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Sir Col
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Binomial Theorem for Primes
« on: Jun 5th, 2004, 9:59am » |
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Consider the following results,
75 = 16807 [equiv] 2 mod 5
35+45 = 1267 [equiv] 2 mod 5
117 = 19487171 [equiv] 4 mod 7
37+87 = 2099339 [equiv] 4 mod 7
Given that p is prime, and a,b[in][bbn], prove that in general, (a+b)p [equiv] ap+bp mod p.
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Grimbal
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Re: Binomial Theorem for Primes
« Reply #1 on: Jun 5th, 2004, 6:34pm » |
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Grimbal
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Re: Binomial Theorem for Primes
« Reply #3 on: Jun 7th, 2004, 3:00am » |
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It seems to me you are obfuscating a much simpler equation. My smile says: I know.
::For p prime, ap [equiv] a mod p.::
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Sir Col
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Re: Binomial Theorem for Primes
« Reply #4 on: Jun 7th, 2004, 4:41am » |
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The problem is not quite that trivial. Fermats Little Theorem only demonstrates that (a+b)p is congruent with a+b modulo p.
However, you are correct, I was partially obfuscating the problem, namely: except for the first and last terms in the expansion of (a+b)n, show that n divides the coefficients of each term iff n is prime.
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towr
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Re: Binomial Theorem for Primes
« Reply #5 on: Jun 7th, 2004, 5:10am » |
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For the original question
::(a+b)p % p [equiv] (a+b) % p [equiv] [a % p + b % p] % p [equiv] [ap % p + bp % p] % p [equiv] [ap + bp] % p ::
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| « Last Edit: Jun 7th, 2004, 5:12am by towr » |
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Wikipedia, Google, Mathworld, Integer sequence DB
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Sir Col
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Re: Binomial Theorem for Primes
« Reply #6 on: Jun 7th, 2004, 12:23pm » |
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Nicely done, towr! I had not anticipated such a simple solution.
Apologies, Grimbal, for doubting your clever insight! 
Okay...
Quote:| Except for the first and last terms in the expansion of (a+b)n, show that n divides the coefficients of each term iff n is prime. |
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And it is not sufficient to use F.L.T. to demonstrate that the sum of coefficients must be divisible by n; although I hadn't realised that until you clever twosome demonstrated it. Consider the following:
(a+b)2 = a2 + 2ab + b2
(a+b)3 = a3 + 3a2b + 3ab2 + b3
(a+b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4
Clearly 2|2, 3|3, and 4|(4+6+4), but 4 does not divide 6.
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