230hw6sol - Math 230 E Fall 2011 Homework 6 Drew Armstrong...

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Unformatted text preview: Math 230 E Fall 2011 Homework 6 Drew Armstrong Problem 1. Without using Fermat’s little Theorem , prove the Freshman’s Binomial Theorem : For all a,b,p ∈ Z with p prime we have ( a + b ) p ≡ a p + b p mod p. Proof. By the Binomial Theorem we have ( a + b ) p = a p + p 1 a p- 1 b + p 2 a p- 2 b 2 + ··· + p p- 2 a 2 b p- 2 + p p- 1 ab p- 1 + b p , so we will be done if we can show that ( p k ) ≡ 0 mod p for all 0 < k < p ; in other words, that p divides the integer ( p k ) . Well, we know that p divides p !, hence it must divide the product p ! = ( p k ) ( k !( p- k )!). However, if 0 < k < p , then k !( p- k )! is a product of integers all of which are strictly smaller than p . If p divides k !( p- k )!, then by Euclid’s Lemma, p divides some number strictly smaller than itself, which is a contradiction to Lemma 2.11(iv) on page 25 of the text. Hence p does not divide k !( p- k )!. But then since p divides the product ( p k ) ( k !( p- k )!), Euclid’s Lemma implies that p divides ( p k ) ....
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This note was uploaded on 01/08/2012 for the course MATH 461 taught by Professor Armstrong during the Fall '11 term at University of Miami.

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230hw6sol - Math 230 E Fall 2011 Homework 6 Drew Armstrong...

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