SOLUTIONS

Solution of Exercise 6

The gcd of 480 and 175 can be determined as follows by use of the Euclid's algorithm

480 = 2 · 175 + 130
175 = 1 · 130 + 45
130 = 2 · 45 + 40
45 = 1 · 40 + 5
40 = 8 · 5

So the gcd of 480 and 175 is 5.

We write

5 = 45 - 40
= 45-(130 - 2·45)
= (175 - 130) - (130 - 2·(175-130))
= 3·175 - 4·130
= 3·175 - 4·(480-2·175)
= 11·175 - 4·480

Idem for 5621 and 219:

5621 = 25 · 219 + 146
219 = 1 · 146 + 73
146 = 2 · 73

So the gcd of 5621 and 219 is 73.

Similarly, we find 73 = 26·219 - 5621.

Idem for 983675 and 105120:

983675 = 9 · 105120 + 37595
105120 = 2 · 37595 + 29930
37595 = 1 · 29930 + 7665
29930 = 3 · 7665 + 6935
7665 = 1 · 6935 + 730
6935 = 9 · 730 + 365
730 = 2 · 365

So the gcd of 983675 and 105120 is 365.

Similarly, we find 365 = -137·983675 + 1282·105120.

Solution of Exercise 7

We show that z gcd(x,y) is the greatest common divisor of zx and zy. It is clear that z gcd(x,y) is a divisor of both zx and zy.

Suppose that d is a common divisor of zx and zy. With the help of the extended Euclidean algorithm we can find integers a and b with gcd(x,y) = (ax + by). So z gcd(x,y) = z(ax + by) = azx + bzy. As d divides zx and zy, it also divides z gcd(x,y). This implies that z gcd(x,y) is greater than or equal to d. In particular, z gcd(x,y) is the greatest common divisor of zx and zy.

Solution of Exercise 15

1. a² - b² = (a + b)(a - b).
2. a³ - b³ = (a - b)(a² + ab + b²).
3. aⁿ - bⁿ = (a - b)(a^n-1 + a^n-2b + ... + ab^n-2 + b^n-1).
4. The expression a + b is a factor of a²ⁿ⁺¹ + b²ⁿ⁺¹ , because a²ⁿ⁺¹ + b²ⁿ⁺¹ = (a + b)(a²ⁿ - a^2n-1b + ... - ab^2n-1 + b²ⁿ).
5. We write 5¹⁰ - 2¹⁰ = (5⁵ + 2⁵)(5⁵ - 2⁵) = (5 + 2)(5⁴ - 5³2 + 5²2² - 5 2³ + 2⁴)(5-2)(5⁴ + 5³2 + 5²2² + 5 2³ + 2⁴) = 7·541·3·1031 = 3·7·11·41·1031.

Solution of Exercise 11

Suppose that there are only finite many primes of the form 4n+3, say p₁, ... , p_r. Note that p_j = -1 mod 4. Consider the product m = p₁ ··· p_r. If m = -1 mod 4, put M = m + 4. If m = 1 mod 4, put M = m + 2. Then M is of the form 4n + 3. There is no p_j dividing M. Since in both cases M = -1 mod 4, M contains a prime divisor p of the form 4n + 3. This is a prime that is not in our list. Contradiction.

Solution of Exercise 1

Dividing c by lcm(a,b) yields

c = q lcm(a,b) + r

where 0 r < lcm(a,b). Now a divides c and lcm(a,b) and then also r. Similarly b divides r and r is a common multiple of a and b. But, as 0 r < lcm(a,b), we find r = 0.

Solution of Exercise 12

For n>2 we see that n - 1 is a real divisor of n² - 1. Only for n = 2 the number n² - 1 is prime.

Solution of Exercise 13

Write 13p + 1 = n² for some integer n. Then 13p = n² - 1 = (n + 1)(n - 1). Since 13p is the prime factorization of (n + 1)(n - 1), there are only two possibilities. First, n + 1 = 13 and n - 1 = 11 = p is prime. Secondly, n - 1 = 13 and n + 1 = 15 = p. But as 15 is not prime, this case cannot occur. We conclude that 144 is the only square of the form 13p + 1.

Solution of Exercise 8

From the extended Euclidean algorithm we know that c = gcd(a,b) = ua + vb for some integers u and v. Since c | a and c | b, we find that c divides xa + yb for all integers x and y. In particular, if xa + yb > 0, then c is less than or equal to xa + yb. So c = min{xa + yb | xa + yb > 0, x, y Z}.

Solution of Exercise 16

Suppose that the cube root of 17 can be written as p/q, for two integers p and q. Then 17q³ = p³. Now ord₁₇( 17q³) = 1 mod 3 and ord₁₇(p³) is 0 mod 3. This is a contradiction.

Solution of Exercise 17

2³² - 1 = (2¹⁶ - 1) · (2¹⁶ + 1) = (2⁸ - 1) · (2⁸ + 1) ·(2¹⁶ + 1)
=(2⁴ - 1) · (2⁴ + 1) · (2⁸ + 1) ·(2¹⁶ + 1)
=(2² - 1) ·(2² + 1) · (2⁴ + 1) · (2⁸ + 1) ·(2¹⁶ + 1)
= 3 · 5 · 17 · 257 · 65537.

The numbers 3, 5, 17, 257 and 65537 are all prime.

Solution of Exercise 2

10³ⁿ = (10²ⁿ + 10ⁿ + 1)(10ⁿ - 1) + 1, so the remainder is 1. The quotient 10²ⁿ + 10ⁿ + 1 is found in two steps: in the first step you find 10³ⁿ = 10²ⁿ (10ⁿ - 1) + 10²ⁿ. Since the remainder is still greater than 10ⁿ - 1, we need a second step: 10³ⁿ = (10²ⁿ + 10ⁿ)(10ⁿ - 1) + 10ⁿ. Finally, in the third step we find 10³ⁿ = (10²ⁿ + 10ⁿ + 1)(10ⁿ - 1) + 1.

Solution of Exercise 3

• If the left-hand wheel turns n times, then each of its cogs moves 24n positions, so each cog of the right-hand wheel moves 24n positions, but this corresponds to 24n/16 complete turns of the right-hand wheel. So we have to look for the smallest positive number n so that 16 divides 24n. Since 16 = 8 · 2 and 24 = 8 · 3, we conclude that 2 | 3n and therefore 2 | n. We find n = 2.
• Reasoning as above, 24n must be divisible by 15 and 16. This implies that 5 | 8n and 2 | 3n, so that 5 | n and 2 | n. The smallest positive solution is therefore lcm(5,2) = 10.

Solution of Exercise 4

Write m = 2n+1 and square both sides: m² = 4n² + 4n+1. Since this can be written as 2(2n² + 2n) + 1, the number is odd. Rewriting it as 4(n² + n) + 1 we see that the number leaves remainder 1 upon division by 4. Since n² + n = n(n + 1) is always divisible by 2 (at least one of the two factors on the right-hand side is), the term 4(n² + n) is divisible by 8. So the number leaves remainder 1 upon division by 8. Since 3² leaves remainder 7 upon divison by 16, we cannot replace 8 by 16.

Solution of Exercise 5

• If b = q₁a and c = q₂b for some integers q₁ and q₂, then c = q₁q₂a.
• If b = q₁a and d = q₂c for some integers q₁ and q₂, then bd = q₁q₂ac.

Solution of Exercise 9

Since 18 and 25 are relatively prime there exist integers x and y such that 1 = 18x + 25y. Multiply both sides of this equality with a:

a = 18ax + 25ax.

Since 18a, x, 25a and x are all integers, the right-hand side is an integer and therefore also the left-hand side, i.e., a is an integer.

Note that the same argument works if we replace 18 and 25 by two relatively prime numbers.

Solution of Exercise 10

Equivalently, we solve 16x + 5y = 3. Since the gcd of 16 and 5 divides 3, the equation does have solutions. The homogeneous equation 16x + 5y = 0 has solutions x = 5n, y = -16n. The extended Euclidean algorithm produces the equality 1 · 16 -3 · 5 = 1. Multiplying both sides by 3 we obtain 3 · 16 -9 · 5 = 3. So the solutions are x = 3 + 5n, y = -9 - 16n.

Solution of Exercise 14

Rewrite the equation as p(q - 4) = 7q. The left-hand side is divisible by p and so the right-hand side is divisible by p. Since p is a prime this implies that p = 7 or p = q. The first possibility leads to q - 4 = q, which is impossible. Hence p = q. This leads to p = q = 11.