Mathematics 4: Number Theory Problem Sheet 3 Workshop 26 Oct 2012 The aim of this workshop is to show that Carmichael numbers are squarefree and have at least 3 distinct prime factors. (1) (Warm-up question.) Show that n > 1 is prime iff an−1 ≡ 1 (mod n) for 1 ≤ a ≤ n − 1. If n is prime, then the result is true by Fermat’s Theorem. If n is composite, then, for a = p, a prime factor of n, the equation pn−1 + kn =1 has no solution, since the LHS is divisible by p. So result is true iff n prime. Recall that a positive integer is said to be squarefree if it is not divisible by the square of any prime number. Recall too that a Carmichael number is a composite number n with the property that for every integer a coprime to n we have an−1 ≡ 1 (mod n). (2) Proving that Carmichael numbers are squarefree. (a) Show that a given nonsquarefree number n can be written in the form n = pℓN for some prime p and integers N and ℓ with ℓ ≥ 2 and gcd(p, N)=1. (b) Show that (1 + pN)n−1 6≡ 1 (mod p2). (c) Deduce that Carmichael numbers are squarefree. (a) Take a prime p such that p2 | n. Then pℓ||n for some ℓ ≥ 2. So n = pℓN say, where N = n/pℓ is coprime to p. (b) Now (1+pN)n−1 ≡ 1+ pN(n − 1) (mod p2) ≡ 1 − pN (mod p2) 6≡ 1 (mod p2), by the Binomial Theorem and because p2 | n and gcd(N,p)=1. (c) Now take a =1+ pN. Then an−1 6≡ 1 (mod p2) by (b), so an−1 6≡ 1 (mod n). Hence, as gcd(a, n)=1, n is not a Carmichael number. (3) Proving that Carmichael numbers have at least 3 distinct prime factors. (a) Let p and q be distinct primes. Prove that if gcd(a, pq) = 1 then alcm(p−1,q−1) ≡ 1 (mod pq). 1 2 (b) Now let g be a primitive root (mod p) and h be a primitive root (mod q). Using g and h, apply the Chinese Remainder Theorem to specify an integer a whose order (mod pq) is (exactly) lcm(p − 1, q − 1). (c) Now suppose that p is the larger of the primes p and q. Calculate pq − 1 (mod p − 1) ∈{0, 1,...,p − 2}. Deduce that p − 1 ∤ pq − 1. (d) Use the above to show that there is an a with gcd(a, pq) = 1 and apq−1 6≡ 1 (mod pq). (e) Deduce from the above that a Carmichael number must have at least 3 distinct prime factors. (a) By Fermat’s Theorem we have ap−1 ≡ 1 (mod p) and aq−1 ≡ 1 (mod q), so that, taking appropriate powers, alcm(p−1,q−1) ≡ 1 (mod p) and alcm(p−1,q−1) ≡ 1 (mod q). Hence alcm(p−1,q−1) ≡ 1 (mod pq). (b): We simply choose a so that a ≡ g (mod p) and a ≡ h (mod q). This is possible, by CRT. Since g has order p − 1 modulo p and h has order q − 1 modulo q, we have that a has order p − 1 modulo p and order q − 1 modulo q. Hence a has order lcm(p − 1, q − 1) modulo pq. (c): Now pq − 1=(p − 1)q + q − 1 ≡ q − 1 (mod p − 1) 6≡ 0 (mod p − 1), as 0 < q − 1 <p − 1. Hence p − 1 ∤ pq − 1. (d): From (b) there is an a whose order (mod pq) is lcm(p−1, q−1), so that if gcd(a, p)=1 then from (a) we have that ak ≡ 1 (mod pq) iff k is a multiple of lcm(p−1, q−1). But pq−1 is not a multiple of lcm(p− 1, q − 1), since q − 1 ∤ pq − 1. So apq−1 6≡ 1 (mod pq). (e): Now the number a in (d) is clearly coprime to both p and q, i.e., gcd(a, pq)= 1. Hence, from (d), pq is not a Carmichael number. As Carmichael numbers are not prime and are not divisible by the square of any prime, they must have at least 3 prime factors. (4) (Cool-down question.) Suppose that a, k, ℓ, m, n ∈ N with ak ≡ 1 (mod m) and aℓ ≡ 1 (mod n). Prove that (a) alcm(k,ℓ) ≡ 1 (mod lcm(m, n)); (b) agcd(k,ℓ) ≡ 1 (mod gcd(m, n)). ′ (a) Let kk′ = lcm(k,ℓ). Then ak − 1 divides akk − 1 = alcm(k,ℓ) − 1, so that alcm(k,ℓ) ≡ 1 (mod m). Similarly, alcm(k,ℓ) ≡ 1 (mod n). Hence alcm(k,ℓ)− ep fp 1 is divisible by both m and n. Writing m = p p and n = p p , we see that, for each p prime, alcm(k,ℓ)−1 is divisibleQ by pmax(ep,fQp), and max(ep,fp) lcm(k,ℓ) so is divisible by p p = lcm(m, n). So a ≡ 1 (mod lcm(m, n)). Q 3 (b) From ak ≡ 1 (mod m) and aℓ ≡ 1 (mod n) and the fact that G = gcd(m, n) divides both m and n we have ak ≡ 1 (mod G) and aℓ ≡ 1 (mod G). Next, by the Extended Euclidean Algorithm we can find integers k1 and ℓ1 such that kk1 + ℓℓ1 = gcd(k,ℓ). Hence agcd(k,ℓ) = akk1+ℓℓ1 =(ak)k1 · (aℓ)ℓ1 ≡ 1k1 · 1ℓ1 ≡ 1 (mod G). Handin: due Friday, week 7, 2 Nov, before 12.10 lecture. Please hand it in at the lecture The squarefree part of n You are expected to write clearly and legibly, giving thought to the presentation of your answer as a document written in mathematical English. (5) (a) Show that every positive integer n can be written uniquely in the form n = 2 n1n2, where n1 is squarefree. Let us denote n1 by g(n), the squarefree part of n. (b) Prove that g(n) is a multiplicative function. (c) Find the Euler product for Dg(s). (d) Prove that Dg(s)= ζ(2s)ζ(s − 1)/ζ(2s − 2). 2k+1 k 2 2k k 2 (a) From p = p · (p ) and p = 1 · (p ) we see that we can take n1 to be the product of primes that divide n to an odd power, as then n1 2 is squarefree and n/n1 is a square, = n2 say. [2 marks] (b) For gcd(m, n)=1, g(m)g(n) = m1n1 with gcd(m1, n1)=1, so that 2 mn = m1n1(m2n2) , with m1n1 squarefree, giving g(mn)= m1n1 = g(m)g(n). [2 marks] (c) Now g(p2k+1)= p, g(p2k)=1 so that ∞ g(pk) p 1 = + pks pks pks Xk=1 kXodd k Xeven =(p/ps +1)(1+1/p2s +1/p4s + . ) =(1+1/ps−1)/(1 − 1/p2s). s−1 2s Hence Dg(s)= p(1+1/p )/(1 − 1/p ). [3 marks] (d) Then Q s−1 2s Dg(s)= (1+1/p )/(1 − 1/p ) Yp = (1 − 1/p2s−2)/((1 − 1/p2s)(1 − 1/ps−1)) Yp =ζ(2s)ζ(s − 1)/ζ(2s − 2). 4 [3 marks] Problems on congruences (6) Let m1,...,mn be pairwise relatively prime. Show that as x runs through the integers x = 1, 2, 3,...,m1m2 ··· mn, the n-tuples (x mod m1, x mod m2,...,x n mod mn) run through all n-tuples in i=1{0, 1,...,mi − 1}. Q ′ If for x, x ∈{1, 2, 3,...,m1m2 ··· mn} the n-tuples (x mod m1, x mod m2,...,x mod mn) and ′ ′ ′ (x mod m1, x mod m2,...,x mod mn) ′ ′ ′ were equal, then the n-tuple (x−x mod m1, x−x mod m2,...,x−x mod mn) ′ would be 0, so that x − x would be divisible by m1m2 ··· mn. This is impossible as they differ by less than this number. Hence all the n-tuples (x mod m1, x mod m2,...,x mod mn) for x =1, 2, 3,...,m1m2 ··· mn are different. As there are only m1m2 ··· mn of them, they must run through them all. (7) Show that the equation xy ≡ 2 (mod 19) has a solution in integers {x, y} iff x is congruent to a primitive root mod 19. Deduce that then y is uniquely specified mod 18. Now (easily checked) 2 is a primitive root (mod 19), so if x is not a primitive root, then xy certainly isn’t. On the other hand, if x is a primitive root, then the powers xy with gcd(y, 18) = 1 give all primitive roots, including 2. Also, if gcd(y, 18) > 1 then xy is not a primitive root. As x18 ≡ 1 (mod 19), y is uniquely specified (mod 18). (8) Wilson’s Theorem. This states that, for a prime p, we have (p − 1)! ≡ −1 (mod p). Prove Wilson’s Theorem in (at least!) two different ways. p−1 [Suggestions: (i) Factorize x − 1 over Fp. (ii) Try to pair up a ∈{1,...,p − 1} with its multiplicative inverse.] (i) Since, from Fermat’s Theorem, xp−1−1 ≡ (x − 1)(x − 2) . (x − (p − 1)) (mod p), Wilson’s Theorem follows on putting x =0. (ii) For any a ∈ {1, 2,...,p − 1} there’s an a′ ∈ {1, 2,...,p − 1} with aa′ ≡ 1 (mod p). Further, a′ = a iff a =1 or p − 1. Hence, in (p − 1)!, the numbers forming this product can be cancelled in pairs, apart from 1 and p − 1.