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linux
Posts:
9
Registered:
4/4/07


Re: Congruence involving binomial coefficients
Posted:
Feb 25, 2013 3:35 AM


"quasi" <quasi@null.set> a écrit dans le message de news: f8vji8tc9tmgnqv86eabjn34rbfotvotgp@4ax.com... >  (9) lemma > > If n is an odd positive integer such that q = (n1)/2 is an odd > prime, then A(n,k) = 0 mod q for all even integers k with > 2 <= k <= q1. > > Proof outline: > > The verification for k = 2 is immediate. Proceeding by induction > on k, assume k is even with 2 < k <= q1 and simplify the > difference A(n,k)  A(n,k2). It's not hard to show (not hard > but a little tedious) that, assuming 2 < k <= q1, the difference > is congruent to 0 mod q, hence the truth of the lemma follows > from its truth for k=2. > > quasi
(9) Note that if q is prime et k is an integer with 2 <= k <= q1, it is easy to prove that Binomial(2q+1,k) an Binomial(q,k) are divisible by q without induction. For this , the hypothesis 2q+1 prime is not necessary.



