Let $p$ be a prime and define $f(x),g(x) \in \mathbb{Z}_p[x]$ by $f(x) = x^{p-1} - 1$ and $g(x) = (x-1)(x-2) \cdots (x-(p-1))$.
Prove that $f(x) = g(x)$. Hint: what is the degree of $f(x) - g(x)$, and how many zeroes does it have?
Use part a to prove that $(p-1)! \equiv -1 \pmod{p}$. Hint: if you are having a problem with signs, remember that most primes are odd.
Prove the converse: if $(n-1)! \equiv -1 \pmod{n}$ then $n$ is prime.
Let $K$ be a field, let $\alpha_1,\ldots,\alpha_n$ be distinct elements of $K$, and let $\beta_1,\ldots,\beta_n$ be elements of $K$ (not necessarily distinct.)
For every $i \in \{1,\ldots,n\}$, find a polynomial $f_i(x) \in K[x]$ such that $\deg f_i(x) = n-1$ and $$f_i(\alpha_j) = \begin{cases} 1 & \text{if }i = j\\ 0 & \text{if }i \ne j. \end{cases}$$ Hint: for a first attempt, do not worry about the case $i = j$.
Find a polynomial $f(x) \in K[x]$ such that $\deg f(x) = n-1$ and $f(\alpha_i) = \beta_i$ for all $i \in \{1,\ldots,n\}$.
A correction: this part should say "$\deg f(x) \le n-1$" instead of "$\deg f(x) = n-1$".
Now suppose that $K$ is $\mathbb{Z}$ instead of a field. Find an example of $\alpha_1,\ldots,\alpha_n$ and $\beta_1,\ldots,\beta_n$ for which there is no polynomial $f(x)$ as in part b.
There are only two problems this week.