Download NUMBERS (MA10001): PROBLEM SHEET 5, SOLUTIONS 1. Using

NUMBERS (MA10001): PROBLEM SHEET 5, SOLUTIONS
1. Using Euclid’s algorithm, calculate the greatest common divisor d = hcf(a, b) in each
of the following cases:
(i) a = 533, b = 117;
(ii) a = 53669, b = 64821; (iii) a = 7244433, b = 3798601.
These are straightforward, especially with a calculator:
(i) 533 = 4 × 117 + 65, 117 = 65 + 52, 65 = 52 + 13, 52 = 4 × 13 so hcf(533, 117) = 13
(ii) 64821 = 53669 + 11152, 53669 = 4 × 11152 + 9061, 11152 = 9061 + 2091. 9061 =
4 × 2091 + 697, 2091 = 3 × 697, so hcf(53669, 64821) = 697.
(iii) 7244433 = 2 × 3798601 − 352769 (note the use of the form a = mb − r here so as to get
a smaller r – not essential), 3798601 = 10 × 352769 + 270911, 352769 = 270911 + 81858,
270911 = 3 × 81858 + 25337, 81858 = 3 × 25337 + 5847, 25337 = 4 × 5847 + 1949,
5847 = 3 × 1949, so hcf(7244433, 3798601) = 1949.
2. Suppose a, b, c ∈ Z with a, b not both zero.
Prove that the equation xa + yb = c has an integer solution x, y ∈ Z if and only if hcf(a, b)
divides c.
If d = hcf(a, b) and xa + yb = c then d|a and d|b so d|(xa + yb) = c. Conversely, if d|c then
c = dz for some z ∈ Z: take λ, µ ∈ Z such that λa + µb = d and then put x = λz, u = µz.
Prove that hcf(na, nb) = n hcf(a, b) for all n ∈ N.
For instance: we can take λa + µb = hcf(a, b) so λna + µnb = n hcf(a, b). Therefore
hcf(na, nb)|n hcf(a, b). But obviously n hcf(a, b)|na (and nb) so n hcf(a, b) is a common
factor of na and nb, so it divides hcf(na, nb). So the two numbers must be equal, because
each divides the other and they are positive.
Prove that, if d = hcf(a, b) and d|c, then xa + yb = c has infinitely many integer solutions.
an
bn
d , y − d with n
an
bn
d and d are integers.
[Hint: let x, y be an integer solution and consider x +
These obviously work, the point being just that
∈ Z.]
Does the equation 53669x + 64821y = 1394 have any integer solutions? If so, find one.
We showed earlier that hcf(64821, 53669) = 697 and 1394 = 2 × 697 so the answer should
be yes. The rest of the calculation is a bit tedious: 697 = 9061 − 4 × 2091 = 9061 −
4 × (11152 − 9061) = −4 × 11152 + 5 × 9061 = −4 × 11152 + 5 × (53669 − 4 × 11152) =
5 × 53669 − 24 × 11152 = 5 × 53669 − 24 × (64821 − 53669) = −24 × 64821 + 29 × 53669.
We want twice that, so the right numbers are −48 and 58.
3. List all the primes up to and including 43. Show that the consecutive odd numbers
1949 and 1951 are both prime. (Note that 452 = 2025.)
You may well know them. 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43. It is easy to
check that none of these divide 1949 or 1951 and that’s enough.
Prove that 3, 5, 7 are the only three consecutive odd numbers that are also prime.
One of n, n + 2 and n + 4 is divisible by 3.
Prove that 3 is the only prime number of the form 22n − 1, n ∈ N.
22n − 1 = (2n − 1)(2n + 1), so we have to have 2n − 1 = 1.
4. Let a, b, c, d ∈ N. Write m = hcf(a, b) and n = hcf(c, d). Show that the highest common
factor of the four numbers a, b, c and d is given by hcf(m, n). Find all the primes that
divide all four of the numbers numbers 8029, 5957, 4277 and 2821.
hcf(m, n) obviously is a common divisor of all four. If λa + µb = m, λ0 c + µ0 d = n and
αm + βn = hcf(m.n) then hcf(m, n) = αλa + αµb + βλ0 c + βµ0 d, so any common factor of
a, b, c, d divides hcf(m, n).
hcf(8029, 5957) = 259; hcf(4277, 2821) = 91; and hcf(259, 91) = 7 so the answer to the
last part is 7 only.
GKS, 13/11/08