# 12-240/Homework Assignment 2

• 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 $7$?