Chapter.1: Induction Monday, July 1 Fermat s Little Theorem Evaluate the following: 1. 1 (mod ) 1 ( ) 1 1 (mod ). (mod 7) ( ) 8 ) 1 8 1 (mod ). 77 (mod 19). 18 (mod 1) 77 ( 18 ) 1 1 (mod 19) 18 1 (mod ) and 18 (mod ), so solving the simultaneous equations (by whatever method you like) gives 18 (mod 1).. (mod 1) (mod ) and (mod 7), so solving the two equations gives (mod 1).. 100 (mod ) 100 1 (mod ) and 100 1 (mod 11), so solving the two equations gives 100 1 (mod ). (Hard) A composite number n is called a Carmichael number b n 1 1 (mod n) for every number b such that gcd(b, n) 1 (their existence is unfortunate, since it means that we cannot use FLT to tell for certain whether a number is prime). Prove: There is one and only one Carmichael number of the form p q, where p and q are prime numbers. We know that if n pq is a Carmichael number and gcd(b, n) 1 then b pq 1 1 (mod pq) b pq 1 1 (mod ) b pq 1 1 (mod p) b pq 1 1 (mod q) Using Fermat s Little Theorem on the last three equations in turn gives us pq 1 p 1 pq 1 q 1 pq 1 1
The first just tells us that p and q must be odd. Then since pq 1 pq q + q 1 q(p 1) + q 1 (and similarly pq 1 p(q 1) + p 1), we can conclude p 1 q 1 q 1 p 1 Suppose (without loss of generality) that p < q. Then since q 1 p 1 < q 1, we know that either q 1 p 1 or (q 1) p 1. The first possibility would give q p, contradicting the given that p was prime. Therefore (q 1) p 1. We can then substitute this into the first statment: p 1 q 1 q + ((q 1))+ (p 1)+, so (p 1) 9 p + 1/, or p 9p + 1, or p 9p + 1 (p ) p + 9. Since p p + 9 means that p p + 9, we must have p 11. Since p, checking the other cases, 7, and 11 show that p 11 is the only option. Therefore q 17, and the only Carmichael number of the form pq is 11 17 1. Induction 1. Prove that 1 + + + + n n(n + 1)(n + 1) for n 0. Base case: it works for n 0 since 0 0(0 + 1)(0 + )/. Inductive step. Suppose that the formula works for n. Then (1 + + + n ) + (n + 1) n(n + 1)(n + 1)/ + n + n + 1 ( ) n(n + 1). Prove that 1 + + + + n for n 0. Base case: it works for n 0. Inductive step: suppose it works for n. Then n + n + n + n + 1n + n + 9n + 1n + (n + 1)(n + )(n + ) (1 + + + n ) + (n + 1) n (n + 1) + n + n + n + 1 n + n + n + n + 1n + 1n + n + n + 1n + 1n + (n + 1) (n + ). Prove that 1 1! +! + + n n! (n + 1)! 1 for n 1. Base case: it works for n 1. Inductive step: suppose it works for n. Then (1 1! +! + + n n!) + (n + 1) (n + 1)! (n + 1)! 1 + [(n + ) (n + 1)! (n + 1)!] (n + )! 1
. Find a closed form for n k1 ( 1)k k and prove that it is correct. The first few terms are 1,,, 10, 1,..., so guess that the formula is ( 1) n n(n + 1)/. Base case: The formula works for n 1. Inductive step: suppose that it works for n. Then ( 1) k k k1 n ( 1) k k + ( 1) (n + 1) k1. For what integers is n n true? Prove it. True for n 0, n 1, but also for n 10. Base case: 10 10 1000 10. Inductive step: suppose that n n. Then ( 1) n n(n + 1)/ + ( 1) (n + n + 1) ( 1) n + n + n n ( 1) n + n + (n + 1)(n + ) ( 1) n n + n n + n n + 10n n + n + n + 1 (n + 1) The step n + n n + 10n relied on the fact that n 10.
From to many 1. Given that ab ba, prove that a n b ba n for all n 1. (Original problem had a typo.) Base case: a 1 b ba 1 was given, so it works for n 1. Inductive step: if a n b ba n, then a b a(a n b) aba n baa n ba.. Given that ab ba, prove that a n b m b m a n for all n, m 1 (let n be arbitrary, then use the previous result and induction on m). Base case: if m 1 then a n b ba n was given by the result of the previous problem. Inductive step: if a n b m b m a n then a n b m+1 a n b m b b m a n b b m ba n b m+1 a n.. Given: if a b (mod m) and c d (mod m) then a + c b + d (mod m). Prove: if a i b i (mod m) for i 1,,..., n, then n a i n b i (mod m). Base case: When n the formula a + c b + d (mod m) was already given. Inductive step: Supposing the formula works for n, we get n a i ( a i ) + a n b i + b. (Calculus) Suppose we know that d dx x 1 and that for any functions f and g, (fg) f g + fg. Prove that d dx xn nx n 1 for all n 1. d Base case: when n 1, dx x1 1 1 x 0. d Inductive step: If dx xn nx n 1, then. Prove: n A i n A i. b i d dx x (x x n ) x x n + (x n ) x x n + nx n 1 x (n + 1)x n Base case: When n A B A B is given by one of DeMorgan s Laws.
Inductive step: Suppose the formula works for n. Then A i n A i A n A i A n A i A A i Recursion 1. Define a sequence a n by a 0 1, a 1 and a n a n 1 + a n for n. Find a. Prove that a n n+ + ( 1) n. a 0 a 1 a a a a a 1 11 1 8 Base case for proof by induction: The formula works for n 0 and n 1. Inductive step: Suppose that the formula works for n AND n + 1. Then a n+ a + a n n+ + ( 1) n+ + ( 1) n n+ + ( 1) n+ + n+ + ( 1) n Note that this time we needed to use the formula for both a and a n, so we needed to prove two base cases.. Define a sequence a n by a 0 1, a n a n 1 + 1 if n 1. Find a non-recursive formula for a n and prove that it is correct. The sequence goes 1,, 7, 1, 1,... guess that it is equal to 1. Prove the base case: it works for n 0. Inductive step: If it works for n, then a a n + 1 ( 1) + 1 n+ + 1 n+ 1.
. Prove: gcd(f, f n ) 1 for all n 0. Proof: gcd(f 0, f 1 ) gcd(0, 1) 1 for the base case n 0. Inductive step: use the fact that gcd(a, b) gcd(a b, b). Then if the proposition holds for n, we have gcd(f n+, f ) gcd(f n+ f, f ) gcd(f n, f ) 1.. Prove that f 1 + f + + f n f n f for n 1. Base case: it works for n 1 since f 1 1 1 f 1 f. Inductive step: if the formula holds for n, then (f 1 + f + + f n) + f f n f + f f f (f n + f ) f f n+. Prove that f 1 + f + + f n 1 f n for n 1. (Original problem had a typo.) Base case: f 1 1 f when n 1. Inductive step: if the formula holds for n, then. Show that f f n 1 f n ( 1) n for n 1. (f 1 + f + + f n 1 ) + f f n + f f n+ Base case: when n 1, we have f f 0 f 1 0 1 ( 1) 1. Inductive step: If the formula holds for n then f n+ f n f (f n + f )f n f f n + f f n f f n + f (f n f ) f n + f ( f n 1 ) (f f n 1 f n) ( 1) n ( 1) 7. Prove that f n (α n β n )/, where α (1 + )/ and β (1 )/. (Hint: both α and β satisfy the equation x x + 1). Proof: It holds for f 0 and f 1, base cases n 0 and n 1. Inductive step: if it holds for n AND n + 1 then f n+ f + f n α β + αn β n αn (α + 1) β n (β + 1) αn+ β n+
8. Prove that f m+n f m 1 f n + f m f. (fix n arbitrarily, then use induction on m) Base case: when m 1 the formula becomes f f 0 f n + f 1 f, which is true because f 0 0 and f 1 1. Inductive step: Suppose the formula holds for m. Then f (m+1)+n f m+() f m 1 f + f m f n+ f m 1 f + (f m f + f m f n ) (f m 1 + f m )f + f m f n f m f n + f m+1 f f (m+1) 1 f n + f m+1 f 9. Prove (now using induction on n) that f m f mn for all n 1. Base case: When n 1 this is just f m f m, which is clearly true. Inductive step: suppose f m f mn. Then f m() f mn+m f mn 1 f m + f mn f m+1. Since f m and (by the inductive hypothesis) f mn are both divisible by f m, the linear combination (and therefore f m() ) is also divible by f m. 10. Prove that gcd(f m, f n ) f gcd(m,n). Let n qm + r. Since f m f qm (from the previous problem), we know that gcd(f m, f n ) gcd(f m, f qm+r ) gcd(f m, f qm 1 f r + f qm f r+1 ) gcd(f m, f qm 1 f r ), Then since f m f qm but gcd(f qm, f qm 1 ) 1, we can conclude that gcd(f m, f n ) gcd(f m, f qm 1 f r ) gcd(f m, f r ). This allows us to use a process similar to the Euclidean Algorithm and continue until we hit the greatest common divisor. In particular, this means that if p is a prime number, then f p shares a common divisor with f n if and only if p n (and if p n then f p f n ). In particlar, we know that f, so (since is prime), the even Fibonacci numbers will be precisely those of the form f k for k Z. 7