Hints and answers

1.9 HINTS AND ANSWERS
1.27
1.28
Establish the values of k for which the binomial coefficient p Ck is divisible by p
when p is a prime number. Use your result and the method of induction to prove
that np − n is divisible by p for all integers n and all prime numbers p. Deduce
that n5 − n is divisible by 30 for any integer n.
An arithmetic progression of integers an is one in which an = a0 + nd, where a0
and d are integers and n takes successive values 0, 1, 2, . . . .
(a) Show that if any one term of the progression is the cube of an integer then
so are infinitely many others.
(b) Show that no cube of an integer can be expressed as 7n + 5 for some positive
integer n.
1.29
Prove, by the method of contradiction, that the equation
xn + an−1 xn−1 + · · · + a1 x + a0 = 0,
in which all the coefficients ai are integers, cannot have a rational root, unless
that root is an integer. Deduce that any integral root must be a divisor of a0 and
hence find all rational roots of
(a) x4 + 6x3 + 4x2 + 5x + 4 = 0,
(b) x4 + 5x3 + 2x2 − 10x + 6 = 0.
Necessary and sufficient conditions
1.30
1.31
1.32
Prove that the equation ax2 + bx + c = 0, in which a, b and c are real and a > 0,
has two real distinct solutions IFF b2 > 4ac.
For the real variable x, show that a sufficient, but not necessary, condition for
f(x) = x(x + 1)(2x + 1) to be divisible by 6 is that x is an integer.
Given that at least one of a and b, and at least one of c and d, are non-zero,
show that ad = bc is both a necessary and sufficient condition for the equations
ax + by = 0,
cx + dy = 0,
1.33
to have a solution in which at least one of x and y is non-zero.
The coefficients ai in the polynomial Q(x) = a4 x4 + a3 x3 + a2 x2 + a1 x are all
integers. Show that Q(n) is divisible by 24 for all integers n ≥ 0 if and only if all
of the following conditions are satisfied:
(i) 2a4 + a3 is divisible by 4;
(ii) a4 + a2 is divisible by 12;
(iii) a4 + a3 + a2 + a1 is divisible by 24.
1.9 Hints and answers
1.1
1.3
1.5
1.7
1.9
√
√
(b) The roots are 1, 18 (−7 + 33) = −0.1569, 18 (−7 − 33) = −1.593. (c) −5 and
7
1
are the values of k that make f(−1) and f( 2 ) equal to zero.
4
are all positive. Therefore f (x) > 0 for all x > 0.
(a) a = 4, b = 38 and c = 23
16
(b) f(1) = 5, f(0) = −2 and f(−1) = 5, and so there is at least one root in each
of the ranges 0 < x < 1 and −1 <√
x < 0. (x7 + 5x6 ) + (x4 − x3 ) + (x2 − 2)
is positive definite for −5 < x < − 2. There are therefore no roots in this
range, but there must be one to the left of x = −5.
(a) x2 + 9x + 18 = 0; √
(b) x2 − 4x = 0; (c) x2 − 4x + 4 = 0; (d) x2 − 6x + 13 = 0.
(a) Use sin(π/4) = 1/ 2. (b) Use results (1.20) and (1.21).
(a) 1.339, −2.626. (b) No solution because 62 > 42 + 32 . (c) −0.0849, −2.276.
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PRELIMINARY ALGEBRA
1.11
1.13
1.15
1.17
1.19
1.21
1.23
1.25
1.27
1.29
1.31
1.33
Show that the equation is equivalent to sin(5θ/2) sin(θ) sin(θ/2) = 0.
Solutions are −4π/5, −2π/5, 0, 2π/5, 4π/5, π. The solution θ = 0 has multiplicity
3.
(a) A circle of radius 5 centred on (−3, −4).
(b) A hyperbola with ‘centre’ (3, −2) and ‘semi-axes’ 2 and 3.
(c) The expression factorises into two lines, x + 2y − 3 = 0 and 2x + y + 2 = 0.
(d) Write the expression as (x + y)2 = 8(x − y) to see that it represents a parabola
passing through the origin, with the line x + y = 0 as its axis of symmetry.
5
9
4
4
(a)
+
,
(b) −
+
.
7(x − 2) 7(x + 5)
3x 3(x − 3)
1
x+1
2
x+2
−
,
(b) 2
+
.
(a) 2
x +4 x−1
x + 9 x2 + 1
(a) 10, (b) not defined, (c) −35, (d) −21.
Look for factors common to the n = N sum and the additional n = N + 1 term,
so as to reduce the sum for n = N + 1 to a single term.
Write 32n as 8m − 7.
Use the half-angle formulae of equations (1.32) to (1.34) to relate functions of
θ/2k to those of θ/2k+1 .
p
Ck nk + 1. Apply
Divisible for k = 1, 2, . . . , p − 1. Expand (n + 1)p as np + p−1
1
5
the stated result for p = 5. Note that n − n = n(n − 1)(n + 1)(n2 + 1); the product
of any three consecutive integers must divide by both 2 and 3.
By assuming x = p/q with q = 1, show that a fraction −pn /q is equal to an
integer an−1 pn−1 + · · · + a1 pq n−2 + a0 q n−1 . This is a contradiction, and is only
resolved if q = 1 and the root is an integer.
(a) The only possible candidates are ±1, ±2, ±4. None is a root.
(b) The only possible candidates are ±1, ±2, ±3, ±6. Only −3 is a root.
f(x) can be written as x(x + 1)(x + 2) + x(x + 1)(x − 1). Each term consists of
the product of three consecutive integers, of which one must therefore divide by
2 and (a different) one by 3. Thus each term separately divides by 6, and so
therefore does f(x). Note that if x is the root of 2x3 + 3x2 + x − 24 = 0 that lies
near the non-integer value x = 1.826, then x(x + 1)(2x + 1) = 24 and therefore
divides by 6.
Note that, e.g., the condition for 6a4 + a3 to be divisible by 4 is the same as the
condition for 2a4 + a3 to be divisible by 4.
For the necessary (only if) part of the proof set n = 1, 2, 3 and take integer
combinations of the resulting equations.
For the sufficient (if) part of the proof use the stated conditions to prove the
proposition by induction. Note that n3 − n is divisible by 6 and that n2 + n is
even.
40