Lecture 14 Enoch Cheung October 15 2013 1 Claim For all n N 1 n has a prime factorization We will prove this by strong induction on n Base case n 2 is prime so 2 is its prime factorization Inductive case Suppose k n k has a prime factorization then consider n If n is prime n is the prime factorization Otherwise n is not prime so n ab where 1 a b n By inductive hypothesis a b has prime factorizations a p1 p2 pk and b q1 q2 ql Then n ab p1 p2 pk q1 q2 ql is a prime factorization 2 Claim n N n 12 a b Z 0 n 4a 5b Base cases Four cases 12 4 3 13 4 2 5 14 4 5 2 15 5 3 Inductive hypothesis For some n 12 suppose 12 k n a b Z 0 k 4a 5b Then suppose n 15 so we are not in one of our base cases then 12 n 4 n so by inductive hypothesis a b Z 0 such that n 4 4a 5b Therefore n 4 a 1 5b 3 Suppose I have variable proposition P m n defined on N N and I know 1 P 1 1 holds 2 m N P m 1 P m 1 1 and 3 m n N P m n P m n 1 For what values of m n N N can I conclude P m n holds By induction on m using 1 as base case and 2 as inductive step we can conclude that m N P m 1 holds Now for a chosen m we can induct on n using what we just showed as a base case and 3 as our inductive step we can show that n N P m n Therefore we have shown that m n N P m n 4 Define the sequence a0 2 a1 2 an 2an 1 8an 2 for n 2 Prove by induction that an 4n 2 n for all n Z 0 Base cases a0 2 1 1 40 2 0 a1 2 4 2 41 2 1 Inductive case Suppose for some n 2 that k n ak 4k 2 k then an 2an 1 8an 2 2 4n 1 2 n 1 8 4n 2 2 n 2 2 4n 1 2 n 2 4n 1 2 2 n 2 2 4n 1 2 1 2 n 4n 2 n 5 Consider the following equation 4x4 2y 4 z 4 In this problem you will prove that this equation has no solution x y z N3 by descent a AFSOC that x y z N3 is such a solution and suppose further that this solution has the smallest value of x amongst all solutions b z is even because z 4 2 2x4 y 4 is even and z even z 2 even z 4 even c y is even because z is even so let z 2k for k Z then 2y 4 2k 4 4x4 so y 4 23 k 4 2x4 which is even so y is even d x is even because if y z are even let y 2l and z 2k for k l Z then 4x4 z 4 2y 4 2k 4 2 2l 4 4 22 k 4 4 23 l4 so x4 22 k 4 23 l4 is even so x is even e Note therefore that a b c x2 y2 z2 N3 is also a solution because x y z are even and 4x4 2y 4 z 4 4x4 2y 4 z4 x y z 4 4 2 4 4 4 4 2 2 2 2 2 1 f Therefore since x2 x so x y z is not a solution with x being smallest amongst all solutions which is a contradiction to a so there are no solutions in N3 g Note that if x y z Z3 is a solution then x y z Z 0 3 is a solution since x 4 x4 etc Therefore we showed that x N so x Z 0 either so x 0 Therefore the only possible 3 0 3 solution is inside Z with x 0 Thus suppose 0 y 2z 2 Z is such a solution then 4 4 2y z so taking the positive square roots we have 2y z because clearly both sides must be positive so we can take the square root and 2y 2 and z 2 must be positive so we are using the 2 positive square root so 2 yz 2 is a contradiction because 2 is irrational Therefore y 0 so z 4 4 04 2 04 0 which means that the only solution is x y z 0 0 0 Z3 the trivial solution 2