Solutions to Homework Set 2

1.2.5that if GCD(a, c) =1andGCD(b, c) = 1, then GCD(ab, c) =1.Prove4Proof.LetS = {common divisors ofa andb}T = {common divisors ofa and(a + b)}We will show that these two sets coincide.Suppose s ∈ S. Then there exist x, y ∈ Z such thata = xsb = ysThus,a + b = sx + sy = s(x + y) ,and so a + b is divisible by s. Soany s ∈ S is also an element of T.Suppose t ∈ T. Then there exist u, v ∈ Z such thata = uta + b = vt .Hence,b = vt − ut = t(v − u) ,and so b is also divisible by t. So any element t ∈ T is also an element of S.Thus, S = T. SoGCD(a, b) = Max(S) = Max(T) =GCD(a, a + b) .Proof.Suppose GCD(a, c) = 1 and GCD(b, c) = 1. Then by Theorem 1.3, there exists integers u, v, x, y such that= ua + vc1= xb + yc1But then1 = 1· 1= (ua + vc)(xb + yc)= (ux) ab +(uay + vxb + vyc) cThus,(4)1=u ′ (ab) +v ′ cwithu ′ = uxv ′ = uay + vxb + vyc .

of all, it is clear that if c and b have no common factors, and t is a factor of b, then c and t have noFirstfactors. Put another way; if t | c and GCD(c, b) = 1 then GCD(t, b) =1.common1.2.6Prove that if a, b, u, v ∈ Z are such that au + bv = 1, then GCD(a, b) =1.(a)note that the condition au + bv = 1 implies that a and b cannot both be zero. According to CorollaryFirstan integer d is the greatest common divisor of a and b if and only if1.4,d | a and d | b(i)if c | a and c | b, then c | d.(ii)now note that the right hand side is divisible by d but the left hand side is not, since d is presummedButbe greater than 1. Hence we have a contradiction unless d =1.to5Now let t be any common divisor of ab and c. Then, by definition, there exists s, t ∈ Z such thatab = rtc = stSo we can rewrite (4) as1=u ′ rt + v ′ st =(u ′ r + v ′ s) t ;from which it is clear that t | 1. Hence, t = ±1. Hence the greatest common divisor of ab and c is 1.∗∗∗∗∗∗∗∗Here is an alternative proof.Now suppose that d = GCD(c, ab). Then d ≥ 1 and there exist integers x and y such thatc = xdab = yd .Since d | c and GCD(a, c) = 1, by the remark above above, we have GCD(t, a) =1.Similarly, t divides c and GCD(b, c) = 1 implies GCD(t, b) =1.Now we apply Theorem 1.5.t | ab and GCD(t, a) =1 ⇒ t | b.But GCD(t, b) = 1. Hence t =1.Proof.Suppose now that d = GCD(a, b) > 1. Thena = sdb = td .But then we have1=sdu + tdv = d(su + tv)(b) Show by example that if au + bv = d>0, then GCD(a, b) neednotbed.Example.

1.3.1p be an integer other than 0, ±1. Prove that p is prime if and only if for each a ∈ Z, eitherGCD(a, p) =1LetProof.⇒∈{±1, ±p}. Hence either GCD(a, p) =1orGCD(a, p) =|p|. In the latter case we have p | a. So eithersp) =1orp | a.GCD(a,p has the property that for every integer a, either GCD(a, p) =1orp | a. If p is not prime thenAssumeexist s, t ∈ Z such thattheret is a divisor of p, GCD(t, p) =t ≠ 1. Therefore, p | t. But this is impossible since |t| < |p|. Hence pSincebe prime.must1.3.2p be an integer other than 0 ± 1 with this property: Whenever b and c are integers such that p | bc,Letp is an integer ≠0, ±1 such that whenever p | bc then p | b or p | c. Let s be a divisor of p. ThenSuppose= sq for some integer q and we havep6Take a =3,u =1,b =3,v = 1. Thenau + bv =5butGCD(3, 2) = 1 .Problems from Section 1.3or p | a.If p is prime then the only divisors of p are ±1 and±p.So if s is a common divisor of a and p, then⇐(Proof by Contradiction)p = stand1 < |s| ≤|t| < |p| .then p | c or p | b. Prove that p is prime.Proof.sq | bc ⇒ sq | b or sq | cIn particular, taking b = s and c = q, wehavesq | s or sq | q .But this implies eithers = ±1 and q = ±pors = ±p and q = ±1 .Hence the only divisors of p are ±1 and ±p; and so p is prime.

1.3.3that if every integer integer n>1 can be written in one and only one way in the formProven = p 1 p 2 ···p rTheorem 1.11, we know that there exists a prime factorization of n that is unique up to changes in theByof factors and flips in the sign of pairs of factors. The statement above just removes the remainingorderin the conclusion of Theorem 1.11. All prime factors are now required to be positive, so thereambiguitycannot flip the sign of terms; and the order of factors is fixed to coincide with their normal ordering asone1.3.4that if p is prime and p | a n , then p n | a n .Proveto Corollary 1.9, if p | a n , then p | a since a n = a · a · a ·····a. Butthen a = pq for some q ∈ Z.Accordinga n = p n q n ,andsop n divides a n .Hence1.3.5Prove that there exist no nonzero integers a, b such that a 2 =2b 2 .(a)a = p 1 p 2 ···p rthe two integers a 2 and 2b 2 have, respectively, an even number and an odd number of prime factors,Since 2 can not equal 2b 2 .a√√rational; i.e., 2= a bis 2a, b ∈ Z, b ≠ 0. Thenwith√= a 2bas we have just seen cannot be satified by any non-zero integers a and b. Hence we have a contradiction.which √ 2 can not be rational.So7where the p i are positive primes such that p1 ≤ p2 ≤···≤p r .Proof.integers.Proof.Proof.According to the Fundamental Theorem of Arithmetic, a and b have prime factorizations of the formb = q 1 q 2 ···q s .But thena 2 = p 1 p 1 p 2 p 2 ···p r p r (2r prime factors)2b 2 = 2q 1 q 1 q 2 q 2 ···q s q s (2s + 1 prime factors)(b) Prove that √ 2 is irrational.Proof.Supposean integer. Squaring both sides of this equation we getis 2 = a 2 2b