PROBLEM SET # 1 SOLUTIONS
C HAPTER 1: E UCLIDEAN A LGORITHM
1.1
Euclidean Algorithm.
1. Use the Euclidean algorithm to find the greatest common divisors of:
a) 1084 and 412
Solution. Running the Euclidean algorithm, we find
1084 = 2 · 412 + 260
412 = 1 · 260 + 152
260 = 1 · 152 + 108
152 = 1 · 108 + 44
108 = 2 · 44 + 20
44 = 2 · 20 + 4
20 = 5 · 4.
Thus, the greatest common divisor of 1084 and 412 is 4.
b) 1979 and 531
Solution. Running the Euclidean algorithm, we find
1979 = 3 · 531 + 386
531 = 1 · 386 + 145
386 = 2 · 145 + 96
145 = 1 · 96 + 49
96 = 1 · 49 + 47
49 = 1 · 47 + 2
47 = 23 · 2 + 1
2 = 2 · 1.
Thus, the greatest common divisor of 1979 and 531 is 1.
c) 305 and 185
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Solution. Running the Euclidean algorithm, we find
305 = 1 · 185 + 120
185 = 1 · 120 + 65
120 = 1 · 65 + 55
65 = 1 · 55 + 10
55 = 5 · 10 + 5
10 = 2 · 5.
Thus, the greatest common divisor of 305 and 185 is 5.
2. Use the calculations from the previous exercise to express the following in continued fraction form:
1084
a)
412
Solution. Using the calculations above, we see that
1
1084
=2+
.
1
412
1
1+
1+
1
1+
2+
1
1
2+ 1
5
b)
1979
531
Solution. Using the calculations above, we see that
1979
1
=3+
.
1
185
1 + 2+
1
1+
1
1+
1+
1
1
23+ 1
2
c)
305
185
Solution. Using the calculations above, we see that
305
1
=1+
.
185
1 + 1+ 1 1
1+
1
5+ 1
2
3. If a | b and b | c, show a | c.
Solution. Since a | b, we have b = ma for some m ∈ Z. Since b | c, we have c = nb
for some n ∈ Z. We then have c = nb = n(ma) = (nm)a, and hence a | c.
√
4. Use the continued fraction algorithm to show 3 is irrational.
2
√
√
Solution. We begin with√α0 = 3 ≈
1.7.
Then
bα
c
=
1
and
so
β
=
3 − 1.
0
0
√
We then
√ bα1 c = 1. Therefore
√ have α1 = 1/(√ 3 − 1) = ( 3 + 1)/2 ≈√1.4, and so
β1 = (√3 + 1)/2 − 1 =√( 3 − 1)/2, and so α2 = 2/( 3 − 1) = 3 + 1 ≈ 2.7. Therefore
β2 = ( 3 + 1) − 2 = 3 − 1 = β0 , and so α3 = α1 , and we see
√ the process will now
loop forever.
√ Thus, the continued fraction algorithm for 3 repeats indefinitely,
and hence 3 cannot be rational.
√
3. Note
5. Compute the fourth convergent
of
the
continued
fraction
expansion
of
√
how well it approximates 3.
Solution. By our calculations above, the continued fraction algorithm for
duces the following data:
√
3 pro-
i bαi c
√ βi
√ αi+1
0 1
( √3 + 1)/2
√ 3−1
1 1 ( √3 − 1)/2 √ 3 + 1
2 2
( √3 + 1)/2
√ 3−1
3 1 ( √3 − 1)/2 √ 3 + 1
4 2
3−1
( 3 + 1)/2
..
..
..
..
.
.
.
.
Thus, the fourth convergent in the continued fraction algorithm for
7
1
= = 1.75.
1+
1
4
1 + 2+ 1
1
√
Since 3 ≈ 1.732050808 . . ., this quite a good approximation.
1.2
√
3 is
Fundamental Theorem of Arithmetic.
1. Find all integer solutions of 13853x + 6951y = gcd(13853, 6951).
Solution. We first compute the required gcd, using the Euclidean algorithm:
13853 = 1 · 6951 + 6902
6951 = 1 · 6902 + 49
6902 = 140 · 49 + 42
49 = 1 · 42 + 7
42 = 6 · 7.
Thus, gcd(13853, 6951) = 7. Moreover, the second-to-last convergent in the continued fraction algorithm for 13853/6951 is
1
283
1+
=
.
1
142
1 + 140+ 1
1
Observe that 13853 · 142 = 1967126 and 6951 · 283 = 1967133, and so
13853 · (−142) + 6951 · (283) = 7 = gcd(13853, 6951).
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One integer solution to the given equation is therefore x = −142, y = 283. By
Theorem 2, all integer solutions to this equation are then of the form x = −142 +
993k, y = 283 − 1979k for k ∈ Z.
2. Find all integer solutions of 15750x + 9150y = gcd(15750, 9150).
Solution. We first compute the required gcd, using the Euclidean algorithm:
15750 = 1 · 9150 + 6600
9150 = 1 · 6600 + 2550
6600 = 2 · 2550 + 1500
2550 = 1 · 1500 + 1050
1500 = 1 · 1050 + 450
1050 = 2 · 450 + 150
450 = 3 · 150.
Thus, gcd(15750, 9150) = 150. Moreover, the second-to-last convergent in the continued fraction algorithm for 15750/9150 is
1+
1
1+
=
1
2+
1+
1
1
1+ 1
2
31
.
18
Observe that 15750 · 18 = 283500 and 9150 · 31 = 283650, and so
15750 · (−18) + 9150 · (31) = 150 = gcd(15750, 9150).
One integer solution to the given equation is therefore x = −18, y = 31. By Theorem 2, all integer solutions to this equation are then of the form x = −18 + 61k, y =
31 − 105k for k ∈ Z.
3. Show that 427x + 259y = 13 has no integer solutions.
Solution. We compute gcd(427, 259) using the Euclidean algorithm:
427 = 1 · 259 + 160
259 = 1 · 168 + 91
168 = 1 · 91 + 77
91 = 1 · 77 + 14
77 = 5 · 14 + 7
14 = 2 · 7.
Thus, gcd(427, 259) = 7, and since 7 - 13, the given equation cannot have any
integer solutions.
4. Let a and b be two integers. Use the Fundamental Theorem of Arithmetic to show
that any common divisor of a and b divides gcd(a, b).
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Solution. Let d be any common divisor of a and b. Then a = md and b = nd for
some m, n ∈ Z. By the Fundamental Theorem of Arithmetic, there exist integers
x, y ∈ Z such that ax + by = gcd(a, b). Combining this equation with our previous
statement, we see that (md)x + (nd)y = gcd(a, b), or equivalently d(mx + ny) =
gcd(a, b). Thus, d divides gcd(a, b).
5. Let a, b and d be any integers. Use the Fundamental Theorem of Arithmetic to
show gcd(ad, bd) = gcd(a, b)d.
Solution. First observe that we automatically have gcd(a, b)d | ad and gcd(a, b)d |
bd, so by the previous exercise gcd(a, b)d | gcd(ad, bd). Conversely, by the Fundamental Theorem of Arithmetic we can write ax + by = gcd(a, b) for some integers
x, y ∈ Z. It follows that (ad)x+(bd)y = gcd(a, b)d, and hence gcd(ad, bd) | gcd(a, b)d.
Thus, the numbers gcd(ad, bd) and gcd(a, b)d are the same up to sign, and since they
are both positive, they must therefore be equal.
1.4
Efficiency of the Euclidean Algorithm.
1. Let
√
√
5)n − (1 − 5)n
√
fn =
.
2n 5
Show that fn+2 = fn+1 +fn . In other words, show that fn satisfy the same recursion
relation as the Fibonacci numbers. What else do you need to check in order to
verify that fn are indeed the Fibonacci numbers?
(1 +
Solution. We first directly compute
√
√
√
√
(1 + 5)n+1 − (1 − 5)n+1 (1 + 5)n − (1 − 5)n
√
√
fn+1 + fn =
+
2n+1 5
2n 5
√
√
√
√
(1 + 5)n+1 − (1 − 5)n+1 + 2(1 + 5)n − 2(1 − 5)n
√
=
2n+1 5
√
√
√
√
(1 + 5)n ((1 + 5) + 2) − (1 − 5)n ((1 − 5) + 2)
√
=
2n+1 5
√
√
√
√
(1 + 5)n (3 + 5) − (1 − 5)n (3 − 5)
√
=
.
2n+1 5
√
√
√
√
Now observe that (1 + 5)2 = 2(3 + 5) and (1 − 5)2 = 2(3 − 5). Substituting
these into the above expression then gives
√ 2
√ 2
√
√
(1 + 5)n · (1+2 5) − (1 − 5)n · (1−2 5)
√
fn+1 + fn =
2n+1 5
√
√
(1 + 5)n+2 − (1 − 5)n+2
√
=
2n+2 5
= fn+2 .
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Finally, to verify fn really does agree with the Fibonacci sequence, simply observe
that f0 = 0 and f1 = 1 (by simply plugging in n = 0 and n = 1 into the given
formula).
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