Euclidean algorithm

In previous post, we have learned about Bézout's Lemma. Today, we will learn about Euclidean algorithm. This algorithm is used to determine the coefficients in the Bézout's equation.

Let us recall Bézout's lemma. 

Bézout's Lemma. Let $a$ and $b$ be two integers and $d$ their greatest common divisor. Then there exist two integers $x$ and $y$ such that $$d = a ~x + b ~y.$$

We use Euclidean algorithm to calculate the greatest common divisor $d$ of the two numbers $a$ and $b$, and determine the two values of $x$ and $y$ in Bézout's equation $$d = a ~x + b ~y.$$

We will see that Euclidean algorithm is motivated from a very simple and natural idea.

Bézout's lemma

Today we will learn a very beautiful result in number theory, the Bézout's lemma, it is stated as follows.

Bézout's Lemma. Let $a$ and $b$ be two integers and $d$ their greatest common divisor. Then there exist two integers $x$ and $y$ such that $$d = a x + b y.$$

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 $a = b \pmod{n}$.