Question

Prove that \((p-3)! \equiv \dfrac{p-1}{2} mod p\)
Please, help

Answer 1

p is prime, p >2

Answer 2

By Wilson, we have (p-1)! =-1 (mod p)
hence (p-1)(p-2) (p-3)! = -1 (mod p)
and p-1 =1 mod p
p-2 = 2 mod p
hence 2(p-3)! =-1 mod p
then I am stuck.

Answer 3

it should be
p-1 = -1 (mod p)
p-2 = -2 (mod p)

Answer 4

yes,

Answer 5

Fortunately, the result is the same and I stuck still!! :)

Answer 6

(p-3)! = (p-1)/2 (mod p)
is equivalent to
2(p-3)! = p-1 (mod p)

Answer 7

What if I * (-1) both sides and get -2 (p-3)! = 1 mod p
that is (p-3)! is inverse of 2 mod p? then?

Answer 8

then you can use wilson's theorem to get
(p-1)! =-1 (mod p)
(p-1)(p-2) (p-3)! = -1 (mod p)
(-1)(-2)(p-3)! = -1 (mod p)
2(p-3)! = -1 (mod p)

Answer 9

@jim_thompson5910 Yes, that was what I did above.

Answer 10

but how to go further?

Answer 11

well I think that's enough to show that they are equivalent

`(p-3)! = (p-1)/2 (mod p)` is equivalent to `2(p-3)! = p-1 (mod p)`

`(p-1)! =-1 (mod p)` turns into `2(p-3)! = -1 (mod p)`


by the transitive property of equivalence, we know `(p-3)! = (p-1)/2 (mod p)` is equivalent to `(p-1)! =-1 (mod p)`


so because `p-1)! =-1 (mod p)` is true, this makes `(p-3)! = (p-1)/2 (mod p)` true too

Answer 12

aah it drove me crazy :)

Answer 13

first: we don't have (p-3)! = (p-1)/2 mod p
How ?

Answer 14

I don't understand what you're asking

Answer 15

below the line : "well I think it's enough....."
you stated : (p-3)! = (p-1)/2 mod p. My question is there, how?

Answer 16

that's what you're given

Answer 17

That is the statement we need to prove, not given information.

Answer 18

I'm saying that the thing you need to prove connects to `(p-1)! =-1 (mod p)` (in the work shown above)

Answer 19

You meant we go back ward? A = B
B leads to C
and if C true then B true then A = B hold, right?

Answer 20

what we need to prove: `(p-3)! = (p-1)/2 (mod p)`


that's equivalent to saying "we need to prove `2(p-3)! = (p-1) (mod p)` is true"


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from above `(p-1)! =-1 (mod p)` turns into `2(p-3)! = -1 (mod p)`



since `(p-1)! =-1 (mod p)` is true, this means `2(p-3)! = -1 (mod p)` is true


so `2(p-3)! = -1 (mod p)` has been proven true
leading to `(p-3)! = (p-1)/2 (mod p)` being proven true