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21441, HW03, 09.12.2003-09.19.2003, solutions 1. Find the smallest x ≥ 0 such that x ≡ 2 (mod 3),x ≡ 1 (mod 5), and x ≡ 1 (mod 7) simultaneously. Here’s how to do it using a technique you already know. The solutions are the numbers of the form x = 3a + 2 = 5b + 1 = 7c + 1. Solving 3a − 5b = −1 and 5b − 7c = 0 simultaneously gives a = 23 + 35k, b = 14 + 21k, and c = 10 + 15k where k is arbitrary. Thus the solutions are the numbers x = 71 + 105k. So x = 71 is the smallest solution greater than or equal to zero. Here is the method given in the proof of the Chinese Remainder The- orem. To solve x ≡ ai (mod mi ) for 1 ≤ i ≤ r where the mi are pairwise relatively prime, solve Mi bi ≡ 1 (mod mi ) for 1 ≤ i ≤ r where Mi = m/mi and m = m1 m2 · · · mr . Then x ≡ r Mi bi ai i=1 (mod m). So m = 105, M1 = 35, M2 = 21, M3 = 15. We have 1 ≡ M1 b1 ≡ 35b1 ≡ 2b1 (mod 3) or b1 ≡ 2 (mod 3). We also have 1 ≡ M2 b2 ≡ 21b2 ≡ b2 (mod 5) and 1 ≡ M3 b3 ≡ 15b3 ≡ b3 (mod 7). Thus x ≡ M1 b1 a1 + M2 b2 a2 + M3 b3 a3 ≡ 35 · 2 · 2 + 21 · 1 · 1 + 15 · 1 · 1 ≡ 71 (mod 105). Thus x = 71 is the smallest solution greater than or equal to zero. 2. (a) Show that 61! + 1 ≡ 63! + 1 ≡ 0 (mod 71). The point here is to reduce these calculations to something hope- fully a lot simpler than 60 to 70 multiplications. We use the following fact: if (a, m) = 1, then ab ≡ ac (mod m) ⇔ b ≡ c (mod m). 61! + 1 ≡ 63! + 1 (mod 71) iﬀ 61! ≡ 63! (mod 71) iﬀ 1 ≡ 62 · 63 (mod 71). But 62 · 63 ≡ (−9)(−8) ≡ 72 ≡ 1 (mod 71). Since 71 is a prime we have 70! ≡ −1 (mod 71) by Wilson’s theorem. Thus 63! + 1 ≡ 0 (mod 71) iﬀ 63! ≡ 70! (mod 71) iﬀ 1 ≡ 64(65)(66)(67)(68)(69)(70) ≡ −7! (mod 71). But 7! ≡ 5040 ≡ −1 (mod 71). (b) Prove the converse of Wilson’s Theorem: if n ≥ 2 and (n − 1)! ≡ −1 (mod n), then n is prime. Suppose n is not prime. Then n = ab where 1 < a, b < n and so n = ab|(n − 1)!. Thus (n − 1)! ≡ −1 (mod n). 1 3. Show that 2730|(n13 − n) for all n. Note 2730 = 2 · 3 · 5 · 7 · 13, where all the factors are realtively prime, so n13 ≡ n (mod 2730) iﬀ n13 ≡ n (mod a) for a = 2, 3, 5, 7, and 13. By Fermat’s Little theorem, we have n13 ≡ n (mod 13), n13 ≡ n7 n6 ≡ nn6 ≡ n (mod 7), n13 ≡ (n5 )2 n3 ≡ n2 n3 ≡ n (mod 5), n13 ≡ (n3 )4 n ≡ n4 n ≡ n3 n2 ≡ nn2 ≡ n (mod 3), and n13 ≡ (n2 )6 n ≡ n6 n ≡ (n2 )3 n ≡ n3 n ≡ (n2 )2 ≡ n2 ≡ n (mod 2). 4. Show that if p is a prime and p is not 2 or 5 then p divides inﬁnitely many of the integers in the sequence: 9, 99, 999, 9999, . . .. The members of the sequence are the numbers an = 10n − 1 for n ≥ 1. Thus the claim is that for every prime p ≥ 7, 10n ≡ 1 (mod p) has inﬁnitely many solutions n ≥ 1. Indeed if n = k(p − 1) for any k ≥ 1, then 10n = (10p−1 )k ≡ 1k = 1 (mod p) by Fermat’s theorem. 5. Prove that p|(p − 1)!(1 + 1/2 + 1/3 + · · · + 1/(p − 1)) if p ≥ 3 is a prime. Here’s one solution. For every number 1 ≤ a ≤ p − 1 there is a unique number 1 ≤ a ≤ p − 1 such that aa ≡ 1 (mod p). (The equation ax ≡ 1 (mod p) has a unique solution x mod p.) Thus (p − 1)!(1 + 1/2 + · · · + 1/(p − 1)) ≡ (p − 1)!(1 + 2 + · · · + p − 1) = (p − 1)!(1 + 2 + · · · + p − 1) (mod p). But if n is odd, (n − 1)/2 is an integer and so 1 + 2 + · · · + n − 1 = n((n − 1)/2) ≡ 0 (mod n). Here is a way that mimics the proof of Wilson’s theorem using La- grange’s theorem.. Suppose p ≥ 3 is a prime, then g(x) = (x − 1)(x − 2) · · · (x − (p − 1)) = xp−1 − σ1 xp−2 + · · · + σp−3 x2 − σp−2 x + σp−1 where the σk are all integers. For example: σp−1 = (p − 1)! and σp−2 = (p − 1)!(1 + 1/2 + · · · + 1/(p − 1)). Since g(x) has the roots 1, 2, . . . , p − 1 mod p so does h(x) = (xp−1 − 1) − g(x), by Fermat’s theorem. Since deg(h) ≤ p − 2 and h has ≥ p − 1 roots mod p, h must be identically zero mod p by Lagrange’s Theorem. Thus σk ≡ 0 (mod p) for 1 ≤ k ≤ p − 2. In particular p|σp−2 as claimed. In fact for p ≥ 5, we have p2 |σp−2 (this is called Wolstenholme’s the- orem). We have (p − 1)! = g(p) = pp−1 − σ1 pp−2 + · · · + σp−3 p2 − σp−2 p + (p − 1)! or σp−2 p = pp−1 − σ1 pp−2 + · · · + σp−3 p2 Since p|σk for 1 ≤ k ≤ p − 3, p3 divides every term on the right hand side of this equation and so p3 |pσp−2 . Thus p2 |σp−2 . 2 6. Problem 2.3.14 (page 72). Let f (x) = x3 + 2x − 3. By testing all possible roots, we ﬁnd that f (x) has 3 roots mod 9 and 2 roots mod 5. In fact f (x) ≡ 0 (mod 9) iﬀ x ≡ 1, 2, or −3 (mod 9)) and f (x) ≡ 0 (mod 5) iﬀ x ≡ 1 or −2 (mod 5). Since f (x) ≡ 0 (mod 45) iﬀ f (x) ≡ 0 (mod 9) and f (x) ≡ 0 (mod 5), we see that x is a root of f (x) mod 45 iﬀ x ≡ 1, 2, or −3 (mod 9) and x ≡ 1 or −2 (mod 5). Thus by the Chinese remainder theorem, f (x) has 2 · 3 = 6 roots mod 45. f (x) ≡ 0 (mod 45) ⇔ f (x) ≡ 0 (mod 9) and f (x) ≡ 0 (mod 5) ⇔ (x ≡ 1, 2, or − 3 (mod 9)) and (x ≡ 1 or − 2 (mod 5)) ⇔ (x ≡ 1 (mod 9) and x ≡ 1 (mod 5)) or (x ≡ 1 (mod 9) and x ≡ −2 (mod 5)) or (x ≡ 2 (mod 9) and x ≡ 1 (mod 5)) or (x ≡ 2 (mod 9) and x ≡ −2 (mod 5)) or (x ≡ −3 (mod 9) and x ≡ 1 (mod 5)) or (x ≡ −3 (mod 9) and x ≡ −2 (mod 5)) ⇔ x≡1 (mod 45) or x ≡ 28 (mod 45) or x ≡ 11 (mod 45) or x ≡ 38 (mod 45) or x≡6 (mod 45) or x ≡ 33 (mod 45) So the roots of f (x) mod 45 are 1, 6, 11, 27, 33, 38. (Alternatively you could have tried all 45 residues to see which were roots, but this way may be easier.) 3

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