FoldUnfold Table of Contents The von-Mangoldt Function The von-Mangoldt Function Definition: The von-Mangoldt Function is the function $\Lambda : \mathbb{N} \to \mathbb{R}$ defined for all $n \in \mathbb{N}$ by $\displaystyle{\Lambda (n) = \left\{\begin{matrix} \ln p & \mathrm{if} \: n = p^k\\ 0 & \mathrm{otherwise} \end{matrix}\right.}$. Sometimes the notation "$\log … [Read more...]

## The Prime Counting Function

FoldUnfold Table of Contents The Prime Counting Function The Prime Counting Function Definition: The Prime Counting Function is the function $\pi : \mathbb{R} \to \mathbb{N}$ defined for all $x \in \mathbb{R}$ by $\displaystyle{\pi (x) = \sum_{p \leq x} 1}$. The prime counting function adds $1$ for every prime less than or equal to $x$. Note that the prime counting … [Read more...]

## Solutions to x^2 – dy^2 = N

FoldUnfold Table of Contents Solutions to x^2 - dy^2 = N Solutions to x^2 - dy^2 = N THeorem 1: Let $d > 0$ and $d$ not be a perfect square. Let $\frac{h_n}{k_n}$ be the $n^{\mathrm{th}}$ convergent of the continued fraction expansion of $\sqrt{d}$. If $|N| and $(s, t)$ is a positive solution in lowest terms to $x^2 - dy^2 = N$ then $(s, t) = (h_n, k_n)$ for some $n … [Read more...]

## Solutions to x^2 – dy^2 = 1 and x^2 – dy^2 = -1

FoldUnfold Table of Contents Solutions to x^2 - dy^2 = 1 and x^2 - dy^2 = -1 Solutions to x^2 - dy^2 = 1 Solutions to x^2 - dy^2 = -1 Solutions to x^2 - dy^2 = 1 and x^2 - dy^2 = -1 Recall from the Pell's Equation page that Pell's equation is $x^2 - dy^2 = N$ where $d, N \in \mathbb{Z}$. We noted that Pell's equation has at most finitely many solutions if $d or $d$ is a … [Read more...]

## Pell’s Equation

FoldUnfold Table of Contents Pell's Equation Pell's Equation Definition: Pell's Equation is the equation $x^2 - dy^2 = N$ where $d, N \in \mathbb{Z}$. Of course, we will only be interested in integer solutions $(x, y)$ to the equation. Proposition 1: If $d then Pell's equation has at most finitely many solutions. Proof: Suppose that $d . Then $-d > 0$. Since $N$ … [Read more...]