1. To divide a given number into two cubes such that the sum of their sides is a given number.
Given number 370, given sum of sides 10.
Sides of cubes 5+x, 5-x, satisfying one condition.
Therefore
30x^2 + 250 = 370,
x = 2,
and the cubes are 7^3, 3^3, or 343, 27.
2. To find two numbers such that their difference is a given number, and also the difference of their cubes is a given number.
Difference 6, difference of cubes 504.
Numbers x+3, x-3.
Therefore
18x^2 + 54 = 504
x^2 = 25
x = 5.
The sides of the cubes are 8, 2 and the cubes 512, 8.
3. To multiply one and the same number into a square and its side respectively so as to make the latter product a cube and the former product the side of the cube.
Let the square be x^2. Its side being x, let the number be 8/x.
Hence the products are 8x, 8, and
(8x)^3 = 8.
Therefore
2 = 8x
x = ¼
and the number to be multiplied is 32.
The square is 1/16 and its side ¼.
4. To add the same number to a square and its side respectively and make them the same [ i.e. make the first product a square of which the second product is the side].
Square x^2, with side x.
Let the number added to x^2 be such as to make a square, say 3x^2.
Therefore
3x^2 + x = side of 4x^2 = 2x,
x = 1/3.
The square is 9/25, its side 3/5, and the number 21/25.
5. To add the same number to a square and its side and make them the opposite.
Square x^2, the number a square number of times x^2 minus x, say 4x^2 – x.
Hence
5x^2 – x = side of 4x^2 = 2x,
x = 3/5.
The square is 9/25, its side 3/5, and the number 21/25.
6. To add the same square number to a cube and a square and make them the same.
Let the cube be x^3 and the square any square number of times x^2, say 9x^2.
We want now a square which when added to 9x^2 makes a square. Take two factors of 9, say 9 and 1, subtract 1 from 9, take half the difference and square. This gives 16.
Therefore 16x^2 is the square to be added.
Next
x^3 + 16x^2 = a cube = 8x^3, say;
x = 16/7.
The cube is therefore 4096/343, the square 2304/49, and the added square number 4096/49.
7. To add the same square number to a cube and a square respectively and make them the opposite.
For brevity call the cube (1), the second square (2) and the added square (3).
Now, since (2) + (3) = a cube, suppose (2) + (3) = (1).
Since a^2+b^2+/-2ab is a square, suppose (1) = (a^2+b^2), (3) = 2ab, so that the condition that (1) + (3) = square is satisfied.
But (3) is a square, and, in order that 2ab may be a square, we put a = x, b = 2x.
Suppose then (1) = x^2+(2x)^2 = 5x^2, (3) = 2.x.2x = 4x^2; therefore
(2) = x^2, by subtraction.
But 5x^2 is a cube; therefore
x =5.
The cube (1) = 125, the square (2) = 25, the square (3) = 100.
8. To add the same number to a cube and its side and make them the same.
Added number x, cube 8x^3, say. Therefore second sum = 3x, and this must be the side of 8x^3 + x.
That is, 8x^3 + x = 27x^3, and 19x^2 = x, or 19x^2 = 1.
But 19 is not a square. Hence we must find, to replace it, some square number. Now 19x^2 arises from 27x^3 – 8x^3, where 27 is the cube of 3, and 8 the cube of 2. And the 3x comes from the assumed side 2x, by increasing the coefficient by unity.
Thus we must find two consecutive numbers such that their cubes differ by a square.
Let them be y, y+1.
Therefore 3y^2+3y+1 = square = (1-2y)^2, say, and y = 7.
Going back to the beginning, we assume added number = x, side of cube = 7x.
The side of the new cube is then 8x, and
343x^3 + x = 512x^3
x^2 = 1/169
x = 1/13.
The cube is 343/2197, its side 7/13, and the added number 1/13.
9. To add the same number to a cube and its side and make them the opposite.
Suppose the cube is 8x^3, its side being 2x, and the added number is 27x^3-2x. (The coefficients 8, 27 are chosen as cube numbers.)
Therefore 35x^3 – 2x = side of cube 27x^3 = 3x, or 35x^2 = 5.
This gives no rational value.
But 35 = 27 + 8, and 5 = 3 + 2.
Therefore we have to find two cubes such that their sum has to the sum of their sides the ratio of a square to a square.
Let sum of sides = any number, 2 say, and side of first cube = z, so that the side of the other cube is 2-z. Therefore 8 – 12z + 6z^2 must be twice a square.
That is, 4 – 6z + 3z^2 = square = (2 – 4z)^2, say; z = 10/13, and the sides are 10/13, 16/13.
Neglecting the denominator and the factor 2 in the numerators, we take 5, 8 for the sides.
Starting afresh, we put for the cube 125x^3 and for the number to be added 512x^3 – 5x; we thus get
637x^3 – 5x = 8x
x = 1/7.
The cube is 125/343, its side 5/7, and the added number 267/343.
10. To find two cubes the sum of which is equal to the sum of their sides.
Let the sides be 2x, 3x.
This gives 35x^3 = 5x; but this equation give an irrational result.
We have therefore, as in the last problem, to find two cubes the sum of which has to the sum of their sides the ratio of a square to a square.
These are found, as before, to be 5^3, 8^3.
Assuming then 5x, 8x as the sides of the required cubes, we obtain the equation
637x^3 = 13x
x = 1/7.
The cubes are 125/343, 512/343.
11. To find two cubes such that their difference is equal to the difference of their sides.
Assume 2x, 3x as the sides.
This gives 19x^3 = x, and x is irrational.
We have therefore to find two cubes such that their difference has to the difference of their sides the ratio of a square to a square. Let them be (z+1)^3, z^3, so that the difference of the sides may be a square, namely 1.
Therefore 3z^2 + 3z + 1 = square = (1 - 2z)^2, say, and z = 7.
Starting afresh, assume 7x, 8x as the sides; therefore
169x^3 = x,
x = 1/13.
The sides of the two cubes are therefore 7/13, 8/13.
12. To find two numbers such that the cube of the greater + the less = cube of the less + the greater.
Assume 2x, 3x for the numbers.
Therefore 27x^3 + 2x = 8x^3 + 3x, or 19x^2 = x, and x is irrational.
But 19 is the difference of two cubes, and 1 the difference of their sides. Therefore, as in the last problem, we have to find two cubes such that their difference has to the difference of their sides the ratio of a square to a square.
The sides of these cubes are found, as before, to be 7, 8.
Starting afresh, we assume 7x, 8x for the numbers; then
343x^3 + 8x = 512x^3 + 7x,
x = 1/13.
The numbers are 7/13, 8/13.
13. To find two numbers such that either, or their sum, or their difference added to unity gives a square.
Take for the first number any square less 1; let it be, say, 9x^2 + 6x. But the second + 1 = a square; and first + second + 1 also = a square. Therefore we must find a square such that the sum of that square and 9x^2 + 6x = a square.
Take factors of the difference 9x^2 + 6x, say 9x + 6, x; the square of half the difference between these factors = 16x^2 + 24x + 9.
Therefore, if we put for the second number this expression minus 1, or 16x^2 + 24x + 8, three conditions are satisfied.
The remaining condition gives difference + 1 = square, or
7x^2 + 18x + 9 = square = (3 – 3x)^2, say.
Therefore x = 18, and (3024, 5624) is a solution.
14. To find three square numbers such that their sum is equal to the sum of their differences.
Sum of differences = (greatest) – (middle) + (middle) – (least) + (greatest) – (least) = twice difference of greatest and least.
This is equal to the sum of all three, by hypothesis.
Let the least square be 1, the greatest x^2+2x+1; therefore twice difference of greatest ad least = sum of the three = 2x^2 + 4x.
But least + greatest = x^2+2x+2, so that middle = x^2+2x-2.
Hence
x^2 + 2x – 2 = square = (x – 4)^2, say, and
x = 9/5.
The squares are (196/25, 121/25, 1) or (196, 121, 25).
15. To find three numbers such that the sum of any two multiplied into the third is a given number.
Let (first + second) x (third) = 35, (second + third) x (first) = 27 and (third + first) x second = 32.
Let the third be x.
Therefore (first + second) = 35/x.
Assume first = 10/x, second = 25/x; then 250/(x^2) + 10 = 27 and 250/(x^2) + 25 = 32.
These equations are inconsistent; but they would not be if 25 – 10 were equal to 32 – 27 or 5.
Therefore we have to divide 35 into two parts (to replace 25 and 10) such that their difference is 5. The parts are 15, 20 [cf. I. 1].
We take therefore 15/x for the first number, 20/x for the second, and we have
300/(x^2) + 15 = 27, or
300/(x^2) + 20 = 32
x = 5.
(3, 4, 5) is a solution.
16. To find three numbers such that their sum is a square, while the sum of the square of each added to the next following number gives a square.
Let the middle number be any number of x’s, say 4x; we have therefore to find what square + 4x gives a square. Split 4x into two factors, say 2x, 2, and take the square of half their difference, (x – 1)^2. This is the square required.
Thus the first number is x – 1.
again, 16x^2 + third number = a square. Therefore, if we subtract 16x^2 from a square, we shall have the third number. Take as the side of this square the side of 16x^2, or 4x, plus 1.
Therefore third number = (4x + 1)^2 – 16x^2 = 8x + 1.
Now the sum of the three numbers = a square; therefore 13x = a square = 169y^2, say.
The numbers are then 13y^2 – 1, 52y^2, 104y^2 + 1.
Lastly, (third)^2 + first = a square.
Therefore
10816y^4 + 221y^2 = a square, or
10816y^2 + 221 = a square = (104y + 1)^2, say.
Therefore
y = 220/208 = 55/52.
(36621/2704, 157300/2704, 317304/2704) or (2817/208, 3025/52, 3051/26) is a solution.
17. To find three numbers such that their sum is a square, while the square on any one minus the next following also gives a square.
The solution is precisely similar to the last.
The middle number is assumed to be 4x. The square which exceeds this by a square is (x+1)^2, and we therefore take x+1 for the first number.
For the third number we take 16x^2 – (4x – 1)^2 or 8x – 1.
The sum of the numbers being a square,
13x = a square = 169y^2, say.
The numbers are then 13y^2 + 1, 52y^2, 104y^2 – 1,
Lastly, since
(third)^2 – first = a square,
10816y^4 – 221y^2 = a square, or
10816y^2 – 221 = a square = (104y – 1)^2, say.
Thus
y = 111/104, and
(170989/10816, 640692/10816, 1270568/10816) or (13153/832, 12321/208, 12217/104) is a solution.
18. To find two numbers such that the cube of the first added to the second gives a cube, and the square of the second added to the first gives a square.
First number x. Therefore second is a cube number minus x^3, say 8–x^3.
Therefore x^6-16x^3+64+x = a square = (x^3+8)^2, say, whence
32x^3 = x, or
32x^2 = 1.
This gives an irrational result; x would however be rational if 32 were a square.
But 32 comes from 4 times 8. We must therefore substitute for 8 in our assumptions a cube which when multiplied by 4 gives a square. If y^3 is the cube,
4y^3 = a square = 16y^2, say;
whence
y = 4.
Thus we must assume x, 64-x^3 for the numbers.
Therefore
x^6 – 128x^3 + 4096 + x = a square = (x^3 + 64)^2, say;
whence
256x^3 = x
x = 1/16.
The numbers are 1/16, 262143/4096.
19. To find three numbers indeterminately such that the product of any two increased by 1 is a square.
Take for the product of first and second some square minus 1, say x^2+2x; this satisfies one condition.
Let second = x, so that first = x+2.
Now product of second and third + 1 = a square; let the square be (3x+1)^2, so that product of second and third = 9x^2+6x.
Therefore third = 9x+6.
Also product of third and first + 1 = a square; therefore 9x^2 + 24x + 13 = a square.
Now if 13 were a square, and the coefficient of x were twice the product of the side of this square and the side of the coefficient of x^2, the problem would be solved indeterminately.
But 13 comes from 2.6 + 1, the 2 in this from twice 1, and the 6 from twice 3. Therefore we want two coefficients (to replace 1, 3) such that the product of their doubles + 1 = a square, or four times their product + 1 = a square.
Now four times the product of any two numbers plus the square of their difference gives a square. Thus the requirement is satisfied by taking as coefficients any two consecutive numbers, since the square of their difference is 1. [The assumption of two consecutive numbers for the coefficients simultaneously satisfies the second of the two requirements indicated in the italicised sentence above.]
Beginning again, we take (x + 1)^2 – 1 for the product of the first and second and (2x + 1)^2 – 1 for the product of second and third.
Let the second be x, so that first = x + 2, third = 4x + 4.
[Then product of first and third + 1 = 4x^2 + 12x + 9, and the third condition is satisfied.]
Thus the required indeterminate solution is (x+2, x, 4x+4), so a solution is, e.g. (3, 1, 8).
20. To find four numbers such that the product of any two increased by unity is a square.
For the product of the first and second take a square minus 1, say (x+1)^2 – 1 = x^2 + 2x.
Let first = x, so that second = x+2.
For the product of first and third take (2x+1)^2 – 1, or 4x^2 + 4x, the coefficient of x being the number next following the coefficient (1) taken in the first case, for the reason shown in the last problem; thus third number = 4x+4.
Similarly take (3x+1)^2 – 1, or 9x^2 + 6x, for the product of first and fourth; therefore fourth = 9x+6.
And product of third and fourth + 1
= (4x + 4)(9x + 6) + 1 = 36x^2 + 60x + 25, which is a square.
lastly, product of second and fourth + 1
= 9x^2 + 24x + 13 = square = (3x – 4)^2, say;
which gives
x = 1/16.
All the conditions are now satisfied, and (1/16, 33/16, 17/4, 105/16) is the solution.
21. To find three numbers in proportion and such that the difference of any two is a square.
Assume x for the least, x+4 for the middle (in order that the difference of middle and least may be a square), x+13 for the greatest (in order that the difference of greatest and middle may be a square).
If now 13 were a square, we should have an indeterminate solution satisfying three of the conditions. We must therefore replace 13 by a square which is the sum of two squares. Any rational right-angled triangle will furnish what is wanted, say 3, 4, 5;
we therefore put for the numbers x, x+9, x+25.
The fourth condition gives
x(x + 25) = (x + 9)^2,
x = 81/7.
Thus (81/7, 144/7, 256/7) is a solution.
22. To find three numbers such that their solid content added to any one of them gives a square.
Assume continued product x^2+2x, first number 1, second number 4x+9, so that two conditions are satisfied.
The third number is then (x^2+2x)/(4x+9).
This cannot be divided out unless x^2:4x = 2x:9 or, alternately, x^2:2x = 4x:9; but it could be done if 4 were half of 9.
Now 4x comes from 6x-2x, and the 6x in this from twice 3x; the 9 comes from 3^2.
Therefore we have to find a number m to replace 3 such that
2m – 2 = (1/2)m^2
m^2 = 4m – 4,
m = 2.
We put therefore for the second number (x+2)^2 – (x^2+2x), or 2x+4; the third number is then (x^2+2x)/(2x+4) or (1/2)x.
Lastly the third condition requires
x^2 + 2x + (1/2)x = a square = 4x^2, say.
Therefore
x = 5/6,
and (1, 17/3, 5/12) is a solution.
23. To find three numbers such that their solid content minus any one gives a square.
First number x, solid content x^2+x; therefore product of second and third = x+1.
Let the second be 1, so that the third is x+1.
The two remaining conditions require that x^2+x-1, and x^2-1 shall both be squares.
The difference = x = (1/2).2x, say;
thus
(x + ¼)^2 = x^2 + x – 1, and
x = 17/8.
The numbers are (17/8, 1, 25/8).
24. To divide a given number into two parts such that their product is a cube minus its side.
Given number 6. First part x; therefore second = 6-x, and 6x-x^2 = a cube minus its side.
Form a cube from a side of the form mx-1, say 2x-1, and equate 6x-x^2 to this cube minus its side.
Therefore
8x^3 – 12x^2 + 4x = 6x – x^2.
Now, if the coefficient of x were the same on both sides, this would reduce to a simple equation, and x would be rational.
In order that this may be the case, we must put m for 2 in our assumption, where 3m – m = 6 (the 6 being the given number in the original hypothesis). Thus m = 3.
We therefore assume
(3x – 1)^3 – (3x – 1) = 6x – x^2,
27x^3 – 27x^2 + 6x = 6x – x^2
x = 26/27.
The parts are (26/27, 136/27).
25. To divide a given number into three parts such that their continued product gives a cube the side of which is equal to the sum of the differences of the parts.
Given number 4.
Since the product is a cube, let it be 8x^3, the side of which is 2x.
Now (second part) – (first) + (third) – (second) + (third) – (first) = twice difference between third and first.
Therefore difference between third and first = half sum of differences = x.
Let the first be any multiple of x, say 2x; therefore the third = 3x.
Hence second = 8x^3/6x^2 = 4x/3; and, if the second had lain between the first and third, the problem would have been solved.
Now the second came from dividing 8 by 2.3, and the 2 and 3 are not two numbers at random, but consecutive numbers. Therefore we have to find two consecutive numbers such that, when 8 is divided by their product, the quotient lies between the numbers. Assume m, m+1; therefore 8/(m^2+m) lies between m and m+1.
Therefore
8/(m^2 + m) + 1 > m + 1, so that
m^2 + m + 8 > m^3 + 2m^2 + m, or
8 > m^3 + m^2.
I form a cube such that it has m^3, m^2 as terms, that is, the cube (m + 1/3)^3, which is greater than m^3+m^2, and I put
(m + 1/3)^3 = 8;
therefore,
m + 1/3 = 2, and
m = 5/3.
Assume now for first number 5x/3; the third is 8x/3, and the second is 9x/5.
Multiplying throughout by 15, we take 25x, 27x, 40x, and the product of these numbers is a cube the side of which is the sum of their differences.
The sum =
92x = 4, by hypothesis,
therefore
x = 1/23,
and the parts required are (25/23, 27/23, 40/23).
26. To find two numbers such that their product added to either gives a cube.
Let the first number be of the form m^3x, say 8x.
Second x^2-1. Therefore one condition is satisfied, since
8x^3 – 8x + 8x = a cube.
Also
8x^3 – 8x + x^2 – 1 = a cube = (2x – 1)^3, say.
Therefore
13x^2 = 14x
x = 14/13.
The numbers are 112/13, 27/169.
27. To find two numbers such that their product minus either gives a cube.
Let the first be of the form m^3x, say 8x, and the second x^2+1 (since 8x^3 + 8x – 8x = a cube).
Also 8x^3 + 8x – x^2 – 1 must be a cube, “which is impossible”.
Accordingly we assume for the first number an expression of the form m^3x+1, say 8x+1, and for the second number x^2 (since 8x^3 + x^2 – x^2 = a cube).
Also
8x^3 + x^2 – 8x – 1 = a cube = (2x – 1)^3, say.
Therefore
x = 14/13,
and the numbers are 125/13, 196/169.
28. To find two numbers such that their product +/- their sum gives a cube.
Assume the first cube (product + sum) to be 64, and the second (product – sum) to be 8.
Therefore twice sum of numbers = 64 – 8 = 56, and the sum = 28, while the product + the sum = 64; therefore the product = 36.
Therefore we have to find two numbers such that their sum is 28, and their product 36. If 14+x, 14-x are the numbers, we have
196 – x^2 = 36, or
x^2 = 160;
and, if 160 were a square, we should have a rational solution.
Now 160 arises from 14^2 – 36, and 14 = ½.28 = ¼.56 = ¼ (difference of two cubes); also 36
= ½ (sum of the cubes).
Therefore we have to find two cubes such that
(1/4 of their difference)^2 – ½ their sum = a square.
Let the sides of the cubes be (z+1), (z-1); therefore ¼ of the difference = (3z^2)/2 + ½, and the square of this is (9z^4)/4 + (3z^2)/2 + ¼;
½ the sum of the cubes is z^3 + 3z;
therefore (9z^4)/4 + (3z^2)/2 + ¼ - z^3 –3z = a square, or
9z^4 + 6z^2 + 1 – 4z^3 – 12z = a square = (3z^2 + 1 – 6z)^2, say;
whence
32z^3 = 36z^2, and
z = 9/8.
The sides of the cubes are therefore 17/8, 1/8, and the cubes 4913/512, 1/512.
Put now product of numbers + their sum = 4913/512, and product – sum = 1/512.
Therefore the sum = 2456/512, and their product 2457/512.
Now let the first number = x + half sum = x + 1228/512, and the second = half sum – x = 1228/512 – x;
therefore
1507984/262144 – x^2 = 2457/512,
262144x^2 = 250000.
x = 500/512,
and (27/8, 91/64) is a solution.
29. To find four square numbers such that their sum added to the sum of their sides makes a given number.
Given number 12.
Now x^2 + x + ¼ = a square.
Therefore the sum of four squares + the sum of their sides + 1 = sum of four other squares = 13, by hypothesis.
Therefore we have to divide 13 into four squares; then if we subtract ½ from each of their sides, we shall have the sides of the required squares.
Now 13 = 4 + 9 = (64/25 + 36/25) + (144/25 + 81/25),
and the sides of the required squares are 11/10, 7/10, 19/10, 13/10, and the squares themselves being 121/100, 49/100, 361/100, 169/100.
30. To find four squares such that their sum minus the sum of their sides is a given number.
Given number 4.
Now x^2 – x + ¼ is a square.
Therefore (the sum of four squares) – (sum of their sides) + 1 = the sum of four other squares = 5, by hypothesis.
Divide 5 into four squares, as 9/25, 16/25, 64/25, 36/25.
The sides of these squares plus ½ in each case are the sides of the required squares.
Therefore sides of the required squares are 11/10, 13/10, 21/10, 17/10, and the squares themselves 121/100, 169/100, 441/100, 289/100.
31. To divide unity into two parts such that, if given numbers are added to them respectively, the product of the two sums gives a square.
Let 3, 5 be the numbers to be added; x, 1-x the parts of 1.
Therefore
(x + 3)(6 – x) = 18 + 3x – x^2 = a square = 4x^2, say;
thus
18 + 3x = 5x^2, which does not give a rational result.
Now 5 comes from a square + 1; and in order that the equation may have a rational solution, we must substitute for the square taken (4) a square such that
(the square + 1).18 + (3/2)^2 = a square.
Put
(m^2 + 1).18 + 9/4 = a square, or
72m^2 + 81 = a square = (8m + 9)^2, say, and
m = 18,
m^2 = 324.
Hence we must put
(x + 3)(6 – x) = 18 + 3x – x^2 = 324x^2.
Therefore
325x^2 – 3x – 18 = 0,
x = 78/325 = 6/25,
and (6/25, 19/25) is a solution.
32. To divide a given number into three parts such that the product of the first and second +/- the third gives a square.
Given number 6.
Suppose third part = x, second = any number less than 6, say 2; therefore first part = 4-x.
The two remaining conditions require that 8 – 2x +/- x = a square, or 8 – x, 8 – 3x are both squares.
This does not give a rational result since the ratio of the coefficients of x is not a ratio of a square to a square.
But the coefficients of x are 2-1 and 2+1; therefore we must find a number y to replace2 such that
(y + 1)/(y – 1) = ratio of square to square = 4/1, say.
Therefore
y + 1 = 4y – 4, and
y = 5/3.
Now put second part = 5/3; therefore first = 13/3 – x.
Therefore
65/9 – 5x/3 +/- x = a square.
That is, 65 – 6x, 65 – 24x are both squares, or 260 – 24x, 65 – 24x are both squares.
The difference = 195 =15.13; we put therefore
¼ (15 – 13)^2 = 65 – 24x, and
x = 8/3.
Therefore the required parts are (5/3, 5/3, 8/3).
33. To find two numbers such that the first with a fraction of the second is to the remainder of the second in a given ratio, and also the second with the same fraction of the first is to the remainder of the first in a given ratio.
Let the first with the fraction of the second = 3 times the remainder of the second, and the second with the same fraction of the first = 5 times the remainder of the first.
Let the second = x+1, and let the part of it received by the first = 1; therefore the first = 3x – 1 (since 3x – 1 + 1 = 3x).
Since the second plus the fraction of the first = 5 times the remainder of the first, and the second + the first = 6 times the remainder of the first.
And first + second = 4x; therefore remainder of first = 2x/3, and hence the second receives from the first 3x – 1 – 2x/3 or 7x/3 – 1.
We have therefore to secure that 7x/3 – 1 is the same fraction of 3x – 1 that 1 is of x + 1.
This requires that
(7x/3 – 1)(x + 1) = (3x – 1).1;
therefore
(7x^2)/3 + 4x/3 – 1 = 3x – 1, and
x = 5/7.
Accordingly the numbers are 8/7, 12/7; and 1 is 7/12 of the second.
Multiply by 7 and the numbers are 8, 12, and the fraction is 7/12; but 8 is not divisible by 12: so multiply by 3, and (24, 36) is a solution.
Lemma to the next problem.
To find two numbers indeterminately such that their product together with their sum is a given number.
Given number 8.
Assume the first number to be x, the second 3.
Therefore 3x + x + 3 = given number = 8; x = 5/4, and the numbers are 5/4, 3.
Now 5/4 arises from (8 – 3)/(3 + 1), where 3 is the assumed second number.
We may accordingly put for the second number (instead of 3) any (undetermined) number whatever; then, substituting this for 3 in the above expression, we have the corresponding first number.
For example, we may take x – 1 for the second number; the first is then 9 – x divided by x, or 9/x – 1.
34. To find three numbers such that the product of any two together with the sum of those two makes a given number.
Necessary condition. Each number must be 1 less than some square.
Let (product + sum) of first and second = 8.
Let (product + sum) of second and third = 15.
Let (product + sum) of third and first = 24.
By the first equation, if we divide (8 – second) by (second + 1), we have the first number.
Let the second number be x – 1.
Therefore the first = (9 – x)/x = 9/x – 1.
Similarly the third number = 16/x – 1.
The third equation remains, which gives
144/x^2 – 1 = 24, and
x = 12/5.
The numbers are 11/4, 7/5, 17/3.
Lemma to the following problem.
To find two numbers indeterminately such that their product minus their sum is a given number.
Given number 8.
First number x, second 3 suppose; therefore (product – sum) = 3x – x – 3 = 2x – 3 = 8, and x = 11/2. The first number is therefore 11/2, the second 3.
But 11/2 comes from (8 + 3)/(3 – 1), and we may put for 3 any number whatever.
E.g. put the second number = x + 1; the first is then x + 9 divided by x, or 1 + 9/x.
35. To find three numbers such that the product of any two minus the sum of those two is a given number.
Necessary condition. Each of the given numbers must be 1 less than some square.
Let (product – sum) of first and second = 8.
Let (product – sum) of second and third = 15.
Let (product – sum) of third and first = 24.
By the first equation, if we divide (8 + second) by (second – 1), we have the first number.
Assuming x + 1 for the second number, we have 1 + 9/x for the first.
Similarly 1 + 16/x is the third number, and two conditions are satisfied.
The third gives
144/x^2 – 1 = 24, and
x = 12/5.
The numbers are 19/4, 17/5, 23/3.
Lemma to the following problem.
To find two numbers indeterminately such that their product has to their sum a given ratio.
Let the given ratio be 3:1, the first number x, the second 5.
Therefore 5x = 3(5 + x), x = 15/2; and the numbers are 15/2, 5.
But 15/2 arises from 15 divided by 2, while the 15 is the second number multiplied by the given ratio, and the 2 is the excess of the second number over the ratio.
Putting therefore x (instead of 5) for the second number, we have, for the first number, 3x divided by x – 3.
The numbers are therefore 3x/(x – 3), x.
36. To find three numbers such that the product of any two bears to the sum of those two a given ratio.
Let product of first and second be 3 times their sum.
Let product of second and third be 4 times their sum.
Let product of third and first be 5 times their sum.
Let second number be x; the first is therefore 3x/(x – 3), by the Lemma, and similarly the third is 4x/(x – 4).
Lastly
(3x/(x – 3)).(4x/(x – 4)) = 5((3x/(x – 3)) + (4x/(x – 4))), or
12x^2 = 35x^2 – 120x.
Therefore
x = 120/23,
and the numbers are 120/17, 120/23, 120/7.
37. To find three numbers such that the product of any two has to the sum of the three a given ratio.
Let product of first and second = 3 times sum of the three.
Let product of second and third = 4 times sum of the three.
Let product of third and first = 5 times sum of the three.
First seek three numbers such that the product of any two has to an arbitrary number (say 5) the given ratio.
Then product of first and second = 15; and, if x be the second, the first is 15/x.
The product of second and third = 20; therefore third = 20/x.
It follows that 20. 15/x^2 = 25.
And if the ratio of 20.15 to 25 were that of a square to a square, the problem would be solved.
Now 15 = 3.5, and 20 is 4.5, the 3 and 4 being fixed by the original hypothesis, but 5 being an arbitrary number.
We must therefore find a number m (to replace 5) such that 12m^2/5m = ratio of a square to a square. Thus
12m^2.5m = 60m^2 = a square = 900m^2, say; and
m = 15.
Let then the sum of the three numbers be 15.
Product of first and second is therefore 45, and first = 45/x.
Similarly, third = 60/x.
Therefore
45.60/x^2 = 75, and
x = 6.
Therefore the numbers are 15/2, 6, and 10, and the sum of these = 47/2.
Now, if this sum were 15 instead, the problem would be solved.
Put therefore for the sum of the three numbers 15x^2, and for the numbers themselves 15x/2, 6x, 10x.
Therefore
47x/2 = 15x^2, so that
x = 47/30,
and 47/4, 47/5, 47/3 is a solution.
38. To find three numbers such that their sum multiplied into the first gives a triangular number, their sum multiplied into the second a square, and their sum multiplied into the third a cube.
Let the sum be x^2, and let the numbers be m/x^2, n/x^2, p/x^2, where m, n, p are a triangular number, a square and a cube respectively; say first number = 6/x^2, second 4/x^2, third 8/x^2.
But the sum is x^2; therefore 18/x^2 = x^2, or 18 = x^4.
Therefore we must replace 18 by some fourth power.
But 18 = sum of a triangular number, a square and a cube.
Let x^4 be the required fourth power, which must therefore be the sum of a triangular number, a square and a cube.
Let the square be x^4 – 2x^2 + 1; therefore the triangular number + the cube = 2x^2 – 1.
Let the cube be 8; therefore the triangular number is 2x^2 – 9.
But 8 times a triangular number + 1 = a square; therefore
16x^2 – 71 = a square = (4x – 1)^2, say; thus
x = 9,
the triangular number is 153, the square 6400 and the cube 8.
Assume then as the numbers 153/x^2, 6400/x^2, 8/x^2.
Therefore
6561/x^2 = x^2, or
x^4 = 6561, and
x = 9.
Thus (17/9, 6400/81, 8/81) is a solution.
39. To find three numbers such that the difference of the greatest and the middle has to the difference of the middle and the least a given ratio, and also the sum of any two is a square.
Ratio 3:1. Since middle number + least = square, let the square be 4.
Therefore middle > 2; let it be x+2, so that least = 2-x.
Therefore difference of greatest and middle = 6x, whence the greatest = 7x+2.
Therefore 8x+4, 6x+4 are both squares.
Take the difference 2x, split it into factors, say x/2, 4, and proceed by the rule; therefore x = 112.
But I cannot take 112 from 2; therefore x must be found to be < 2, so that 6x+4 < 16.
Thus there are to be three squares 8x+4, 6x+4 and 4 (the 4 arising from 2.2), and the difference of the greatest and middle is 1/3 of the difference of the middle and least.
We have therefore to find three squares having this property and such that the least = 4 and the middle < 16.
Let side of middle square be z+2; therefore excess of middle over least = z^2+4z, whence excess of greatest over middle = (z^2)/3 + 4z/3, and therefore the greatest
= (4z^2)/3 + 16z/3 + 4.
This must be a square; therefore, multiplying by 9, we have
12z^2 + 48z + 36 = a square, or
3z^2 + 12z + 9 = a square = (mz – 3)^2, say.
It follows that z = (6m + 12)/(m^2 – 3), which must be < 2.
Therefore
6m + 12 < 2m^2 – 6, or
2m^2 > 6m + 18.
“When we solve such an equation, we multiply half the coefficient of x into itself – this gives 9 – then multiply the coefficient of x^2 into the units – 2.18 = 36 – add this last number to the 9, making 45, and take the side [square root] of 45, which is not less than 7; add half the coefficient of x – making a number not less than 10 – and divide the result by the coefficient of x^2; the result is not less than 5.”
We may therefore put m = 3/2 + 7/2, or 5, and we thus have
3z^2 + 12z + 9 = (3 – 5z)^2.
Therefore z = 21/11, and the side of the middle square is 43/11, the square itself being 1849/121.
Turning to the original problem, we put
6x + 4 = 1849/121, and
x = 1365/726,
which is less than 2.
The greatest of the required numbers = 7x+2 = 11007/726, the middle = x+2 = 2817/726, and the least = 2-x = 87/726.
The denominator not being a square, we can make it a square by dividing out by 6; then multiplying by 4 (another square) the result is
7338/484, 1878/484, 58/484,
or in its lowest terms, this solution is 3669/242, 939/242, 29/242.
40. To find three numbers such that the difference of the squares of the greatest and the middle numbers has to the difference of the middle and the least a given ratio, and also the sum of each pair is a square.
Ratio 3:1.
Let greatest + middle number = the square 16x^2; therefore greatest > 8x^2: let it be 8x^2 + 2; hence middle = 8x^2 – 2.
And, since greatest + middle > greatest + least, 16x^2 > (greatest + least) > 8x^2; let greatest + least = 9x^2, say; therefore least = x^2 – 2.
Now difference of squares of greatest and middle = 64x^2, and difference of middle and least = 7x^2.
But 64 is not equal to 3.7 or 21.
Now 64 comes from 32.2; therefore we must find a number m (in place of 2) such that 32m = 21. Therefore m = 21/32.
Assume now greatest number = 8x^2 + 21/32, middle = 8x^2 – 21/32, least = x^2 – 21/32. [And difference of squares of greatest and middle = 21x^2 = 3.7x^2]
The only condition left is: middle + least = square, that is
9x^2 – 42/32 = a square = (3x – 6)^2, say.
Therefore
x = 597/576, and
(42625/4608, 36577/4608, 15409/36864) is a solution.