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1
Joseph Muscat 2015
Diophantus of Alexandria
Arithmetica
Book I
[email protected]
27 November 2015
1
Notation
Diophantus’s Arithmetica 1 is a list of about 128 algebraic problems with solutions. In it he introduced algebraic manipulations on equations including
a symbol for one unknown (probably following other authors in Alexandria).
Although most of his methods were known to his Babylonian and Egyptian
predecessors, his symbols allowed for clearer solutions and to consider more
complicated problems. His notation was as follows:
unknown
square
cube
etc.
∈
reciprocal
fraction
units
equals
subtract
Diophantus
ς`
∆
K
∆∆
∆K
ς`×
×
β γ or βγ
M
ισ
Modern
x
x2
x3
x4
x5
1/x
3
2
=
−
∈
Like all Greeks at the time, Diophantus used the (extended) Greek alphabet
to denote numbers, namely, ὰ 1, β̀ 2,. . . , θ̀ 9, ὶ 10, κ̀ 20, . . . , up to ϡ’ 900. So,
with his notation, a quadratic equation like 2x2 − 3x + 1 = 4 would be written
as
∆β̀ Mὰ ς`γ̀ ισ Mδ̀
Note that juxtaposed terms like ∆β̀ Mὰ are assumed to be added. For ease
of reading, the modern equivalent symbols are used here, except for addition,
squares and cubes, so the above equation is written as
2∆ + 1 − 3x = 4.
Only positive numbers “made sense” to the Greeks (because they denoted
‘physical’ quantities like length and area), so equations had to be ‘balanced’;
x2 + 3x = x was not a ‘realistic’ equation.
1 http://www.e-rara.ch/doi/10.3931/e-rara-9423
Joseph Muscat 2015
2
2
Problems
Problem 1
ence (40).
To split a given number (100) in two parts having a given differ-
Problem 2
(3:1).
To split a given number (60) in two parts having a given ratio
Problem 3
To split a given number (80) in two parts, the larger of which
has a given ratio (3:1) to the sum of the smaller and a given number (4).
Problem 4
To find two numbers in a given ratio (5:1) and having a given
difference (20).
Problem 5
To split a given number (100) in two parts such that given fractions of them (1/3, 1/5) have a given sum (30).
Problem 6
To split a given number (100) in two parts such that given fractions of them (1/4, 1/6) have a given difference (20).
Problem 7
To find a number such that when two given numbers (100,20)
are subtracted from it, the remaining parts have a given ratio (1:3).
Problem 8
To find a number such that when two given numbers (100,20)
are added to it, the sums have a given ratio (3:1).
Problem 9
To find a number which when subtracted from two given numbers
(100,20), the remaining parts have a given ratio (6:1).
Problem 10
To find a number such that when added to a given number (20)
and subtracted from another (100), the results have a given ratio (1:4).
Problem 11
To find a number such that when a given number (20) is added
to it, and another (100) subtracted from it, the results have a given ratio (3:1).
Problem 12
To split a given number (100) twice into two pairs such that the
first of each pair have a given ratio (2:1) as do the seconds (3:1).
Problem 13
To split a given number (100) as three pairs such that the second
number of each pair and the first of the next pair (in cyclic order) have given
ratios (3:1, 2:1, 4:1).
Joseph Muscat 2015
Problem 14
(3:1).
3
To find two numbers whose product and sum have a given ratio
Problem 15
To find two numbers such that when a given number (30) is
transferred from the second to the first, and when another number (50) is transferred from the first to the second, the resulting pairs have given ratios (2:1,
3:1).
Problem 16
(20,30,40).
To find three numbers such that the sum of any two are given
Problem 17
To find four numbers such that the sums of any three of them
are given (20,22,24,27).
Problem 18
To find three numbers such that the sum of any two is greater
than the third by given amounts (20,30,40).
Problem 19 is the same as 18, solved with a different method.
Problem 20
To find four numbers such that the sum of any three is greater
than the fourth by given amounts (20,30,40,50).
Problem 21 is the same as 20, solved with a different method.
Problem 22
To split a given number (100) in three parts such the first two
together have a given ratio to the third (3:1) and the last two together have a
given ratio to the first (4:1).
Problem 23
To find three numbers such that the third exceeds the second
by a given fraction (1/3) of the first; the second exceeds the first by a fraction
(1/3) of the third; and the first exceeds a fraction (1/3) of the second by a given
amount (10).
Problem 24 is the same as 23, solved with a different method.
Problem 25
To find three numbers such that if each transfers a given fraction
of itself (1/3,1/4,1/5) to the next (cyclically), the results are all equal.
Problem 26
To find four numbers such that if each transfers a given fraction
of itself (1/3,1/4,1/5,1/6) to the next (cyclically), the results are all equal.
Problem 27
To find three numbers such that if each receives a given fraction
of the sum of the remaining numbers (1/3,1/4,1/5), the results are all equal.
Joseph Muscat 2015
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Problem 28
To find four numbers such that if each receives a given fraction of
the sum of the remaining numbers (1/3,1/4,1/5,1/6), the results are all equal.
Problem 29
To find a number which when multiplied by two given numbers
(200,5) makes one a square and the other its side.
Problem 30
To find two numbers whose sum and product are given (20,96).
Problem 31
(20,208).
To find two numbers whose sum and sum of squares are given
Problem 32
given (20,80).
To find two numbers whose sum and difference of squares are
Problem 33
(4,96).
To find two numbers whose difference and product are given
Problem 34
To find two numbers in a given ratio (3:1) such that their sum
and the sum of their squares have a given ratio (1:5).
Problem 35
To find two numbers in a given ratio (3:1) such that their difference and the sum of their squares have a given ratio (1:10).
Problem 36
To find two numbers in a given ratio (3:1) such that their sum
and the difference of their squares have a given ratio (1:6).
Problem 37
To find two numbers in a given ratio (3:1) such that their difference and the difference of their squares have a given ratio (1:12).
Problem 38
To find two numbers in a given ratio (3:1) such that the smaller
square and the larger number have a given ratio (6:1).
Problem 39
To find two numbers in a given ratio (3:1) such that the smaller
square and its side have a given ratio (6:1).
Problem 40
To find two numbers in a given ratio (3:1) such that the smaller
square and their sum have a given ratio (2:1).
Problem 41
To find two numbers in a given ratio (3:1) such that the smaller
square and their difference have a given ratio (6:1).
Problem 42 restates problems 38 to 41 for the larger square.
Joseph Muscat 2015
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Problem 43
To find a number, together with two given numbers (3,5), such
that when any two are added together then multiplied with the third, the result
is three numbers which have equal differences.
Following is a sample of problems in the other books:
Book II
Problem 8
To split a given square (16) in two squares.
Problem 13
To find a number whose differences from two given numbers
(9,21) are both squares.
Book III
Problem 9
To find three squares at equal intervals.
Book IV
Problem 21
To find four numbers such that the product of any two added
to one gives a square.
Book V
Problem 1
To find three numbers in geometric proportion such that when
a given number (12) is subtracted from them, they form squares.
Book VI
Problem 16
To find the sides of a right triangle of given area (60) and
perimeter (40).
3
Solutions
Solution 1
Numbers
sum
subtract
divide
solution
x
x + 40
2x + 40 = 100
2x = 60
x = 30
30
70
Solution 2
Numbers
sum
divide
solution
x 3x
4x = 60
x = 15
15 45
Solution 3
Numbers
sum
subtract
divide
solution
x 3x + 4
4x + 4 = 80
4x = 76
x = 19
19
61
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Joseph Muscat 2015
Solution 4
Numbers
difference
divide
solution
x 5x
4x = 20
x=5
5
25
Solution 5
Numbers
sum
subtract
divide
solution
5x 90 − 3x
2x + 90 = 100
2x = 10
x=5
25
75
Numbers
sum
subtract
divide
solution
6x
4x + 80
10x + 80 = 100
10x = 20
x=2
12
88
Solution 6
Solution 7
Number
remainders
condition
add/subtract
divide
x
x − 100
x − 20
x − 20 = 3x − 300
2x = 280
x = 140
Solution 8
Number
sums
condition
add
divide
x
x + 100
x + 20
x + 100 = 3x + 60
2x = 40
x = 20
Solution 9
Number
remainders
condition
add
divide
x
100 − x
20 − x
100 − x = 120 − 6x
5x = 20
x=4
Solution 10
Number
new numbers
condition
add
divide
x
x + 20
100 − x = 4x + 80
5x = 20
x=4
100 − x
7
Joseph Muscat 2015
Solution 11
Number
new numbers
condition
add
divide
x
x + 20
x + 20 = 3x − 300
2x = 320
x = 160
x − 100
Solution 12
x
2x
100 − x = 300 − 6x
5x = 200
x = 40
40,60
First pair
Second pair
condition
add
divide
solution
100 − x
100 − 2x
20,80
Solution 13
First pair
Second pair
Third pair
sum
add
divide
solution
x
4x − 100
100 − 4x
25x − 300 = 100
25x = 400
x = 16
(16,84),(28,72),(36,64)
24x − 300
200 − 8x
4x
Solution 14
Try numbers
condition
Modify
condition
add
solution
x
2x = 3x + 6
x
4x = 3x + 12
x = 12
12
2
no solution
4
4
Solution 15
2nd number
after first transfer
2nd number
1st number
original 1st number
after second transfer
1st number
2nd number
condition
add
divide
solution
x
x − 30
2x − 60
2x − 90
2x − 140
x + 50
x + 50 = 6x − 420
5x = 470
x = 94
94
98
8
Joseph Muscat 2015
Solution 16
Sum of numbers
numbers
condition
add
divide
solution
x
x − 40
2x − 70 = 20
2x = 90
x = 45
5
x − 30
x − 20
15
25
Solution 17
Sum of numbers
numbers
sum of first three
add
divide
solution
x
x − 27
3x − 73 = 20
3x = 93
x = 31
4
x − 24
x − 22
x − 20
7
9
11
Solution 18
Sum of numbers
2nd and 3rd together
condition
add
numbers (similarly)
sum
multiply
add
solution
x
x−1st
x−1st = 1st + 40
Twice 1st= x − 40
2× x − 20
x = 2× 3x − 45
2x = 3x − 90
x = 90
25
2× x − 15
2× x − 10
30
35
Another solution:
3rd number
1st and 2nd
3rd and 2nd
so, 1st and 3rd and twice 2nd
twice 2nd
2nd
1st
condition
add
divide
solution
x
3rd + 20
1st + 30
1st + 3rd + 50
50
25
x−5
2x − 5 = 65
2x = 70
x = 35
30
25
35
9
Joseph Muscat 2015
Solution 20
Sum of numbers
sum of 2nd, 3rd, 4th
condition
add
numbers (similarly)
sum
add
solution
x
x−1st
x−1st = 1st + 50
Twice 1st = x − 50
2× x − 25
2x − 70 = x
x = 70
10
2× x − 20
2× x − 15
2× x − 10
15
20
25
Solution 22
Smallest number
sum of others
condition
add
divide
solution
x
100 − x
100 − x = 4x
5x = 100
x = 20
20
55
25
Solution 23
Smallest number
2nd number
2nd minus 1st
3rd number
condition
add
divide
solution
x + 10
3x
2x − 10
6x − 30
9x − 90 = x + 10
8x = 100
x = 2× 25
22 12
37 21
45
Solution 25
1st number
2nd number
2nd after transfer
transfer from 3rd to 1st
3rd
condition
solution
3x
4 units
x+3
x + 3 + x − 3x
15 − 5x
x + 3 = 15 − 5x − (3 − x) + 1
5x = 10
x=2
6
4
5
10
Joseph Muscat 2015
Solution 26
1st number
2nd number
2nd after transfer
transfer 4th to 1st
4th
transfer 3rd to 4th
3rd
condition
add
divide
solution
3x
4 units
x+3
3−x
18 − 6x
6x − 12
30x − 60
24x − 47 = x + 3
23x = 50
x = 50/23
150
92 120 114
Solution 27
First number
transfer from 2nd and 3rd
total sum
1st (or 2nd) after transfer
four 2nd, and rest
three 2nd
2nd
3rd
fifth of 1st and 2nd
condition
solve
numbers
solution
x
1 unit
x+3
x+1
4x + 4
3x + 1
x + 13
8
3 −x
2
1
5 x + 15
41
3
15 − 5 x = x + 1
x = 13
12
13
12
17
12
19
12
13
17
19
Solution 28
First number
transfer from rest
total sum
1st after transfer
four 2nd, and rest
three 2nd
2nd
similarly, 3rd,4th
1st,2nd,3rd together
sum
solve
solution
x
1
x+3
x+1
4x + 4
3x + 1
x + 13
x + 12
x + 35
3x + 65
4x + 43
30 = x + 3
x = 47
90
47
77
92 101
11
Joseph Muscat 2015
Solution 29
Number
products
condition
divide
divide
x
5x
200x = 25∆
∆ = 8x
x=8
200x
Solution 30
x + 10
100 − ∆ = 96
∆=4
x=2
12
Numbers
product
add
root
solution
10 − x
8
Solution 31
Numbers
squares
sum of squares
divide
root
solution
x + 10
∆ + 20x + 100
2∆ + 200 = 208
∆=4
x=2
12
10 − x
∆ + 100 − 20x
8
Solution 32
Numbers
difference of squares
divide
solution
10 − x
x + 10
40x = 80
x=2
12
8
Solution 33
Numbers
product
add
root
solution
x+2
∆ − 4 = 96
∆ = 100
x = 10
12
x−2
x
4x
10∆ = 20x
x=2
2
3x
10∆
8
Solution 34
Numbers
sums
ratio
divide
solution
6
Solution 35
Numbers
sum and difference
ratio
divide
solution
x
10∆
10∆ = 20x
x=2
2
3x
2x
6
12
Joseph Muscat 2015
Solution 36
Numbers
sum,diff
ratio
divide
solution
x
8∆
8∆ = 24x
x=3
3
3x
4x
Numbers
differences
ratio
divide
solution
x
2x
8∆ = 24x
x=3
3
3x
8∆
9
Solution 37
9
Solution 38
Numbers
condition
divide
solution
x
∆ = 18x
x = 18
18
3x
Numbers
condition
divide
solution
x
∆ = 6x
x=6
6
3x
Numbers
condition
divide
solution
x
∆ = 8x
x=8
8
3x
Numbers
condition
divide
solution
x
∆ = 12x
x = 12
12
3x
54
Solution 39
18
Solution 40
24
Solution 41
36
13
Joseph Muscat 2015
Solution 43
Number
sum of pairs
multiplied by third
Possibility 1
difference
Possibility 2
Possibility 3
solutions
x
x+5
3x + 15
3x + 15
2x = 3x − 15
x = 15
3x + 15
5x − 15 = 15 − 3x
8x = 30
x = 15/4
8x
15 − 5x = 2x
x = 15/7
60
105/4
150/7
x+3
5x + 15
5x + 15
8
8x
8x
8x
5x + 15
3x + 15
5x + 15
90
135/4
180/7
120
30
120/7
Book II
Solution 8
Squares
try square on x − 4
try square on 2x − 4
solution
∆
16 − ∆
∆ + 16 − 8x = 16 − ∆
2∆ = 8x
x=4
4∆ + 16 − 16x = 16 − ∆
5∆ = 16x
x = 16/5
256/25
144/25
Solution 13
1st square
number
try 2nd square on x − 4
solution
∆
9−∆
∆ + 16 − 8x
∆ + 16 − 8x = ∆ + 12
8x = 4
x = 1/2
143/16
Book III
Solution 9
1st and 2nd sides
squares
3rd, try square on, say, 8 − x
solve
solution
x
x+1
∆
∆ + 2x + 1 ∆ + 4x + 2
∆ + 64 − 16x = ∆ + 4x + 2
x = 31/10
961
1681
2401
14
Joseph Muscat 2015
Book IV
Solution 21
1st number
square on x + 1
∴ 2nd number
square on 2x + 1
∴ 3rd number
square on 3x + 1
∴ 4th number
2nd times 3rd plus 1
equals square on 2x + 3
3rd times 4th plus 1
equals square on 6x + 5
2nd times 4th plus 1
square on 3x − 4
solve
solution
x
∆ + 2x + 1
x+2
4∆ + 4x + 1
4x + 4
9∆ + 6x + 1
9x + 6
4∆ + 12x + 9
36∆ + 60x + 25
9∆ + 24x + 13
9∆ − 24x + 16
24x + 13 = −24x + 16
x = 1/16
1/16
33/16
68/16
105/16
Book V
Solution 1
1st number, can take square on
numbers
subtract 12
∴ difference of squares
equals product of sum and difference of sides
sum of sides
difference of sides
smaller side
smaller square
solve
solution
13/2
169/4
13/2 x
∆
13
121/4
∆ − 12
2 x − 12
∆ − 13
2 x
equals product of x and x − 13/2
x
x − 13/2
13/4
13
169
2 x − 12 = 16
x = 361/104
169/4
361/16
130321/10816
Book VI
Solution 16
hypotenuse
sum of sides
squared
solve
sides
product
solve
solution
x
40 − x
∆ + 1600 − 2x = ∆ + 120
x = 17
x + 23/2
529/4 − ∆ = 120
x = 7/2
15
23/2 − x
8
17