Chapter
2
Algebra
2.1 Identities with Squares
Recall that x + y
2
= x + yx + y = x
2
+ y
2
+ 2xy 2.1
If we substitute y by y + z we obtain x + y + z
2
= x
2
+ y
2
+ z
2
+ 2xy + 2xz + 2yz 2.2
If we substitute z by z + w we obtain x + y + z + w
2
= x
2
+ y
2
+ z
2
+ w
2
+ 2xy + 2xz + 2xw + 2yz + 2yw + 2zw 2.3
56 Example The sum of two numbers is 21 and their product −7. Find i the sum of their squares, ii the sum of their
reciprocals and iii the sum of their fourth powers. Solution: If the two numbers are a and b, we are given that a + b = 21 and ab = −7
. Hence a
2
+ b
2
= a + b
2
− 2ab = 21
2
− 2−7 = 455 and
1 a
+ 1
b =
b + a ab
= 21
−7 = −3
Also a
4
+ b
4
= a
2
+ b
2 2
− 2a
2
b
2
= 455
2
− 2−7
2
= 357
57 Example Find positive integers a and b with
È
5 + √
24 = √
a + √
b .
Solution: Observe that 5 +
√ 24 = 3 + 2
√ 2
· 3+ 2 = √
2 + √
3
2
. Therefore
È
5 + 2 √
6 = √
2 + √
3 .
58 Example
Compute
È
1000000100000110000021000003 + 1 without using a calculator.
8
Identities with Squares 9
Solution: Let x = 1 000 000 = 10
6
. Then xx + 1x + 2x + 3 = xx + 3x + 1x + 2 = x
2
+ 3xx
2
+ 3x + 2 .
Put y = x
2
+ 3x . Then
xx + 1x + 2x + 3 + 1 = x
2
+ 3xx
2
+ 3x + 2 + 1 = yy + 2 + 1 = y + 1
2
. Thus
È
xx + 1x + 2x + 3 + 1 =
y + 1 =
x
2
+ 3x + 1 =
10
12
+ 3 · 10
6
+ 1 =
1 000 003 000 001 .
Another useful identity is the difference of squares: x
2
− y
2
= x − yx + y 2.4
59 Example
Explain how to compute 123456789
2
− 123456790 × 123456788 mentally.
Solution: Put x = 123456789. Then 123456789
2
− 123456790 × 123456788 = x
2
− x + 1x − 1 = 1 .
60 Example Shew that
1 + x + x
2
+ ··· + x
1023
= 1 + x1 + x
2
1 + x
4
··· 1 + x
256
1 + x
512
. Solution: Put S = 1 + x + x
2
+ ··· + x
1023
. Then xS = x + x
2
+ ··· + x
1024
. This gives S − xS = 1 + x + x
2
+ ··· + x
1023
− x + x
2
+ ··· + x
1024
= 1 − x
1024
or S1 − x = 1 − x
1024
, from where 1 + x + x
2
+ ··· + x
1023
= S = 1 − x
1024
1 − x .
But 1 − x
1024
1 − x =
1 − x
1024
1 − x
512
1 − x
512
1 − x
256
··· 1 − x
4
1 − x
2
1 − x
2
1 − x =
1 + x
512
1 + x
256
··· 1 + x
2
1 + x ,
proving the assertion.
61 Example Given that
1 √
1 + √
2 +
1 √
2 + √
3 +
1 √
3 + √
4 +
··· + 1
√ 99 +
√ 100
is an integer, find it. Solution: As 1 = n + 1 − n =
√ n + 1 −
√ n
√ n + 1 +
√ n
, we have 1
√ n +
√ n + 1
= √
n + 1 − √
n .
10 Chapter 2
Therefore 1
√ 1 +
√ 2
= √
2 − √
1 1
√ 2 +
√ 3
= √
3 − √
2 1
√ 3 +
√ 4
= √
4 − √
3 ..
. ..
. ..
. 1
√ 99 +
√ 100
= √
100 − √
99 ,
and thus 1
√ 1 +
√ 2
+ 1
√ 2 +
√ 3
+ 1
√ 3 +
√ 4
+ ··· +
1 √
99 + √
100 =
√ 100 −
√ 1 = 9
. Using the difference of squares identity,
x
4
+ x
2
y
2
+ y
4
= x
4
+ 2x
2
y
2
+ y
4
− x
2
y
2
= x
2
+ y
2 2
− xy
2
= x
2
− xy + y
2
x
2
+ xy + y
2
. The following factorisation is credited to Sophie Germain.
a
4
+ 4b
4
= a
4
+ 4a
2
b
2
+ 4b
4
− 4a
2
b
2
= a
2
+ 2b
2 2
− 2ab
2
= a
2
− 2ab + 2b
2
a
2
+ 2ab + 2b
2
62 Example Prove that n
4
+ 4 is a prime only when n = 1 for n ∈ N.
Solution: Using Sophie Germain’s trick, n
4
+ 4 =
n
4
+ 4n
2
+ 4 − 4n
2
= n
2
+ 2
2
− 2n
2
= n
2
+ 2 − 2nn
2
+ 2 + 2n =
n − 1
2
+ 1n + 1
2
+ 1 .
Each factor is greater than 1 for n 1, and so n
4
+ 4 cannot be a prime if n 1.
63 Example Shew that the product of four consecutive integers, none of them 0, is never a perfect square.
Solution: Let n − 1 , n, n + 1, n + 2 be four consecutive integers. Then their product P is
P = n − 1nn + 1n + 2 = n
3
− nn + 2 = n
4
+ 2n
3
− n
2
− 2n .
But n
2
+ n − 1
2
= n
4
+ 2n
3
− n
2
− 2n + 1 = P + 1 P.
As P 6= 0 and P is 1 more than a square, P cannot be a square.
64 Example
Find infinitely many pairs of integers m , n such that m and n share their prime factors and m − 1, n − 1 share
their prime factors.
Practice 11
Solution: Take m = 2
k
− 1 , n = 2
k
− 1
2
, k = 2, 3, . . .. Then m, n obviously share their prime factors and m − 1 = 22
k−1
− 1 shares its prime factors with n − 1 = 2
k+1
2
k−1
− 1.
65 Example
Prove that if r ≥ s ≥ t then
r
2
− s
2
+ t
2
≥ r − s + t
2
2.5 Solution: We have
r − s + t
2
− t
2
= r − s + t − tr − s + t + t = r − sr − s + 2t .
Since t − s ≤ 0, r − s + 2t = r + s + 2t − s ≤ r + s and so
r − s + t
2
− t
2
≤ r − sr + s = r
2
− s
2
which gives r − s + t
2
≤ r
2
− s
2
+ t
2
.
Practice
66 Problem The sum of two numbers is −7 and their product 2. Find i the sum of
their reciprocals, ii the sum of their squares.
67 Problem
Write x
2
as a sum of powers of x + 3 .
68 Problem Write x
2
− 3x + 8 as a sum of powers of x − 1 .
69 Problem
Prove that 3 is the only prime of the form n
2
− 1 .
70 Problem Prove that there are no primes of the form n
4
− 1 .
71 Problem
Prove that n
4
+ 4
n
is prime only for n = 1.
72 Problem Use Sophie Germain’s trick to obtain
x
4
+ x
2
+ 1 = x
2
+ x + 1x
2
− x + 1 ,
and then find all the primes of the form n
4
+ n
2
+ 1 .
73 Problem If a
, b satisfy 2
a + b =
1 a
+ 1
b , find
a
2
b
2
.
74 Problem If cot x + tan x = a, prove that cot
2
x + tan
2
x = a
2
− 2 . .
75 Problem Prove that if a
, b, c are positive integers, then √
a + √
b + √
c− √
a + √
b + √
c ·
√ a −
√ b +
√ c
√ a +
√ b −
√ c
is an integer.
76 Problem By direct computation, shew that the product of sums of two squares is
itself a sum of two squares: a
2
+ b
2
c
2
+ d
2
= ac + bd
2
+ ad − bc
2
2.6
77 Problem Divide x
128
− y
128
by x + yx
2
+ y
2
x
4
+ y
4
x
8
+ y
8
x
16
+ y
16
x
32
+ y
32
x
64
+ y
64
.
78 Problem Solve the system
x + y = 9 ,
x
2
+ xy + y
2
= 61 .
79 Problem Solve the system
x − y = 10 ,
x
2
− 4xy + y
2
= 52 .
80 Problem Find the sum of the prime divisors of 2
16
− 1 .
81 Problem Find integers a
, b with
p
11 + √
72 = a + √
b .
82 Problem Given that the difference
p
57 − 40 √
2 −
p
57 + 40 √
2 is an integer, find it.
83 Problem Solve the equation
p
x + 3 − 4 √
x − 1 +
p
x + 8 − 6 √
x − 1 = 1 .
84 Problem Prove that if a
0, b 0, a + b c, then √
a + √
b √
c
85 Problem Prove that if 1
x 2, then 1
p
x + 2 √
x − 1 +
1
p
x − 2 √
x − 1 =
2 2 − x
.
86 Problem If x
0, from √
x + 1 − √
x = 1
√ x + 1 +
√ x
, prove that
1 2
√ x + 1
√ x + 1 −
√ x
1 2
√ x
. Use this to prove that if n
1 is a positive integer, then 2
√ n + 1 − 2
1 + 1
√ 2
+ 1
√ 3
+ ··· +
1 √
n 2
√ n − 1
12 Chapter 2
87 Problem
Shew that 1 + x1 + x
2
1 + x
4
1 + x
8
··· 1 + x
1024
= 1 − x
2048
1 − x .
88 Problem Shew that
a
2
+ b
2
+ c
2
− ab − bc − ca = 1
2 a − b
2
+ b − c
2
+ c − a
2
.
89 Problem
Prove that if r ≥ s ≥ t ≥ u ≥ v then
r
2
− s
2
+ t
2
− u
2
+ v
2
≥ r − s + t − u + v
2
2.7
90 Problem AIME 1987 Compute
10
4
+ 32422
4
+ 32434
4
+ 32446
4
+ 32458
4
+ 324 4
4
+ 32416
4
+ 32428
4
+ 32440
4
+ 32452
4
+ 324 .
91 Problem Write a
2
+ a + 1
2
as the sum of three squares.
2.2 Squares of Real Numbers
Kategori