9
$\begingroup$

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…

$\endgroup$
0
1
$\begingroup$

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 $\mathbb Z_p$.

$\endgroup$
0
$\begingroup$

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

$\endgroup$
1
  • 2
    $\begingroup$ It would probably be best to expand this answer somewhat. $\endgroup$ – user642796 Apr 28 '15 at 16:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.