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If $p $ is an odd prime and if $1+\frac{1}{2}+\cdots +\frac{1}{p-1}=\frac{a}{b}$ where $a,b $ are integers prove that $p|a$. If $p>3\implies p^2|a$

My Try:

Can the problem be interpreted as a problem in $\mathbb Z_p^*$ where $p$ is an odd prime. Thus we have the sum of inverses of all the elements in $\mathbb Z_p^*$. Also $\mathbb Z_p^*$ is cyclic.

Can all these facts be used in some manner to conclude the answer? Or it is not the way?

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  • $\begingroup$ $p|a$ in $\Bbb Z_p$ means $a=0$ which is not even in $\mathbb Z_p^*$. $\endgroup$ – Gregory Grant May 16 '15 at 7:49
  • $\begingroup$ But you're on the right track, need to show the sum of all elements in $\mathbb Z_p^*$ equals $\overline0$ in $\mathbb Z_p$ $\endgroup$ – Gregory Grant May 16 '15 at 7:55
  • $\begingroup$ See here: [ Wolstenholme's theorem](math.stackexchange.com/questions/1256032/…) $\endgroup$ – hamid kamali May 16 '15 at 8:15
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Hint: $$ 1+\frac{1}{2}+\cdots+\frac{1}{p-1}=1+2+\cdots+(p-1) \mod p. $$

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