1. To find two numbers such that their sum is to the sum of their squares in a given ratio.
Given ratio 1:5.
Let the numbers be x and 2 units. Therefore
x^2 + 4 = 5(x + 2)
x^2 – 5x = 6
(x – 5/2)^2 = 49/4
x = 6.
The required numbers are 6, 2.
2. To find two numbers such that their difference is to the difference of their squares in a given ratio.
Given ratio 1:5.
Let the numbers be x and 1 unit. Therefore
x^2 – 1 = 5(x – 1)
x + 1 = 5
x = 4.
The required numbers are 4, 1.
3. To find two numbers such that their product is to their sum (or their difference) in a given ratio.
Given ratio 2:1.
Let the numbers be x and 4 units. Therefore
4x = 2(x + 4)
x = 4.
The required numbers are 4, 4.
4. To find two numbers such that the sum of their squares is to their difference in a given ratio.
Given ratio 5:1.
Let the numbers be x and 1 unit. Therefore
x^2 + 1 = 5(x – 1)
x^2 – 5x = -6
(x – 5/2)^2 = 1/4
x = 3.
The required numbers are 3, 1.
5. To find two numbers such that the difference of their squares is to their sum in a given ratio.
Given ratio 5:1.
Let the numbers be x and 1 unit. Therefore
x^2 – 1 = 5(x + 1)
x – 1 = 5
x = 6.
The required numbers are 6, 1.
6. To find two numbers having a given difference and such that the difference of their squares exceeds their difference by a given number.
Necessary condition. The square of their difference must be less than the sum of the said difference and the given excess of the difference of the squares over the difference of the numbers.
Difference of numbers 2, the other given number 20.
Lesser number x. Therefore x+2 is the greater, and
4x + 4 = 22
x = 9/2.
The numbers are 9/2, 13/2.
7. To find two numbers such that the difference of their squares is greater by a given number than a given ratio of their difference.
Necessary condition. The given ratio being 3:1, the square of the difference of the numbers must be less than the sum of three times that difference and the given number.
Given number 10, difference of required numbers 2.
Lesser number x. Therefore the greater is x+2, and
4x + 4 = 3.2 + 10
x = 3.
The numbers are 3, 5.
8. To divide a given square number into two squares.
Given square number 16.
x^2 one of the required squares. Therefore 16–x^2 must be equal to a square.
Take a square of the form (mx–4)^2, m being any integer and 4 the number which is the square root of 16, e.g. take (2x-4)^2, and equate it to 16-x^2. Therefore
4x^2 – 16x + 16 = 16 – x^2
5x^2 = 16x
x = 16/5
The required squares are therefore 256/25, 144/25.
9. To divide a given number which is the sum of two squares into two other squares.
Given number 13 = 2^2 + 3^2.
As the roots of these squares are 2, 3, take (x+2)^2 as the first square and (mx-3)^2 as the second (where m is an integer), say (2x-3)^2. Therefore
(x^2 + 4x + 4) + (4x^2 – 12x + 9) = 13
5x^2 – 8x + 13 = 13
x = 8/5.
The required squares are 324/25, 1/25.
10. To find two square numbers having a given difference.
Given difference 60.
Side of one number x, side of the other x plus any number the square of which is not greater than 60, say 3.
Therefore
(x + 3)^2 – x^2 = 60
x = 17/2.
The required squares are 289/4, 529/4.
11. To add the same (required) number to two given numbers so as to make each of them a square.
Given numbers 2, 3; required number x.
Therefore x+2, x+3 must both be squares.
(1) This is called a double-equation.
To solve it, take the difference between the two expressions and resolve it into two factors; in this case let us say 4, ¼.
Then take either (a) the square of half the difference between these factors and equate it to the lesser expression, or (b) the square of half the sum and equate it to the greater.
In this case (a) the square of half the difference is 225/64. Therefore
x + 2 = 225/64
x = 97/64.
Taking (b) the square of half the sum, we have
x + 3 = 289/64
x = 97/64.
(2) To avoid a double-equation, first find a number which when added to 2, or to 3, gives a square.
Take e.g. the number x^2-2, which when added to 2 gives a square. Therefore since this same number added to 3 gives a square,
x^2 + 1 = a square = (x – 4)^2, say
the number of units in the expression (in this case 4) being so taken that the solution may give x^2 > 2.
Therefore
x = 15/8
The required number is 97/64, as before.
12. To subtract the same (required) number from two given numbers so as to make both remainders squares.
Given numbers 9, 21.
Assuming 9-x^2 as the required number, we satisfy one condition, and the other requires that 12+x^2 shall be a square.
Assume as the side of the square x minus some number the square of which > 12, say 4.
Therefore
(x – 4)^2 = 12 + x^2
x = ½.
The required number is then 35/4.
13. From the same (required) number to subtract two given numbers so as to make both remainders squares.
Given numbers 6, 7.
(1) Let x be the required number.
Therefore x-6, x-7 are both squares.
The difference is 1, which is the product of, say, 2 and ½; so by the rule for solving a double equation,
x – 7 = 9/16
x = 121/16.
(2) To avoid a double-equation, seek a number which exceeds a square by 6, say x^2+6.
Therefore x^2-1 must also be a square = (x-2)^2, say.
Therefore
x = 5/4,
and the required number is 121/16.
14. To divide a given number into two parts and to find a square which when added to each of the two parts gives a square number.
Given number 20.
Take two numbers such that the sum of their squares < 20, say 2, 3.
Add x to each and square.
We then have x^2+4x+4, and x^2+6x+9, and if 4x+4, 6x+9 are respectively subtracted, the remainders are the same square.
Let then x^2 be the required square, and we have only to make 4x+4, 6x+9 the required parts of 20. Thus
10x + 13 = 20
x = 7/10.
The required parts are then 68/10, 132/10, and the required square is 49/100.
15. To divide a given number into two parts and to find a square which, when each part is respectively subtracted from it, gives a square.
Given number 20.
Take (x+m)^2 for the required square, where m^2 is not greater than 20, e.g. take (x+2)^2.
This leaves a square if either 4x+4 or 2x+3 is subtracted.
Let these then be the parts of 20. Therefore
6x + 7 = 20
x = 13/6.
The required parts are therefore 76/6, 44/6, and the required square is 625/36.
16. To find two numbers in a given ratio and such that each when added to an assigned square gives a square.
Given square 9, given ratio 3:1.
If we take a square of side mx+3 and subtract 9 from it, the remainder may be taken as one of the numbers required.
Take e.g., (x+3)^2-9, or x^2+6x, for the lesser number.
Therefore 3x^2+18x is the greater number, and 3x^2+18x+9 must be made a square = (2x-3)^2, say.
Therefore
x = 30.
The required numbers are 1080, 3240.
17. To find three numbers such that, if each give to the next following a given fraction of itself and a given number besides, the results after each has given and taken may be equal.
First gives to the second 1/5 of itself + 6, second to third 1/6 of itself + 7, third to first 1/7 of itself + 8.
Let first and second be 5x, 6x respectively.
When second has taken x+6 from first it becomes 7x+6, and when it has given x+7 to third it becomes 6x-1.
But first when it has given x+6 to second becomes 4x-6; and this too when it has taken 1/7 of third + 8 must become 6x-1.
Therefore 1/7 of third + 8 = 2x+5
Third = 14x–21.
Next, third after receiving 1/6 of second + 7 and giving 1/7 of itself + 8 must become 6x-1.
Therefore
13x – 19 = 6x – 1
x = 18/7.
The required numbers are 90/7, 108/7, 105/7.
18. To divide a given number into three parts satisfying the conditions of the preceding problem.
Given number 80.
Let first give to second 1/5 of itself + 6, second to third 1/6 of itself + 7, and third to first 1/7 of itself + 8.
Let numbers be 5x, 6y and 7z respectively
Then
4x – 6 + z + 8 = 5y – 7 + x + 6 = 6z – 8 + y + 7
x = (26y – 18)/19
z = (17y – 3)/19.
Thus the general solution is
5(26y–18)/19, 6y, 7(17y–3)/19.
Adding the three together and equating to 80 gives:
y(5.26 + 6.19 + 7.17) – 5.18 – 7.3 = 80.19
y = 1631/363.
The required numbers are 9440/363, 9786/363, 9814/363.
19. To find three squares such that the difference between the greatest and the middle has to the difference between the middle and the least a given ratio.
Given ratio 3:1.
Assume the least square = x^2, the middle = x^2+2x+1.
Therefore the greatest = x^2+8x+4 = square = (x+3)^2, say.
Thus
x = 5/2.
The squares are 121/4, 49/4, 25/4.
20. To find two numbers such that the square of either added to the other gives a square.
Assume for the numbers x, 2x+1, which by their form satisfy one condition.
The other condition gives
4x^2 + 5x + 1 = square = (2x - 2)^2, say.
Therefore
x = 3/13.
The numbers are 3/13, 19/13.
21. To find two numbers such that the square of either minus the other number gives a square.
x+1, 2x+1 are assumed, satisfying one condition.
The other condition gives
4x^2 + 3x = square = 9x^2, say.
Therefore
x = 3/5.
The numbers are 8/5, 11/5.
22. To find two numbers such that the square of either added to the sum of both gives a square.
Assume x, x+1 for the numbers. Thus one condition is satisfied.
It remains that
x^2 + 4x + 2 = square = (x – 2)^2, say.
Therefore
x = ¼.
The numbers are ¼, 5/4.
23. To find two numbers such that the square of either minus the sum of both gives a square.
Assume x, x+1 for the numbers, thus satisfying one condition.
Then
x^2 – 2x – 1 = square = (x – 3)^2, say.
Therefore
x = 5/2.
The numbers are 5/2, 7/2.
24. To find two numbers such that either added to the square of their sum gives a square.
Since x^2+3x^2, x^2+8x^2 are both squares, let the numbers be 3x^2, 8x^2 and their sum x.
Therefore
121x^4 = x^2
11x^2 = x
x = 1/11.
The numbers are therefore 3/121, 8/121.
25. To find two numbers such that the square of their sum minus either number gives a square.
If we subtract 7 or 12 from 16, we get a square.
Assume then 12x^2, 7x^2 for the numbers, and 16x^2 for the square of their sum.
Hence
19x^2 = 4x
x = 4/19.
The numbers are 192/361, 112/361.
26. To find two numbers such that their product added to either gives a square, and the sides of the two squares added together produce a given number.
Let the given number be 6.
Since x(4x-1)+x is a square, let x, 4x-1 be the numbers.
Therefore 4x^2+3x-1 is a square, and the side of this square must be 6-2x [since 2x is the side of the first square and the sum of the sides of the square is 6].
Since
4x^2 + 3x – 1 = (6 – 2x)^2
we have
x = 37/27.
The numbers are 37/27, 121/27.
27. To find two numbers such that their product minus either gives a square, and the sides of the two squares so arising when added together produce a given number.
Let the given number be 5.
Assume 4x+1, x for the numbers, so that one condition is satisfied.
Also
4x^2 – 3x – 1 = (5 – 2x)^2
x = 26/17.
The numbers are 26/17, 121/17.
28. To find two square numbers such that their product added to either gives a square.
Let the numbers be x^2, y^2.
Therefore x^2y^2+y^2, x^2y^2+x^2 are both squares.
To make the first expression a square we make x^2+1 a square, putting
x^2 + 1 = (x – 2)^2, say.
Therefore
x = ¾
x^2 = 9/16.
We now have to make (9/16)(y^2 + 1) a square [ and y must be different from x].
Put
9y^2 + 9 = (3y – 4)^2, say
so
y = 7/24.
Therefore the numbers are 9/16, 49/576.
29. To find two square numbers such that their product minus either gives a square.
Let x^2, y^2 be the numbers
Then x^2y^2 – y^2, x^2y^2 – x^2 are both squares.
A solution of x^2 – 1 = (a square) is x^2 = 25/16.
We now have to solve
(25/16)y^2 – 25/16 = a square.
Put
y^2 – 1 = (y – 4)^2, say.
Therefore
y = 17/8
The numbers are 289/64, 100/64.
30. To find two numbers such that their product +/- their sum gives a square.
Now m^2 + n^2 +/- 2mn is a square.
Put 2, 3, say, for m, n respectively, and of course
2^2 + 3^2 +/- 2.2.3 is a square.
Assume then product of numbers = (2^2 + 3^2) x^2 or 13x^2, and
sum = 2.2.3x^2 or 12x^2.
The product being 13x^2, let x, 13x be the numbers.
Therefore their sum
14x = 12x^2
x = 7/6.
The numbers are therefore 7/6, 91/6.
31. To find two numbers such that their sum is a square and their product +/- their sum gives a square.
2.2m.m = a square, and (2m)^2 + m^2 +/- 2.2m.m = a square.
If m = 2, 4^2 + 2^2 +/- 2.4.2 = 36 or 4.
Let then the product of the numbers be (4^2 + 2^2)x^2 or 20x^2, and their sum 2.4.2x^2 or 16x^2, and take 2x, 10x for the numbers.
Then
12x = 16x^2
x = ¾.
The numbers are 6/4, 30/4.
32. To find three numbers such that the square of any one of them added to the next following gives a square.
Let the first be x, the second 2x+1, and the third 2(2x+1)+1 or 4x+3, so that two conditions are satisfied.
The last condition gives
(4x + 3)^2 + x = square = (4x – 4)^2, say.
Therefore
x = 7/57
and the numbers are 7/57, 71/57, 199/57.
33. To find three numbers such that the square of any one of them minus the next following gives a square.
Assume x+1, 2x+1, 4x+1 for the numbers, so that two conditions are satisfied.
Lastly,
16x^2 + 7x = square = 25x^2, say
so
x = 7/9.
The numbers are 16/9, 23/9, 37/9.
34. To find three numbers such that the square of any one added to the sum of all three gives a square.
(1/2(m-n))^2 + mn is a square. Take a number separable into two factors (m, n) in three ways, say 12, which is the product of (1, 12), (2, 6) and (3, 4).
The values then of ½(m – n) are 11/2, 2, ½.
Take 11x/2, 2x and x/2 for the numbers, and for their sum 12x^2.
Therefore
8x = 12x^2, and
x= 2/3.
The numbers are 11/3, 4/3, 1/3.
35. To find three numbers such that the square of any one minus the sum of all three gives a square.
(1/2(m+n))^2 – mn is a square. Take, as before, a number divisible into factors in three ways, as 12.
Let then 13x/2, 4x, 7x/2 be the numbers, and their sum 12x^2.
Therefore
14x = 12x^2, and
x = 7/6.
The numbers are 273/6, 28/6, 147/6.