Pythagorean Identity

This year we will start a blog series about mathematical identities. Let us begin today with the Pythagorean Identity $$ a^2 + b^2 = c^2 $$

Go Bugs Go!

How to construct a regular polygon with 15 sides

To feed your curiosity, today we will look at a compass and straightedge construction of the regular polygon with 15 sides and we will show that there is a connection between this construction problem with the measuring liquid puzzle that we have learned from our previous post.

Measuring liquid puzzle

Today we will look at a brainteaser puzzle: "how to measure out exactly 1 liter of water using a 3-liter jug and a 5-liter jug." We will analyse this puzzle to see that, despite its innocent look, this puzzle has a close connection with the linear Diophantine equation.

Power sum and Wolstenholme's theorem

In the last post, we have learned how to derive formulas for the power summation $$S_k(n) = 1^k + 2^k + 3^k + \dots + n^k.$$

Today we will investigate some divisibility properties of the power summation $S_k(n)$. We will prove that if $p$ is a prime number and $k$ is not divisible by $p-1$ then $$S_k(p-1) = 1^k + 2^k + 3^k + \dots + (p-1)^k = 0 \pmod{p}.$$

We will also investigate the following summation $$S_{-k}(n) = \frac{1}{1^k} + \frac{1}{2^k} + \frac{1}{3^k} + \dots + \frac{1}{n^k}.$$

There is a theorem in number theory concerning this summation, it is called Wolstenholme's theorem.

Wolstenholme's Theorem. If $p$ is a prime number $>3$ then $$S_{-1}(p-1) = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{p-1} ~=_{Q} ~0 \pmod{p^2} $$ and $$S_{-2}(p-1) = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \dots + \frac{1}{(p-1)^2} ~=_{Q} ~0 \pmod{p}.$$

We will prove a general result, that is, if $p$ is a prime number and $k$ is not divisible by $p-1$ then $$S_{-k}(p-1) = \frac{1}{1^k} + \frac{1}{2^k} + \frac{1}{3^k} + \dots + \frac{1}{(p-1)^k} ~=_{Q} ~0 \pmod{p}.$$

Power sum

Today we will learn about how to derive the formula for power summation $$S_k(n) = 1^k + 2^k + 3^k + \dots + n^k .$$


Our main tool is the following binomial identity
$$(x+y)^k = x^k + {k \choose 1} x^{k-1} y + {k \choose 2} x^{k-2} y^2 + {k \choose 3} x^{k-3} y^3 + \dots + {k \choose {k-1}} x y^{k-1} + y^k.$$

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 $$d = a x + b y.$$

Another Proof of Wilson's Theorem

In previous posts, we have learned about modulo for rational numbers. To show its application, today, we will use the language of modulo of rationals to give another proof of Wilson's Theorem.

Wilson's theorem is a well known theorem in number theory. It is stated as follows.

Wilson's Theorem. If $p$ is a prime number then $$(p-1)! = -1 \pmod{p}.$$

Modulo for rational numbers II

In the previous post, we have learned about modulo for rational numbers. Today, we will continue on this topic and will learn about some properties of this modulo.

First of all, let us look at some examples and review the definition: $$\frac{14}{5} =_{Q} ~0 \pmod{7}, $$ $$ \frac{16}{55} =_{Q} ~\frac{9}{55} =_{Q} ~\frac{2}{55} \pmod{7},$$ $$\frac{1}{4} =_{Q} ~\frac{8}{4} =_{Q} ~2 \pmod{7}, \dots$$

Definition. Let $n$ be an integer, and $\alpha$, $\beta$ two rational numbers. We say that $\alpha$ is equal to $\beta$ modulo $n$, and write $$\alpha =_{Q} ~\beta \pmod{n}$$ if and only if there exists an integer $k$ co-prime to $n$ such that $k(\alpha - \beta)$ is an integer and $$k(\alpha - \beta) = 0 \pmod{n}.$$

Modulo for rational numbers

In our series on modulo, we have learned about modulo for integer numbers. Let us recall the definition.

Definition. Let $n$, $a$, $b$ be integer numbers. We say that $a$ is equal to $b$ modulo $n$, and write $$a = b \pmod{n}$$ if and only if $a-b$ is a multiple of $n$.

For example, $$8 = 0 \pmod{4},$$ $$9 = 1 \pmod{4},$$ $$-5 = -1 = 3 = 7 \pmod{4}, \dots $$

Today, we will learn a new concept - modulo for rational numbers. Before going into the details, let us list here a few examples, so that we may roughly see what modulo for rationals is about.

Examples of modulo for rational numbers: $$\frac{8}{5} =_{Q} ~0 \pmod{4}, $$ $$ -\frac{12}{55} =_{Q} ~0 \pmod{4},$$ $$\frac{29}{15} =_{Q} ~\frac{25}{15} = \frac{5}{3} \pmod{4}$$

A proof of Wilson's Theorem using polynomial interpolation

In the last posts, we have learned about polynomial interpolation through two formulas, the Newton's interpolation formula and the Lagrange's interpolation formula. Both formulas can be used to prove the Wilson's theorem.

In this post, we will present a proof of the Wilson's theorem based on the Newton's interpolation formula. The other proof, that is, proving Wilson's theorem based on the Lagrange's interpolation formula, we leave it as a homework exercise.

Wilson's Theorem is a well known theorem in number theory. It says that if $p$ is prime number then the number $(p−1)!+1$ is a multiple of $p$.

Mathematical induction II

Today we will use induction to solve some more problems. 

Problem 4. Prove that $$1 \times 2 \times 3 + 2 \times 3 \times 4 + \dots + n (n+1)(n+2) = \frac{1}{4} n(n+1)(n+2)(n+3).$$

Mathematical induction

Today, we will learn about mathematical induction. We usually use induction to prove a certain statement to be correct for all natural numbers.

Let us use $P(n)$ to denote a statement that involves a natural number $n$. To prove that $P(n)$ is correct for all natural number $n$, an induction proof will have the following steps

Step 1: is called the initial step. We will prove that the statement $P(n)$ is correct for the case $n=0$.

Step 2: is called the induction step. This is the most important step. In this step,

• we assume that for any $0 \leq n \leq k$, the statement $P(n)$ is correct;
• with this assumption, we will prove that the statement $P(n)$ is also correct for the case $n=k+1$.

With these two steps, by the mathematical induction principle, we conclude that the statement $P(n)$ must be correct for all natural number $n$.

Pascal's triangle

Today, we will look at a famous number pattern, the Pascal's number triangle.

Pascal's triangle

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.