1. Section 4.1 - Problem 4
If a is even, we have a ≡ 0 (mod 4) or a ≡ 2 (mod 4). As 02 = 0 and 22 = 4 ≡ 0 (mod 4) in
this case a2 ≡ 0 (mod 4).
Suppose a is odd. If a ≡ 1 (mod 4), then a2 ≡ 12 (mod 4). If a ≡ 3 (mod 4), then a2 ≡ 9 ≡
1 (mod 4) as well.
2. Section 4.1 - Problem 22 (5th edition). Problem 30 (6th edition)
For the base case we have 40 = 1 ≡ 1 + 3(0) (mod 9).
For the induction step, we assume we have the result 4n ≡ 1 + 3n (mod 9) and prove for
n + 1. This becomes
4n+1 = 4 · 4n
≡ 4(1 + 3n) (mod 9)
= 4 + 12n
≡ 4 + 3n (mod 9)
= 1 + 3(n + 1)
Thus 4n+1 ≡ 1 + 3n (mod 9) and our result holds for all integers n ≥ 0 by induction.
3. Section 4.1 - Problem 28 (5th edition). Problem 36 (6th edition)
We repeatedly square 2 and reduce modulo 47.
22 = 4
24 = 16
28 = 256 ≡ 21 (mod 47)
216 ≡ 212 ≡ 18 (mod 47)
232 ≡ 182 ≡ 42 (mod 47)
264 ≡ 422 ≡ 25 (mod 47)
2128 ≡ 252 ≡ 14 (mod 47)
a) The least positive residue will be between 0 and 46 inclusive.
As calculated above we have 232 ≡ 18 (mod 47), so 18 is the least positive residue.
b) We have 247 = 232 28 24 22 21 ≡ 18 · 21 · 16 · 4 · 2 ≡ 21 (mod 47).
c) We have 2200 = 2128 264 28 ≡ 14 · 25 · 21 ≡ 18 (mod 47).
4. Section 4.2 - Problem 2
a) As (7, 3) = 1, there is exactly one solution modulo 7. We have 3x − 7y = 2. A particular
solution to this Diophantine equation is x0 = 3, y0 = 1. This is x ≡ 3 (mod 7) is the unique
solution to the original congruence.
b) As (9, 6) = 3, which divides 3, there are 3 incongruent solutions. These are obtained by
solving the Diophantine equation 6x − 9y = 3 or equivalently, 2x − 3y = 1. A particular
solution is x0 = 2 and y0 = 1, a general solution is x = 2 + 3t, y = 1 + 2t, as t ranges over
the integers. The three solutions to the original congruence correspond to t = 0, 1, 2; they
are x ≡ 2 (mod 9) or x ≡ 5 (mod 9) or x ≡ 8 (mod 9).
c) As (21, 17) = 1 there is exactly one solution modulo 21. The Diophantine equation we
need to consider is 17x − 21y = 14. We only need one particular solution; we obtain it by
performing the Euclidean algorithm to compute (21, 17), and then using back substitution.
21 = 17 + 4
17 = 4 · 4 + 1.
Thus 1 = 17 − 4 · 4 = 17 − (21 − 17) · 4 = 17 · 5 − 21 · 4, so 14 = 17(5 · 14) − 21(4 · 14). The
solution to the original congruence we are seeking is thus x ≡ 5 · 14 ≡ 70 ≡ 7 (mod 21).
d) As (25, 15) = 5 which does not divide 9, there are no solutions to 15x ≡ 9 (mod 25).
5. Section 4.2 - Problem 6
The congruence 12x ≡ c (mod 30) will have solutions if and only if (30, 12) = 6 divides c.
This occurs if and only if c = 0, 6, 12, 18 or 24.
6. Section 4.2 - Problem 8
a)
b)
c)
d)
These inverses can be calculated by trial and error.
7
9
8
6
7. Section 4.2 - Problem 10
(i) An integer a will have an inverse modulo 14 if and only if (a, 14) = 1. Thus the integers
with inverses are a ∈ {1, 3, 5, 9, 11, 13}.
(ii) Clearly 1−1 ≡ 1 (mod 14), and 3 and 5 are inverses of each other. The other invertible
integers will have inverse the negative of these.
1−1 ≡ 1 (mod 14)
3−1 ≡ 5 (mod 14)
5−1 ≡ 3 (mod 14)
9−1 ≡ (−5)−1 ≡ −3 ≡ 11 (mod 14)
11−1 ≡ (−3)−1 ≡ −5 ≡ 9 (mod 14)
13−1 ≡ (−1)−1 ≡ 13 (mod 14)
8. Section 4.3 - Problem 2
We wish to solve the congruences
x ≡ 1 (mod 2)
x ≡ 1 (mod 5)
x ≡ 0 (mod 3)
The unique solution of the first two congruences is x = 1 (mod 10). Thus the original system is
equivalent to
x ≡ 1 (mod 10)
x ≡ 0 (mod 3)
We now use the formula
x = 1 · M1 · y1 + 0 · M2 · y2 = M1 y1 ,
−1
where M1 = 3 and y1 = 3
modulo 30.
(mod 10) = 7 (mod 10). Thus x = 3·7 = 21 is the unique solution
9. Section 4.3 - Problem 4
a) First we compute y1 = 17−1 (mod 11) and y2 = 11−1 (mod 17). Both can be found by
solving the Diophantian equation 17y1 + 11y2 = 1. We only need a particular solution, and
one is easy to find by trial and error: y1 = 2, y2 = −3. Now
x = 4 · 17 · 2 + 3 · 11 · (−3) = 37 .
Thus x = 37 is the unique solution, modulo 11 · 17 = 187.
b) The first two equations can be rewritten as x ≡ −1 (mod 2) and x ≡ −1 (mod 3). By the
Chinese Remainder Theorem they are equivalent to x ≡ −1 (mod 6). Thus our system of
congruences is equivalent to
x ≡ −1 (mod 6)
x ≡ 3 (mod 5)
Now x ≡ (−1) · 5 · y1 + 3 · 6 · y2 (mod 30) where y1 = 5−1 ≡ −1 (mod 6) and y2 = 6−1 ≡
1 (mod 5). Thus x ≡ (−1) · 5 · (−1) + 3 · 6 · 1 ≡ 23 (mod 30).
c) x = 6 is the unique solution modulo 2 · 3 · 5 · 7 = 210.
10. Section 4.3 - Problem 22
This problem is equivalent to finding the least positive solution to the following system of
congruences:
x ≡ 3 (mod 17)
x ≡ 10 (mod 16)
x ≡ 0 (mod 15)
As 17, 16 and 15 are pairwise relatively prime, we can use the Chinese Remainder Theorem to
find the general solution modulo M = 15 · 16 · 17 = 4080. Thus solution is given by the formula
x = 3M1 y1 + 10M2 y2 + 0 · M3 · y3 (mod M ),
where M1 = 15 · 16 = 240, M2 = 15 · 17 = 255,
y1 ≡ M1−1 ≡ (15 · 16)−1 ≡ ((−1) · (−2))−1 ≡ 9 (mod 17)
and
y2 ≡ M2−1 ≡ (15 · 17)−1 ≡ −1 (mod 16) .
Thus x = 3 · 240 · 9 + 10 · 255 · (−1) = 3930, modulo 4080. In other words, the number of coins
can be 3930 + 4080n, where n is a non-negative integer. The smallest such number is 3930.