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
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 $p=4$
$(2p-1)! = 7! = 5040$
$p^2 = 16$
$5040 \pmod{16} = 0 \neq 4$
this sequence in $p$ is a bit interesting, it's $s_p = \begin{cases}p,&p \text{ is prime.} \\ 0, &\text{else} \end{cases}$
Sorry I should have put in the original question that p is a prime number! I have to show that it's true for all primes and this is the part I'm struggling with.
Thanks, Alex.