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linux
Posts:
9
Registered:
4/4/07
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Re: Congruence involving binomial coefficients
Posted:
Feb 25, 2013 3:35 AM
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"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 = (n-1)/2 is an odd
> prime, then A(n,k) = 0 mod q for all even integers k with
> 2 <= k <= q-1.
>
> Proof outline:
>
> The verification for k = 2 is immediate. Proceeding by induction
> on k, assume k is even with 2 < k <= q-1 and simplify the
> difference A(n,k) - A(n,k-2). It's not hard to show (not hard
> but a little tedious) that, assuming 2 < k <= q-1, 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 <= q-1, 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.
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