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Let $p\gt 3$, be a prime number and $1+\frac{1}{2} +\frac{1}{3} +...+\frac{1}{p-1} =\frac{a}{b}$ when $a,b\in \mathbb N$ and $gcd(a,b)=1$.

prove that $p^2|a$.

I proved that $p|a$, but I cant prove $p^2|a$, and my idea is as follow:

By multiplying $(p-1)!$, we get:

$(p-1)!+\frac{(p-1)!}{2}+...+\frac{(p-1)!}{p-2}+\frac{(p-1)!}{p-1}=\frac{a(p-1)!}{b}$ $\,\,\,\,$ $(1)$

So, the number $\frac{a(p-1)!}{b}$ should be integer. consider the left hand of the sum of $(1)$ in the ring $Z_p$. Since in $Z_p$ the inverses are unique and what we have is indeed a rearrangement of all the congruence classes $mod\,p$. So:

$(p-1)!+\frac{(p-1)!}{2}+...+\frac{(p-1)!}{p-2}+\frac{(p-1)!}{p-1}\equiv 1+2+...+p-1\equiv \frac{p(p-1)}{2}\equiv 0$$\,\,\pmod p$. And from here, $p|\frac{a(p-1)!}{b}$, and that's mean: $p|a$.

I tried to prove the statement by considering the ring $Z_{p^2}$ but I get nothing.

Thanks in advance…

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2 Answers 2

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Define two polynomials $$f(x)=x^{p-1}-1$$ and $$g(x)=(x-1)(x-2)\dotsm(x-(p-1))$$ and observe that $h(x)=f(x)-g(x)$ is a polynomial of degree at most $p-2$ which all of it's coefficients is divisible by $p$ ($h$ has $p-1$ roots $\{1,2,\dotsc,p-1\}$ in $\mathbb Z_p[x]$, which means that $h(x)=0$ in $\mathbb Z_p[x]$). Write g as $$g(x)=x^{p-1}-s_1x^{p-2}+\dotsb-s_{p-2}p+(p-1)!$$ then $$g(p)=(p-1)!=p^{p-1}-s_1p^{p-2}+\dotsb+(p-1)!$$ Now with observing that all the numbers $s_i$ for $i=1,2,\dotsc,p-2$ are divisible by $p$, because are also coefficients of $h$, we have $$0=p^{p-1}-s_1p^{p-2}+\dotsb-s_{p-2}p\equiv -s_{p-2}p\pmod {p^3}$$ Which means that $s_{p-2}\equiv 0\pmod {p^2}$ as you want.

Note. $\mathbb Z_p[x]$ denotes the ring of all polynomials with coefficients in the field $\mathbb Z_p$.

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This is a consequence/version of Wolstenholme's theorem. A proof is available here.

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  • 2
    $\begingroup$ It would probably be best to expand this answer somewhat. $\endgroup$
    – user642796
    Apr 28, 2015 at 16:16

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