標籤文章彙整：solutions

92年國安局三等考數理組初等數論(一)

Posted by —

2

題目：設 p 是質數，a, b 是整數，如果 \displaystyle\frac{a^p-b^p}{p} 是整數，證明 \displaystyle\frac{a^p-b^p}{p^2} 也是整數。

解答：Suppose we have know the following lemma

If p is prime, then { p \choose k} is divisible by p if and only if nonnegative integer k\neq 0, p

For convenience, let \displaystyle\frac{a^p-b^p}{p} = n and a = b+c , where c is also an integer.

Then we can rewrite n as

\displaystyle\frac{(b+c)^p-b^p}{p} = \frac{\sum_{k=1}^{p-1}{ p \choose k}c^{k}b^{p-k} + c^p}{p}

Since n is an integer, then c^p is divisible by p.

Thus c is also divisible by p , since p is prime.

Then { p \choose k}c^{k}b^{p-k} is divisible by p^2, \forall 1\leq k < p

and c^p is also divisible by p^2 since p \geq 2.

Hence n is divisible by p, that means \displaystyle\frac{a^p-b^p}{p^2} is an integer

Elementary Classical Analysis

Posted by —

0

Elementary Classical Analysis (2nd Edition) by Jerrold E. Marsden, Michael J. Hoffman; Language: English; ISBN: 0716721058

elementary classical analysis

I hope these solutions are correct. :P

  • Solution of Exercise for Chapter 1(update 2006/07/15)
  • Solution of Exercise for Chapter 2(update 2006/07/15) not completed
  • Solution of Exercise for Chapter 3(update 2006/07/15) not completed
  • ERRATA