MAS 6217 Number Theory and Cryptography (Yiu) Fall 2017 Exercise B (Solution)

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1 MAS 6217 Number Theory and Cryptography (Yiu) Fall 2017 Exercise B (Solution) Solution to Problem B1. Let(a n ) be a sequence of numbers defined recursively by a n+1 = a 2 n a n +1, a 1 = 2. Show that for distinct indices i and j,gcd(a i,a j ) = 1. Hence, deduce that there are infinitely many prime numbers. Solution. Note that a n+1 1 = a n (a n 1). Ifj > i, then Rewriting this in the form a j 1 = a j 1 (a j 1 1) =. = a j 1 a j 2 a i (a i 1). a j a j 1 a j 2 (a i 1)a i = 1, we see that a i anda j are relatively prime. Note that a n is always positive. It is equal to 1 if and only ifa n 1 = 1. Sincea 1 = 2,a n > 1 for everyn. The prime divisors of a i anda j are all distinct. This shows that there are infinitely many prime numbers. 1

2 Solution to Problem B2. (a) Prove that d(n) is odd if and only if n is a square. Solution. Letn = k j=1 pa j j be the prime factorization of n. d(n) = k (a j +1) j=1 is odd if and only if each factor 1 + a j is odd. This means a j is even. Writing a j = 2b j, we have 2 k k n = p 2b j j = p b j j. j=1 (b) Find the least number n withd(n) = 100. Answer = j=1

3 Solution to Problem B3. (a) Prove that a power of a prime number cannot be a perfect number. Solution. Letp k,k 1, be a perfect number. a contradiction. p k = 1+p+p 2 +p k 1 < p p k 1 = p k, (b) For a positive integer n, show that if σ(n) is prime, then so isd(n). Solution. Sinceσ is a multiplicative function, ifσ(n) is prime, thennmust be a prime power. Writen = p k, so thatσ(n) = σ(p k ) = 1+p+ p k, and d(n) = 1+k. If1+k = ab, then1+p+ +p k is divisible by 1+p a and is not a prime.

4 Solution to Problem B4. Prove that if n is a perfect number, then 1 d = 2. Solution. Ifnis a perfect number, then d n d = 2n. Writing d = n d, we haved n and n d = 2n = 1 d = 2 d n d n by cancelling n. d n

5 Solution to Problem B5. Two numbers m and n are called amicable if σ(m) = σ(n) = m + n. (a) Verify that 220 and284 are amicable. (b) Define two sequences (p n ) and(q n ) by p n =3 2 n 1, q n =9 2 2n 1 1, n = 1,2,... Show that ifp n 1,p n, andq n are prime numbers, then a = 2 n p n 1 p n, b = 2 n q n are amicable. (c) Make use of (b) to find three pairs of amicable numbers. Solution. (a) σ(220) = σ( ) = σ(2 2 )σ(5)σ(11) = = 504, σ(284) = σ(2 2 71) = σ(2 2 )σ(71) = 7 72 = 504. Therefore, σ(220) = σ(284) = 504 = ; the two numbers are amicable. (b) If p n 1,p n, and q n are prime numbers, they are distinct. a+b = 2 n (p n 1 p n +q n ) = 2 n ((3 2 n 1 1)(3 2 n 1)+9 2 2n 1 1) = 2 n ( n n 3 2 n n 1 1) = 2 n ( n 3 2 n 1 (2+1)) = 2 n 3 2 (2 2n 2 n 1 ) = 2 2n (2 n+1 1); σ(a) = σ(2 n p n 1 p n ) = σ(2 n )σ(p n 1 )σ(p n ) = (2 n+1 1)(p n 1 +1)(p n +1) = (2 n+1 1)(3 2 n 1 )(3 2 n ) = 2 2n (2 n+1 1), σ(b) = σ(2 n q n ) = σ(2 n )σ(q n ) = (2 n+1 1)(q n +1) = (2 n+1 1) 9 2 2n 1 = 2 2n (2 n+1 1). Sinceσ(a) = σ(b) = a+b,aand b are amicable. (c) Three pairs of amicable numbers: n p n 1 p n q n a b σ(a) = σ(b) = a+b

6 Solution to Problem B6. Find all sequences of49 consecutive integers whose squares add up to a square. Solution. Take the49 numbers to be n 24, n 23,...,n 1, n, n+1,...,n+23, n+24. We requiren > 24. The sum of their squares is 49n 2 +2( ) = 49n = 49(n ). Therefore we requiren = y 2 for some integer y. y 2 n 2 = 200 = (y n)(y +n) = 200. y andnmust be of the same parity, andy n andy+n are both even. y n 2 y+n 2 = 50. Sinced(50) = d(2 5 2 ) = 2 3 = 6, we need only consider three possibilities: y n 2 y+n 2 n y (i) (ii) (iii) Since only in case (i) is n > 24, there is only one sequence of 49 consecutive integers whose sum is a square: with n = 49, we have = = Remarks. (1) There is only one sequence of 121 consecutive integers the sum of whose squares is the square of an integer: = (2) There is only two sequences of 169 consecutive integers the sum of whose squares is the square of an integer: = , =

7 Solution to Problem B7. Let p be a prime. Find the highest power of p dividing the product of the first n odd integers (2n 1). Solution. Clearly, this product is not divisible by 2. Letpbe an odd prime. ν 2 (1 3 5 (2n 1)) = 0. ν p (1 3 5 (2n 1)) ( ) (2n)! = ν p (n!) 2 = ν p ((2n)!) 2ν p (n!) = 2n α p(2n) p 1 = 2α p(n) α p (2n), p 1 2 n α p(n) p 1 whereα p (m) is the sum of the digits in the basepexpansion of m.

8 Solution to Problem B8. For a real number x > 1, prove that there is a primepsuch that x < p < 2x. Solution. Letx > 1 be a real number. We need only consider the case whenxis not an integer. x < x < x +1 2 x 2x < 2x. By Bertrand s hypothesis, there is a primepsatisfying x < p < 2 x. Such a primepmust satisfyp x +1 > x andp < 2 x < 2x, and is strictly between x and 2x.

9 Solution to Problem B9. Use Bertrand s hypothesis to prove that for n > 1, H n := n is not an integer. Solution. This is clearly true forn = 2,3. We shall assumen 4. By Bertrand s hypothesis, there is a primepsatisfying n+1 2 < p < 2 n+1 2. For this, n n+1 n+1 2 < p < 2 n Note that p n and2p > n. This shows that ν p (n!) = 1. Remark. The same result can be obtained by Problem B7 with x = n 2. Suppose, for a contradiction, that H n were an integer. Clearing denominators by multiplying by n!, we have n n! H n = 1 2 ˆk n. In this equation, every term, except is divisible by p, a contradiction. k=1 1 2 ˆp n,

10 Solution to Problem B10. Letp n denote then-th prime number. Find the smallest value of n > 1 for which p 1 p 2 p n 1 is not a prime number. n P 1 p 2 p n 1 Prime? = 5 yes = 29 yes = 209 no Find the smallest value ofn > 3 for which n!+1 is a prime number. n n! + 1 Prime? 4 25 = 5 2 False = 11 2 False = False = 71 2 False = False = False = False True Find the smallest value ofn > 7 for which n! 1 is a prime number. n n! 1 Prime? = False = False = False = False True

M381 Number Theory 2004 Page 1

M381 Number Theory 2004 Page 1 M81 Number Theory 2004 Page 1 [[ Comments are written like this. Please send me (dave@wildd.freeserve.co.uk) details of any errors you find or suggestions for improvements. ]] Question 1 20 = 2 * 10 +