# 09-240:HW2

• Note that the numbers ${\displaystyle 1^{6}-1=0}$, ${\displaystyle 2^{6}-1=63}$, ${\displaystyle 3^{6}-1=728}$, ${\displaystyle 4^{6}-1=4,095}$, ${\displaystyle 5^{6}-1=15,624}$ and ${\displaystyle 6^{6}-1=46,655}$ are all divisible by ${\displaystyle 7}$. The following four part exercise explains that this is not a coincidence. But first, let ${\displaystyle p}$ be some odd prime number and let ${\displaystyle {\mathbb {F} }_{p}}$ be the field with p elements as defined in class.
1. Prove that the product ${\displaystyle b:=1\cdot 2\cdot \ldots \cdot (p-2)\cdot (p-1)}$ is a non-zero element of ${\displaystyle {\mathbb {F} }_{p}}$.
2. Let ${\displaystyle a}$ be a non-zero element of ${\displaystyle {\mathbb {F} }_{p}}$. Prove that the sets ${\displaystyle \{1,2,\ldots ,(p-1)\}}$ and ${\displaystyle \{1a,2a,\ldots ,(p-1)a\}}$ are the same (though their elements may be listed here in a different order).
3. With ${\displaystyle a}$ and ${\displaystyle b}$ as in the previous two parts, show that ${\displaystyle ba^{p-1}=b}$ in ${\displaystyle {\mathbb {F} }_{p}}$, and therefore ${\displaystyle a^{p-1}=1}$ in ${\displaystyle {\mathbb {F} }_{p}}$.
4. How does this explain the fact that ${\displaystyle 4^{6}-1}$ is divisible by ${\displaystyle 7}$?