Wilson's Theorem Lemma

Question

I am faced with the following question in my undergraduate Number Theory textbook:

Use the coefficient of x on both sides of (*) to prove that if p $\ge$ 3, then p divides a and $\frac{a}{b}$ = $ 1+ \frac{1}{2}$ + $\frac{1}{3}$ + ... + $\frac{1}{p-1}$.

(*) x^p-1 -1 $\equiv$ (x+1)(x+2)...(x+p-1) (poly mod n)

The equation (*) was given in the proof of Wilson's Theorem.

So far, the only thing I can think of is to reduce (*) to $-1 \equiv (1)(2)(3)...(p-1)$. I am also thinking maybe this has something to do with 1,2,..,p-1 being a complete residue system?

Any help would be super appreciated, thank you!

Answer 1

By saying $-1\equiv (1)(2)\dotsc$ etc. you are looking at the constant term of (*).

Why not look at other terms (e.g. the coefficient before $x$)?

Answer 2

We know that $\prod i \equiv -1 \pmod{p}$

Hint: $\prod_{i\neq j} i \equiv - \frac{1}{j} \pmod{p}$.

Hence, by comparing the coefficent of $x$ in (*),
$$ 0 \equiv \sum_j \prod_{i\neq j} i \equiv \sum_j - \frac{1}{j} \pmod{p}.$$
So $p \mid a$.

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Alternate hint: $ \frac{a}{b} = \sum \frac{1}{j} = \frac{ \sum_{j\neq i} \prod{i} } { \prod i} $.

Hence, by comparing the coefficient of $x$ in (*), $p$ divides the numerator but not the denominator, $p \mid a$.