Hi everyone
I'm having trouble with the following question:
Prove that (2p−1)! ≡ p(mod p^2).
Hint: p divides (2p-1)!.
Any help would be much appreciated.
Thanks, Alex
Let
$\displaystyle a=\prod _{k=1}^{p-1} (p+k)$
$\displaystyle b= \prod _{k=1}^{p-1} k$
$\displaystyle a p b=(2p-1)!$
By Wilson's Theorem, $\displaystyle a=-1$ and $\displaystyle b=-1$ mod p
so $\displaystyle a b-1$ is divisible by p
$\displaystyle a p b - p$ is divisible by $\displaystyle p^2$
$\displaystyle a= \prod _{k=1}^{p-1} (p+k)=(p+1)(p+2)\text{...}(p+(p-1))=1*2*\text{...}*(p-1)=(p-1)!$ (mod p)
and
$\displaystyle b=\prod _{k=1}^{p-1} k=1*2*\text{...}*(p-1)=(p-1)!$ (mod p)
Now, look at $\displaystyle (2p-1)!$.
Break it up into the product of three pieces, a, p, and b so that $\displaystyle (2p-1)!=a p b$
$\displaystyle a=-1$ and $\displaystyle b=-1$ by Wilson's Theorem
so $\displaystyle a b=1$ mod p which means $\displaystyle a b=1+r p$ for some $\displaystyle r$
$\displaystyle (2p-1)!=a p b=p(1+rp)=p+r p^2$
Therefore $\displaystyle (2p-1)!-p$ is divisible by $\displaystyle p^2$