# 14-240/Homework Assignment 2

This assignment is due at the tutorials on Tuesday September 30, or at the Math Aid Centre, Sydney Smith room 1071, at the appropriately labeled mailboxes, by Friday October 3 at 5PM. Here and everywhere, neatness counts!! You may be brilliant and you may mean just the right things, but if the teaching assistants will be having hard time deciphering your work they will give up and assume it is wrong.

Read sections 1.1 through 1.4 in our textbook, and solve the following problems:

• Problems 3a and 3bcd on page 6, problems 1, 7, 18, 19 and 21 on pages 12-16 and problems 8, 9, 11 and 19 on pages 20-21. You need to submit only the underlined problems.
• Note that the numbers $1^6-1=0$, $2^6-1=63$, $3^6-1=728$, $4^6-1=4,095$, $5^6-1=15,624$ and $6^6-1=46,655$ are all divisible by $7$. The following four part exercise explains that this is not a coincidence. But first, let $p$ be some odd prime number and let ${\mathbb F}_p$ be the field with p elements as defined in class.
1. Prove that the product $b:=1\cdot 2\cdot\ldots\cdot(p-2)\cdot(p-1)$ is a non-zero element of ${\mathbb F}_p$.
2. Let $a$ be a non-zero element of ${\mathbb F}_p$. Prove that the sets $\{1,2,\ldots,(p-1)\}$ and $\{1a,2a,\ldots,(p-1)a\}$ are the same (though their elements may be listed here in a different order).
3. With $a$ and $b$ as in the previous two parts, show that $ba^{p-1}=b$ in ${\mathbb F}_p$, and therefore $a^{p-1}=1$ in ${\mathbb F}_p$.
4. How does this explain the fact that $4^6-1$ is divisible by $7$?
You don't need to submit this exercise at all, but you will learn a lot by doing it!
• After September 24, add your name to the Class Photo page!