Elementary Number Theory Problems 3.2 Solution (David M. Burton's 7th Edition) - Q3

This is my solution for Chapter 3.2 Q3 in the book Elementary Number Theory 7th Edition written by David M. Burton.

Background

All theorems, corollaries, and definitions listed in the book's order:

Theorems and Corollaries in Elementary Number Theory (Ch 1 - 3)

All theorems and corollaries mentioned in David M. Burton’s Elementary Number Theory are listed by following the book’s order. (7th Edition) (Currently Ch 1 - 3)

RanblogRan

I will only use theorems or facts that are proved before this chapter. So you will not see that I quote theorems or facts from the later chapters.

Question

Given that $p \not \mid n$ for all primes $p \leq \sqrt[3]{n}$, show that $n \gt 1$ is either a prime or the product of two primes.
[Hint: Assume to the contrary that $n$ contains at least three prime factors.]

Solution

Assume to the contrary that $n$ contains at least three prime factors. We can write $n = p_{1}p_{2}p_{3}d$ where $p_{1}, p_{2}$, and $p_{3}$ are all primes and $d \geq 1$.

We know $p \not \mid n$ for all primes $p \leq \sqrt[3]{n}$, and we know $p_{1}, p_{2}$, and $p_{3}$ can all divide $n$. So, $p_{1}, p_{2}$, and $p_{3}$ are all $\gt \sqrt[3]{n}$. So, $p_{1}p_{2}p_{3} \gt n$. Then $nd \lt p_{1}p_{2}p_{3}d = n$. This means $nd \lt n$, which is a contradiction.

Read More: All My Solutions for This Book

Related Pages

< Chapter 3.2, Q2 Chapter 3.2, Q4 >