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Let $p$ be the prime number bigger than $2$. Prove that $\dfrac{2^p-2}{p}\equiv 1-\dfrac{1}{2}+\dfrac{1}{3}-\dots-\dfrac{1}{p-1} \pmod p$

I do not know where to even start

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$2^p = (1+1)^p$ –  Hurkyl Jul 11 at 5:39
I used that but it was unsuccesfull. $2^p=C_p^0+C_p^1+\dots+C_p^{p-1}+C_p^p$ –  TNT Jul 11 at 5:40
Try writing out what a binomial coefficient divided by $p$ looks like modulo $p$. –  Hurkyl Jul 11 at 5:42
$C_p^k\equiv 0 \pmod p$ for $k=1,..,p-1$ –  TNT Jul 11 at 5:43
Right, but you need to look at $C_p^k/p$ –  Hurkyl Jul 11 at 5:49

1 Answer 1

The reasonable thing to try works. Calculate $2^p-2$ using the Binomial Theorem.

The first term is $p$. Division by $p$ gives $1$. Good start!

Now look at $\binom{p}{2}$. Divide by $p$. We get $\frac{p-1}{1\cdot 2}$, which is congruent to $-\frac{1}{2}$ modulo $p$. (We are taking the reciprocal modulo $p$.)

Let's see whether three times lucky. When we divide $\binom{p}{3}$ by $p$, we get $\frac{(p-1)(p-2)}{1\cdot 2\cdot 3}$, which is congruent to $\frac{1}{3}$ modulo $p$.

The rest is done in exactly the same way. In general, when we divide $\binom{p}{k}$ by $p$, for $k$ in our range, we get $$\frac{(p-1)(p-2)\cdots(p-(k-1))}{k!}.$$ The top is congruent to $(-1)^{k-1}(k-1)!$, and now division by $k!$ modulo $p$ does it.

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