Sols to Assignment 3
18. (a) When a = b, we have a = b · 1 + 0, so quotient = 1 and remainder = 0.
(b) When a = kb, we have a = b · k + 0, so quotient = k and remainder =
0.
(c) When 0 < a < b, we have a = b · 0 + a, so quotient = 0 and remainder
= a.
20. Divide n3 −1 by n+1 to get n3 −1 = (n+1)(n2 −n+1)−2. If n+1 | n3 −1,
then n + 1 must divide 2, so n + 1 = ±1, ±2. Therefore, n = −3, −2, 0, 1.
Each of these values works.
Projects
1. (a) When you divide an integer by 4, the remainder is 0, 1, 2, or 3. If the
integer is odd the remainder is 1 or 3. This means that the integer is of
the form 4k + 1 or 4k + 3.
(b) If both sets were finite, there would be a finite number of odd primes.
Since there is an infinite number of primes, there is an infinite number of
odd primes. (2 is the only even prime). Therefore at least one of these
sets is infinite.
(c) (i) Since
(1 + 4k1 )(1 + 4k2 ) = 1 + 4k1 + 4k2 + 16k1 k2 = 1 + 4k,
the product of two elements of S1 is no longer prime, but remains of the
form 4k+1.
(ii) This is essentially obvious; if one of the pj divided N, it would have to
divide 1.
(iii) N is odd and not divisible by any prime in S3 . Therefore all prime
factors of N come from S1. But by (i), any such number leaves a
remainder of 1 when divided by 4; whereas N leaves a remainder of 3.
So N cannot be a product of only elements of S1.
(iv) By (iii), N is divisible by a prime not in S1 . By (ii), this can’t occur.
So, S3 cannot be finite and must be infinite.
(d) Assume that q 1 = 5, 13, ..., qn are all the primes in S1 . Let
M = (2q1 q2 q3 · · · qn )2 + 1.
If p is a prime and p divides M, then p must be in S1 . So, p = qi for some
i. This is impossible, since when we divide M by any qi the remainder is
1,not 0. Therefore there must be an infinite number of primes in S1 .
(e) First notice that when we divide any integer by 6, the remainder is
0, 1, 2, 3, 4, or 5. If n is an odd integer, the remainder when divided
by 6 must be 1, 3, or 5. Therefore every prime except 2 and 3 is in T1
or T5 . Now, let m = 1 + 6k1 and n = 1 + 6k2 be two integers. Then
mn = 1 + 6k for some integer k, so a product of primes in T1 has the
form 6k + 1. Now, assume that T5 has only a finite number of primes and
let 5, 11, ..., pn be a list of all of them. If N = 6p1 p2 p3 · · · pn − 1, then N
cannot be divisible by any of these p's, so all the prime divisors of
N must be in T1. But, if N is a product only of elements in T1 , then N
would have the form 6k + 1, which is a contradiction. Therefore, T5 must
be infinite.
(f ) Assume that T1 contains a finite number of primes, 7, 13, ...pn . Let
N = 3p1 p2 · · · pn and let M = N 2 + N + 1. If a prime p | M , then p ∈ T1 .
But, when we divide M by 3 or by a prime in T1 , the remainder is 1 and
not 0. Therefore, T1 must contain an infinite number of primes.