
#1
Jun1106, 02:37 AM

P: 36

A question:
Let bin(a,b) denote the binomial coefficient a! / ( b! (a  b)! ). Is it true that bin( 2p, p ) = 2 (mod p) if p is prime and p>=3 ? 



#2
Jun1106, 04:22 AM

Sci Advisor
HW Helper
P: 9,398

Yes, it's fermat's little theorem: x^p=x mod p, for p a prime, hence
(1+x)^2p = (1+x^p)^2 = 1+2x^p+x^{2p} mod p note your requirement on p>=3 is not necessary. 4 choose 2 =6 whcih is congruent to 2 mod 2 as well. 


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