1. Section 3.5 - Problem 8
Let n be a positive integer. By the fundamental theorem of arithmetic we can factor n as
n = pe11 · · · pekk where the pi are distinct primes and ei ≥ 0 for each i. For each ei , by the
division algorithm write ei = 2fi + ri with ri ∈ {0, 1}. Then we have
n = (pf11 · · · pfkk )2 · pr11 · · · prkk
We have (pf11 · · · pfkk )2 is a square. I claim that pr11 · · · prkk is square-free. Suppose it were not
and had a square factor x2 | pr11 · · · prkk , with x 6= 1. Then some prime q divides x and so q 2
divides pr11 · · · prkk . This is a contradiction, as the only possible primes dividing pr11 · · · prkk are
the pi , and as ri ≤ 1, we have p2i does not divide pr11 · · · prkk .
Thus n can be written as the product of a square and a square-free integer.
2. Section 3.5 - Problem 10
Write a = pe11 · · · pekk and b = pf11 · · · pfkk where the pi are distinct primes and ei , fi ≥ 0. As
2
a3 | b2 we have 3ei ≤ 2fi and hence ei ≤ fi ≤ fi for each i. Thus a | b.
3
3. Section 3.5 - Problem 32 (5th edition). Problem 30 (6th edition)
For distinct primes pi we have
min{e1 ,f1 }
· · · pk
max{e1 ,f1 }
· · · pk
(pe11 · · · pekk , pf11 · · · pfkk ) = p1
[pe11 · · · pekk , pf11 · · · pfkk ] = p1
min{ek ,fk }
max{ek ,fk }
We use this to compute the greatest common divisors and least common multiples.
a) (2 · 32 53 , 22 33 72 ) = 2 · 32
(2 · 32 53 , 22 33 72 ) = 22 · 33 · 53 · 72
b) (2 · 3 · 5 · 7, 7 · 11 · 13) = 7
(2 · 3 · 5 · 7, 7 · 11 · 13) = 2 · 3 · 5 · 7 · 11 · 13
c) (28 36 54 1113 , 2 · 3 · 5 · 13) = 2 · 3 · 5
(28 36 54 1113 , 2 · 3 · 5 · 13) = 28 36 54 1113
d) (41101 4743 1031001 , 4111 4347 83111 ) = 4111
(41101 4743 1031001 , 4111 4347 83111 ) = 41101 4347 4743 83111 1031001
4. Section 3.5 - Problem 36 (5th edition). Problem 34 (the 6th edition)
Let us assume that a and b are positive (we will multiply each by ±1 to get all integer solutions
at the end).
Note that (a, b) = 18 = 2 · 32 and [a, b] = 540 = 22 · 33 · 5. Thus the only prime divisors of
a and b are 2, 3 and 5.
Write a = 2d2 3d3 5d5 and b = 2e2 3e3 5e5 , where d2 , d3 , d5 , e2 , e3 , e5 are non-negative integers.
The smaller of the two exponents, min(d2 , e2 ) equals 1 (the exponent of 2 in 18) and max(d2 , e2 )
equals 2 (the exponent of 2 in 540). After possibly interchanging a and b, we may assume d2 = 1
and e2 = 2.
Similarly, min(d3 , e3 ) = 2 (the exponent of 3 in 18), max(d3 , e3 ) = 3 (the exponent of 3 in
540), and min(d5 , e5 ) = 0 (the exponent of 5 in 18), min(d5 , e5 ) = 1 (the exponent of 5 in 540).
Thus (d3 , e3 ) = (2, 3) or (3, 2) and (d5 , e5 ) = (0, 1) or (1, 0). This gives rise to four possibilities:
(i) a = 21 · 32 = 18
b = 22 · 33 · 51 = 540,
(ii) a = 21 · 32 · 5 = 90
b = 22 · 33 = 108,
(iii) a = 21 · 33 = 54
b = 22 · 32 · 5 = 180, and
(iv) a = 21 · 33 · 5 = 270
b = 22 · 32 = 36.
All other possibilities are obtained from these four by interchanging a and b and multiplying
a or b (or both) by −1.
5. Section 3.5 - Problem 60 (50th edition). Problem 56 (6th edition)
We will argue by contradiction. Suppose there are finitely many primes of the form 6k + 5,
where k is a positive integer. Let them be p1 , . . . , pk . Let N = 6p1 · · · pk − 1. We have N > 1,
as p1 = 5 > 0. Let q be an arbitrary prime dividing N . By the division algorithm we have
q = 6a + r for some integer a and r ∈ {0, . . . , 5}.
If r ∈ {0, 2, 4} then q is even and hence is 2. But 2 does not divide N , as 2 divides N + 1.
Thus r 6= 2,
If r = 3, then 3 | q = 3(2a + 1) and thus q = 3. Again, 3 does not divide N , as 3 divides
N + 1. Thus q 6= 3.
If r = 5, then q is of the form 6k + 5. By our assumption p1 , . . . , pk are the only primes of
this form. Thus q = pi for some i. But pi divides N + 1, so does not divide N .
Thus every prime q dividing N is of the form 6a + 1 for some integer a.
For any two integers of this form we have (6a + 1)(6b + 1) = 6(6ab + a + b) + 1, which is also
in this form. As we showed above, N is the product of primes, say, q1 , . . . , qr (not necessarily
distinct), and each of these primes is of the form 6k + 1. Thus N = q1 . . . qr must also be of this
form. This is a contradiction, as N is of the form 6k + 5.
This contradiction shows that there are infinitely many primes of the form 6k + 5.
6. Section 3.5 - Problem 72 (5th edition). Problem 68 (6th edition)
Let pi be one of the primes in the list. Suppose i ≤ m, then we have pi | Q. Supose
pi | (Q + R), then pi | (Q + R − Q) = R. This is a contradiction, as pi does not appear in the
prime factorization of R. Thus pi does not divide Q + R. If m < i ≤ n, then pi | R, and by the
same reasoning we have pi does not divide Q + R. Thus none of the primes in the list divide
Q + R.
If there were finitely many primes p1 , . . . , pn , then as every integer > 1 has a prime factor,
one of the pi would have to divide Q + R. This is a contradiction and so there are infinitely
many primes.
7. Section 3.7 - Problem 2
a) As (3, 4) = 1 there are infinitely many solutions. As x = y = 1 gives a particular solution,
all solutions are given by x = 1 + 4n, y = 1 − 3n, where n is an integer.
b) As (12, 18) = 6 which does not divide 50, there are no solutions
c) First we find a particular solution to 30x + 47y = 1 by using the Euclidean algorithm.
74 = 30 + 17
30 = 17 + 13
17 = 13 + 4
13 = 4 · 3 + 1
Back substitution: 1 = 13−4·3 = 13−(17−13)·3 = 13·4−17·3 = (30−17)·4−17·3 =
30 · 4 − 17 · 7 = 30 · 4 − (47 − 30) · 7 = 30 · 11 − 47 · 7
Thus x0 = 11, y0 = −7 is a particular solution to 30x + 47y = 1. To get a particular solution
to 30x + 47y = −11, multiply both numbers by −11: x1 = −121, y1 = 77.
The general solution is thus x = −121 + 47t, y = 77 − 30t, where t ranges over the integers.
d) As (25, 95) = 5 which divides 970, there are infinitely many solutions. Divide both sides by
5: our equation reduces to 5x + 19y = 194.
x0 = 4, y0 = −1 is a particular solution to 5x + 19y = 1.
x1 = 194 · 4 = 776, y0 = −194 is a particular solution to
As x = 16, y = 6 gives a particular solution, all solutions are given by x = 776 + 19t,
y = −194 − 5t, where t ranges over the integers.
e) Euclidean algorithm:
1001 = 102 · 9 + 83
102 = 83 · 1 + 19
83 = 19 · 4 + 7
19 = 7 · 2 + 5
7=5·1+2
5=2·2+1
Thus (102, 1001) = 1, and the Diophantine equation 102x + 1001y = 1 has infinitely many
solutions. To find a particular solution, use back substitution:
1 = 5 − 2 · 2 = 5 − (7 − 5) · 2 = 7 · (−2) + 5 · 3 = 7 · (−2) + (19 − 7 · 2) · 3 = 19 · 3 − 7 · 8 =
19·3−(83−19·4)·8 = 83·(−8)+19·35 = 83·(−8)+(102−83)·35 = 102·35−83·43 =
102 · 35 − (1001 − 102 · 9) · 43 = 1001 · (−43) + 102 · 422. Thus x = 422, y = −43 gives
a particular solution, and the general solution is x = 422 + 1001t, y = −43 − 102t, where t
ranges over the integers.
8. Section 3.7 - Problem 6
This problem can be posed as finding a nonnegative integer solution to the linear diophantine
equation 63x + 7 = 23y, where x is the number of plantains in a pile, and y is the number of
plantains each traveller receives.
We find a particular solution x = 5, y = 14 and then, noting that (63, 23) = 1, all integer
solutions are given by x = 5+23n, y = 14+63n, where n is an integer. These are both positive
if and only if n ≥ 0. Thus the number of plantains in a pile could be any positive integer of the
form 5 + 23n.
9. Section 3.7 - Problem 8
This problem can be posed as finding a nonnegative integer solution to the linear diophantine
equation 6o + 33g = 549, where o is the number of oranges and g is the number of grapefruits.
Since (18, 33) = 3, and 549 is divisible by 3, we see that there will be infinitely many
solutions. To simplify our equation, let us divide both sides by 3: 6o + 11g = 183.
A particular solution to 6o + 11g = 1 is o = 2, g = −1.
A particular solution to 2o+11g = 183 is thus o = 2·183 = 366, and g = (−1)·183 = −183.
General solution: o = 366 + 11t, g = −183 − 6t, where t is an integer.
To get positive solutions, we need to choose t so that 366 + 11t ≥ 0 and −183 − 6t ≥ 0.
From the first inequality,
366
3
t≥−
= −33 ,
11
11
and from the second,
183
1
t≥−
= −30 .
6
2
The only integer values of t satisfying these inequalities are −31, −32, and −33.
t = −31 correspond to o = 25, g = 3,
t = −32 correspond to o = 14, g = 9,
t = −33 correspond to o = 3, g = 15.
The total number of pieces of fruit purchased in these three cases are
25 + 3 = 28 if t = −31,
14 + 9 = 23 if t = −32, and
3 + 15 = 28 if t = −33.
The smallest number of pieces of fruit purchased is 28: 3 oranges and 15 grapefruits.
10. Section 3.7 - Problem 10
We consider the linear diophantine equation 11` + 8c = T , where ` represents the number of
lobster dinners bought, c represents the number of chicken dinners bought, and T represents the
total bill.
If T = 1, then ` = 3, c = −4 gives a particular solution. As (8, 11) = 1, for an arbitrary
integer T the general solution is given by ` = 3T + 8n, c = −4T − 11n, where n is an
integer. Both solutions are non-negative if 3T + 8n ≥ 0 and c = −4T − 11n ≥ 0, i.e.,
3
4
n ∈ [− T, − T ].
8
11
3
3 4
6
3
4
a) If T = 777, then T = 291 , T = 282 , and the integers n in the interval [− T, − T ]
8
8 11
11
8
11
are −291, −290, . . . , −283. Thus there are 9 possibilities for l = 3 · 777 + 8n and c =
−4 · 777 − 11n:
(1) l = 3 and c = 93 if n = −291,
(2) l = 11 and c = 82 if n = −290,
(3) l = 19 and c = 71 if n = −289,
(4) l = 27 and c = 60 if n = −288,
(5) l = 35 and c = 49 if n = −287,
(6) l = 43 and c = 38 if n = −286,
(7) l = 51 and c = 27 if n = −285,
(8) l = 59 and c = 16 if n = −284, and
(9) l = 67 and c = 5 if n = −283.
4
10
3
4
3
b) If T = 96, then T = 36, T = 34 , and the integers n in the interval [− T, − T ] are
8
11
11
8
11
−36 and −35. Thus there are 2 possibilities for l = 3 · 96 + 8n and c = −4 · 96 − 11n:
(1) l = 0 and c = 12 if n = −36,
(2) l = 8 and c = 1 if n = −35.
3
7 4
1
c) If T = 69, then T = 25 ,
T = 25 , and there are no integers in the interval
8
8 11
11
4
3
[− T, − T ]. It is impossible for the total bill to be $69.
8
11