Construction of regular polygons

There are a few classic problems of ancient mathematics that are easy to state but incredibly difficult to solve. Take for example, the problem of constructing regular polygons and the problem of trisecting an angle by compass and straightedge. It was not until the 18th-19th centuries that mathematicians could finally solve them by employing advanced tools of number theory and algebra. Today, we will look at the regular polygon construction problem.

Regular polygon construction problem. Using compass and straightedge, construct a regular polygon with $n$ sides.

Pythagorean triples

In geometry, there is a well known theorem, called the Pythagorean Theorem, which says that in a right triangle, the square of the hypotenuse is equal to the sum of squares of the other two sides. 

Pythagorean Theorem: $BC^2 = AB^2 + AC^2$

That is why we call the equation $$x^2 + y^2 = z^2$$ the Pythagorean equation and its solution $(x,y,z)$ is called a Pythagorean triple. Of course, we only consider integer solutions.

Today, we will solve the Pythagorean equation and show that this equation has an infinite number of solutions.

Wilson's Theorem

Today, we will study Wilson's Theorem - a theorem concerning prime numbers. Wilson's theorem says that if $p$ is a prime number then the number $(p-1)! + 1$ will be divisible by $p$.

Here, the notation $n!$ denotes $$n! = 1 \times 2 \times 3 \times \dots \times n.$$

For example,

• $1! + 1 = 2$ is divisible by $2$
• $2! + 1 = 3$  is divisible by $3$
• $4! + 1 = 25$  is divisible by $5$
• $6! + 1 = 721$  is divisible by $7$

Some problems on prime numbers

In previous posts, we have learned about prime numbers, and we know from Euclid's Theorem that there exists an infinite number of primes. In this post, we continue to look at prime numbers. Leading mathematicians like Fermat, Euler, Gauss were all fascinated by prime numbers. There are many problems about primes that even the statements are simple, but they still remain unsolved even to this day.

Euclid's theorem on prime numbers

Continuing with our story about prime numbers, today we will prove that there exists an infinite number of primes. This is called the Euclid's theorem on prime numbers. This theorem has a very simple proof but it is probably one of the most beautiful proofs ever in mathematics.

Prime numbers

Today we will learn about prime numbers - a basic building block of arithmetic.

A prime number is a natural number greater than 1 and has no divisors other than 1 and itself. For example, the numbers 2, 3, 5, 7, 11, 13 are prime numbers. The number 9 is not a prime number because it is divisible by 3. The number 2012 is not a prime number because it is divisible by 2.

modulo - Part 6

We recall the definition of modulo. Two numbers $a$ and $b$ are said to be equal modulo $n$ if and only if $a-b$ is a multiple of $n$, and we write $a = b \pmod{n}$. For example, $9 = 1 \pmod{8}$ and $14 = -2 \pmod{8}$. 

In our usual arithmetic, we picture our integer numbers lying on the number line and we do addition and multiplication like this $2 + 7 = 9$, $2 \times 7 = 14$, etc...

our number line

modulo - Part 5

Today we will learn about Fermat's "little" Theorem. We will see that Fermat's little Theorem is very useful in modulo arithmetic. The theorem asserts that for any prime number $p$ and for any number $a$ not divisible by $p$,
$$ a^{p-1} = 1 \pmod{p} . $$

modulo - Part 4

One of the all-time famous mathematicians is Pierre de Fermat. He is a French mathematician and lived in the 17th century.

To mention Fermat, we must mention "his problem" - the Fermat's last problem. The problem that had challenged generations of mathematicians. Probably the reason that his problem is so well-known and attracted so much effort from top mathematicians as well as young school students is that it is stated so simple and that a secondary school student can understand it.

The Fermat's last problem is stated as follows. Prove that for any $n \geq 3$ the following equation does not have non-trivial solutions  

$$ x^n+y^n=z^n $$

Non-trivial solutions here mean non-zero solutions. This is because if  $x$, $y$ or $z$ is equal to 0 then the equation becomes trivial.

modulo - Part 3

Today, we are going to look at some more examples about modulo.

Example 1: Prove that $11 + 2011^{2012} + 2012^{2013}$ is divisible by 13.

modulo - Part 2

Last time, we have learnt about modulo. Two integers $a$ and $b$ are said to be equal modulo $n$, denoted by $a = b \pmod{n}$, iff $a-b$ is divisible by $n$.

For example, $15 = 3 \pmod{4}$ and $99 = -1 \pmod{10}$.

modulo - Part 1

Today we will look at an important concept in number theory -- the concept of modulo. Two integer numbers $a$ and $b$ are said to be equal modulo $n$ iff they have the same remainder when divided by $n$. Or equivalently, iff $(a-b)$ is divisible by $n$. We will write $a = b \pmod{n}$.