Show that $ \sum_{k=1}^{p-1} (k-1)! (p-k)! k^{p-1} \equiv 0 \pmod p\ $ for odd prime $ \ p \ $

Question

Let $ \ p \ $ be an odd prime number . Show that $$ \sum_{k=1}^{p-1} (k-1)! (p-k)! k^{p-1} \equiv 0 \pmod p \ $$

Answer:

Now,

$ \sum_{k=1}^{p-1} (k-1)! (p-k)! k^{p-1} \\ =\sum_{k=1}^{p-1} k! (p-k)! k^{p-2} \\ =\sum_{k=1}^{p-1} (-1)^k k^{p-2} \ \pmod p \\=-1+2^{p-2}-3^{p-2}+......+(-1)^{p-1} (p-1)^{p-2} \pmod p $

But I can not conclude the result.

Help me doing this

Answer 1

The fact that you are using is wrong. What is true is the following $$(k-1)!(p-k)! \equiv (-1)^{k} \pmod p$$

Now using the Fermat's Little Theorem that $k^{p-1} \equiv 1\pmod p$ your have

$$\sum_{k=1}^{p-1} (k-1)! (p-k)! k^{p-1} \equiv \sum_{k=1}^{p-1} (-1)^{k} \equiv 0 \pmod p$$

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The above fact is true as $(p-1)! \equiv -1 \pmod p$ by Wilson's Theorem and then $$(k-1)!(p-k)! \equiv (-1)^{k-1}(p-k+1)(p-k+2)\cdots(p-1)(p-k)! \equiv (-1)^k \pmod p$$