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Chapter31
MichelleBodnar, Andrew Lohr
April 12,2016
Exercise 31.1-1
By the give equation, we ca write c = 1·a+b,with 0 ≥b a. By the
definitio ofremaindersgive just belowthe divisio theorem, this meansthat
bisthe remainderwhe c is dividedbya,that is b=c mod a.
Exercise 31.1-2
Suppose that there are only finitely many primesp ,p ,... ,p . The p =
1 2 k
p p ···p +1isn’tprime,sothere mustbe somep whichdividesit. However,
1 2 k i
p ·(p ···p p ···p )p andp ·(p ···p p ···p +1)p,sop can’t
i 1 i−1 i+1 k 1 i−1 i+1 k i
dividep. Sincethis holds for anychoiceof i,weobtai a contradiction. Thus,
there areinfinitely manyprimes.
Exercise 31.1-3
a|bmeansthereexists k ∈Zsothat k a=b. b|cmeansthereexists k ∈Z
1 1 2
so that k b = c. This means that (k k )a = c. Since the integers are a ring,
2 1 2
k k ∈Z,so,wehavethat a|c.
1 2
Exercise 31.1-4
Letg =gcd(k,p). The g|kandg|p. Sincep isprime,g =p org =1. Since
0k p,g p. Thus,g =1.
Exercise 31.1-5
By Theorem 31.2, since gcd(a,n) = 1, there exist integers p,q so that
pa +qn = 1, so, bpa+bqn = b. Since n|ab, there exists a integer k so that
kn =ab. This means that knp+pqn = (k+q)pn =b. Sincen divides the left
hand side, it mustdividethe righthand side aswell.
Exercise 31.1-6
1
p! p(p−1)···(p−k+1)
p
Observethat = = . Letq=(p−1)(p−2)···(p−
k k!(p−k)! k!
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