MHB Is (p-1)C(a) congruent to (-1)^a mod(p)?

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The discussion focuses on proving that the binomial coefficient (p-1)C(a) is congruent to (-1)^a mod(p) for a prime p and integer a between 1 and p-1. It utilizes Wilson's theorem, which states that (p-1)! is congruent to -1 mod p. The proof involves manipulating the binomial coefficient's formula and applying modular arithmetic to derive the congruence. Induction is suggested as a method to solidify the argument. The conclusion emphasizes the relationship between binomial coefficients and modular arithmetic in the context of prime numbers.
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Let p be prime and a be between 1 and p-1. Show the binomial coefficent (p-1)C(a) satifies

(p-1)C(a) =(-1)^a mod(p).

(p-1)C(a) =$\frac{(p-1)!}{a!(p-1-a)!}$ so we can apply wilson's theorem which says
(p-1)!=-1 (modp)
 
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Re: congruence equation

Poirot said:
Let p be prime and a be between 1 and p-1. Show the binomial coefficent (p-1)C(a) satifies

(p-1)C(a) =(-1)^a mod(p).

(p-1)C(a) =$\frac{(p-1)!}{a!(p-1-a)!}$ so we can apply wilson's theorem which says
(p-1)!=-1 (modp)
Let $x\binom{p-1}{a+1}\equiv \binom{p-1}{a}\pmod{p}$.

Then we have $\frac{x(p-1)!}{(p-a-2)!(a+1)!}\equiv \frac{(p-1)!}{(p-a-1)!a!}\pmod{p}$.

Cancel things out (why can that be done?), you get $x\equiv -1\pmod{p}$

Now apply induction.
 
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