Number theory question involving Wilson's theorem. How to solve??

Question

Given p is a prime number greater than 2, and

$ 1 + \frac{1}{2} + \frac{1}{3} + ... \frac{1}{p-1} = \frac{N}{p-1}$

how do I show, $ p | N $ ???

The previous part of this question had me factor $ x^{p-1} -1$ mod $p$. Which I think is just plainly $(x-1) ... (x-(p-1))$

Answer 1

By Wolstenholme's Theorem, we have that $$\sum_{k=1}^{p-1}\frac{(p-1)!}k\equiv0\pmod{p^2}$$ so $$(p-1)!\left(1+\frac12+\cdots+\frac1{p-1}\right)=np^2$$ for some integer $n$.

From your condition (assuming typo of $(p-1)!$ instead of $p-1$) we get $$\frac{np^2}{(p-1)!}=\frac N{(p-1)!}\implies p^2=\frac Nn$$ and so the result follows.

Answer 2

Presumably, the intended problem was to show, assuming $p$ is an odd prime, that $$1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{p-1} = \frac{N}{(p-1)!}$$ implies $p{\,\mid\,}N$.

It suffices to show the $\text{LHS}$ is congruent to $0$, mod $p$.

But mod $p$, the fractions $$1, \frac{1}{2}, \frac{1}{3},..., \frac{1}{p-1}$$ are just a permutation of $$1,2,3,...,p-1$$ which sums to ${\large{\frac{(p-1)p}{2}}}$, hence is congruent to $0$, mod $p$.

Answer 3

Working $\bmod{p}$: $$ 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{p-1}\\ \equiv 1 + 2^{-1} + 3^{-1} + \cdots + (p-1)^{-1}\\ \equiv 1 + 2 + \cdots + (p-1) \ \text{ reordered} $$ which you have shown is $0\pmod{p}$ when you factored $x^{p-1} -1\pmod{p}$.

So if $\displaystyle 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{p-1} =\frac{N}{p-1}$ then necessarily $N(p-1)^{-1} = 0 \pmod{p}$ so $p$ divides $N$.

Answer 4

HINT.-$x\to\dfrac 1x$ is a bijection of $\mathbb F^*_p$ so $$\sum_1^{p-1} \frac 1x\equiv\sum_1^{p-1} x\equiv 0\pmod p$$