}{1}\right\rfloor + \left\lfloor\frac{n}{2}\right\rfloor + \left\lfloor\frac{n}{3}\right\rfloor + \ldots + \left\lfloor\frac{n}{n}\right\rfloor}$
where $t_k$ is the number of divisors of the natural number $k$.

One can solve this problem by using the principle of mathematical induction or using the fact that the number of integers of the sequence $1,2,3,\ldots , n$ which are divisible by any chosen integer $k$ is equal to $\left\lfloor \frac{n}{k}\right\rfloor$. The second approach of counting suggests an elegant geometric solution.

Consider the equilateral hyperbola $\displaystyle{y = \frac{k}{x}}$ , (of which we shall take only the branches in the first quadrant):

We note all the points in the first quadrant which have integer coordinates as the intersection point of the dotted lines. Now, if an integer $x$ is a divisor of the integer $k$, then the point $(x,y)$ is a point on the graph of the hyperbola $xy=k$. Conversely, if the hyperbola $xy=k$ contains an integer point, then the x-coordinate is a divisor of $k$. Hence the number of integers $t_k$ is  equal to the number of the integer points lying on the hyperbola $xy=k$. The number $t_k$ is thus equal to the number of absciassas of integer points lying on the hyperbola $xy=k$. Now, we make use of the fact that the hyperbola $xy=n$ lies “farther out” in the quadrant than do $xy=1, xy=2, xy=3, \ldots, xy=n-1$. The following implication hold:

The sum $t_1+t_2\ldots + t_n$ is equal to the number of integer points lying under or on the hyperbola $xy=n$. Each such point will lie on a hyperbola $xy=k$, where $k\leq n$. The number of integer points with abscissa $k$ located under the hyperbola is equal to the integer part of the length of the segment $\overline{AB}$ [in figure above $k=3$]. That is $\left\lfloor\frac{n}{k}\right\rfloor$, since $\displaystyle{|\overline{AB}|=\frac{n}{k}}$, i.e. ordinate of point $A$ on hyperbola $xy=n$ for abscissa $k$. Thus, we obtain

$\displaystyle{t_1+t_2+\ldots + t_n = \left\lfloor\frac{n}{1}\right\rfloor + \left\lfloor\frac{n}{2}\right\rfloor + \left\lfloor\frac{n}{3}\right\rfloor + \ldots + \left\lfloor\frac{n}{n}\right\rfloor}$

Caution: Excess of anything is harmful, even mathematics.

# Triangles and Rectangles

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Consider the following question by Bill Sands (asked in 1995):

Are there right triangles with integer sides and area, associated with rectangles having the same perimeter and area?

Try to test your intuition. The solution to this problem is NOT so simple. The solution was published by Richard K. Guy in 1995: http://www.jstor.org/stable/2974502

If you have a simpler solution, please write it in a comment below. Even I don’t understand much of the solution.

# Polynomials and Commutativity

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In high school, I came to know about the statement of the fundamental theorem of algebra:

Every polynomial of degree $n$ with integer coefficients have exactly $n$ complex roots (with appropriate multiplicity).

In high school, a polynomial = a polynomial in one variable. Then last year I learned 3 different proofs of the following statement of the fundamental theorem of algebra [involving, topology, complex analysis and Galois theory]:

Every non-zero, single-variable, degree $n$ polynomial with complex coefficients has, counted with multiplicity, exactly $n$ complex roots.

A more general statement about the number of roots of a polynomial in one variable is the Factor Theorem:

Let $R$ be a commutative ring with identity and let $p(x)\in R[x]$ be a polynomial with coefficients in $R$. The element $a\in R$ is a root of $p(x)$ if and only if $(x-a)$ divides $p(x)$.

A corollary of above theorem is that:

A polynomial $f$ of degree $n$ over a field $F$ has at most $n$ roots in $F$.

(In case you know undergraduate level algebra, recall that $R[x]$ is a Principal Ideal Domain if and only if $R$ is a field.)

The key fact that many times go unnoticed regarding the number of roots of a given polynomial (in one variable) is that the coefficients/solutions belong to a commutative ring (and $\mathbb{C}$ is a field hence a commutative ring). The key step in the proof of all above theorems is the fact that the division algorithm holds only in some special commutative rings (like fields). I would like to illustrate my point with the following fact:

The equation $X^2 + X + 1$ has only 2 complex roots, namely $\omega = \frac{-1+i\sqrt{3}}{2}$ and $\omega^2 = \frac{-1-i\sqrt{3}}{2}$. But if we want solutions over 2×2 matrices (non-commutative set) then we have at least  3 solutions (consider 1 as 2×2 identity matrix and 0 as the 2×2 zero matrix.)

$\displaystyle{A=\begin{bmatrix} 0 & -1 \\1 & -1 \end{bmatrix}, B=\begin{bmatrix} \omega & 0 \\0 & \omega^2 \end{bmatrix}, C=\begin{bmatrix} \omega^2 & 0 \\0 & \omega \end{bmatrix}}$

if we allow complex entries. This phenominona can also be illusttrated using a non-commutative number system, like quaternions. For more details refer to this Math.SE discussion.

# Counting Cycles

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During one of my reading projects in 2015, I read about the Enigma cipher machine. While reading about it, I came to know that the number of possible keys of this machine is 7, 156, 755, 732, 750, 624, 000. One can see the counting procedure at pp. 22 of this document. But the counting procedure was not found to be satisfactory by most members of the audience (during my presentation). My failure to convince the audience that the counting procedure was correct, lead to my distrust in the counting arguments in general. Many times, I still, find the counting procedures controversial.

So, in an attempt to regain the trust, I will present two counting procedures for counting the number of cycles of length $r$ when $n$ objects (colours/beads/numbers) are given.

Procedure A: Using multiplication principle
Step 1: Choose $r$ objects from the $n$ choices.
Step 2: Arrange the selected $r$ objects in a cyclic order.

1. Since the Step 1 and Step 2 are independent of each other but should be performed together, we will multiply the results (i.e. use the multiplication principle). From Step 1 we will get $\binom{n}{r}$ and from Step 2, we will get $(r-1)!$ as per the circular permutation formula. Hence we get:

$\displaystyle{\# r-\text{cycles from } n \text{ objects} =\binom{n}{r}\times(r-1)! = \frac{n!}{r (n-r)!}}$

Procedure B: Using division principle
Step 1: Permute $r$ of the $n$ objects.
Step 2: Realise the mistake that you counted the permutations $r$ extra times because these circular permutations of objects are equivalent since the circle can be rotated.

Since in Step 2 we want to correct the overcounting mistake of Step 1 performed for different objects simultaneously, we will divide the result of Step 1 by the result of Step 2. From Step 1 we will get $^n P_r$ and from Step 2 we will get $r$. Hence we get:

$\displaystyle{\# r-\text{cycles from } n \text{ objects} =\frac{^n P_r}{r} = \frac{n!}{r (n-r)!}}$

I am still not happy with the Procedure B, so if you have a better way of stating it please let me know.

# When the intuition is correct

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Before starting college I read Paul Zeitz’s The Art and Craft of Problem Solving (a must read book along with the books by Arthur Engel and Terence Tao) and after the first example to illustrate Psychological Strategies he writes (pp. 15):

Just because a problem seems impossible does not mean that it is impossible. Never admit defeat after a cursory glance. Begin optimistically; assume that the problem can be solved. Only after several failed attempts should try to prove impossibility. If you cannot do so, then do not admit defeat. Go back to the problem later.

And today I will share a problem posed by August Ferdinand Möbius around 1840:

Problem of the Five Princes:
There was a king in India who had a large kingdom and five sons. In his last will, the king said that after his death the sons should divide the kingdom among themselves in such a way that the region belonging to each son should have a borderline (not just a point) in common with the remaining four regions. How should the kingdom be divided?

The hint is in the title of this blog post. The solution is easy, hence I won’t discuss it here. The reader is invited to write the solution as a comment to this post.

# Ulam Spiral

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Some of you may know what Ulam’s spiral is (I am not describing what it is because the present Wikipedia entry is awesome, though I mentioned it earlier also). When I first read about it, I thought that it is just a coincidence and is a useless observation. But a few days ago while reading an article by Yuri Matiyasevich, I came to know about the importance of this observation. (Though just now I realised that Wikipedia article describes is clearly, so in this post I just want to re-write that idea.)

It’s an open problem in number theory to find a non-linear, non-constant polynomial which can take prime values infinitely many times. There are some conjectures about the conditions to be satisfied by such polynomials but very little progress has been made in this direction. This is a place where Ulam’s spiral raises some hope. In Ulam spiral, the prime numbers tend to create longish chain formations along the diagonals. And the numbers on some diagonals represent the values of some quadratic polynomial with integer coefficients.

Ulam spiral consists of the numbers between 1 and 400, in a square spiral. All the prime numbers are highlighted. ( Ulam Spiral by SplatBang)

Surprisingly, this pattern continues for large numbers. A point to be noted is that this pattern is a feature of spirals not necessarily begin with 1. For examples, the values of the polynomial $x^2+x+41$ form a diagonal pattern on a spiral beginning with 41.

# Finite Sum & Divisibility

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I wish to discuss a small problem from The USSR Olympiad Problem Book (problem 59) about the finite sum of harmonic series. The problem asks us to prove that

$\displaystyle{\sum_{k=2}^{n} \frac{1}{k}}$  can never be an  integer for any value of $n$.

I myself couldn’t think much about how to prove such a statement. So by reading the solution, I realised that how a simple observation about parity leads to this conclusion.

Firstly, observe that among the natural numbers from 2 to $n$ there is exactly one natural number which has the highest power of 2 as its divisor. Now, while summing up the reciprocals of these natural numbers we will get a fraction as the answer. In that fraction, the denominator will be an even number since it’s the least common multiple of all numbers from 2 to $n$. And the numerator will be an odd number since it’s the sum of $(n-2)$ even numbers with one odd number (corresponding to the reciprocal of the number with the highest power of 2 as the factor). Since under no circumstances an even number can completely divide an odd number, denominator can’t be a factor of the numerator. Hence the fraction can’t be reduced to an integer and the sum can never be an integer.