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 $$S_{-k}(p-1) = \frac{1}{1^k} + \frac{1}{2^k} + \frac{1}{3^k} + \dots + \frac{1}{(p-1)^k} ~=_{Q} ~0 \pmod{p}.$$

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: $$\frac{8}{5} =_{Q} ~0 \pmod{4}, $$ $$ -\frac{12}{55} =_{Q} ~0 \pmod{4},$$ $$\frac{29}{15} =_{Q} ~\frac{25}{15} = \frac{5}{3} \pmod{4}$$