1. To find three numbers in geometrical progression such that each of them minus a given number gives a square.

Given number 12.

Find a square which exceeds 12 by a square. “This is easy [II. 10]; 169/4 is such a number.”

Let the first number be 169/4, the third x^2; therefore the middle number = 13x/2.

Therefore x^2 – 12, 13x/2 – 12 are both squares; their difference = x^2 – 13x/2 = x(x – 13/2); half the difference of the factors multiplied into itself = 169/16; therefore, putting

13x/2 – 12 = 169/16,

we have

x = 361/104,

and (169/4, 361/16, 130321/10816) is a solution.

2. To find three numbers in geometrical progression such that each of them when added to a given number gives a square.

Given number 20.

Take a square which when added to 20 gives a square, say 16.

Put for one of the extremes 16, and for the other x^2, so that the middle term = 4x.

Therefore x^2 + 20, 4x + 20 are both squares.

Their difference is x^2 – 4x = x(x – 4), and the usual method gives 4x + 20 = 4, which is absurd, because the 4 ought to be some number greater than 20.

But the 4 = ¼(16), while the 16 is a square which when added to 20 makes a square; therefore to replace 16, we must find some square greater than 4.20 and such that when increased by 20 it makes a square.

Now 81 > 80; therefore putting (m+9)^2 for the required square, we have

(m + 9)^2 + 20 = square = (m – 11)^2, say;

therefore

m = ½,

and the square = (19/2)^2 = 361/4.

Assume now for the numbers 361/4, 19x/2, x^2, and we have x^2 + 20, 19x/2 + 20 both squares.

The difference = x(x – 19/2), and we put

19x/2 + 20 = 361/16

Therefore

x = 41/152, and

(361/4, 41/16, 1681/23104) is a solution.

3. Given one number, to find three others such that any one of them, or the product of any two of them, when added to the given number, gives a square.

Given number 5.

“We have it in the Porisms that if, of two numbers, each, as well as their product, when added to one and the same given number, severally make squares, the two numbers are obtained from the squares of consecutive numbers.”

Take then the squares (x+3)^2, (x+4)^2, and, subtracting the given number 5 from each, put for the first number x^2 + 6x + 4, and for the second x^2 + 8x + 11, and let the third be twice their sum minus 1, or

4x^2 + 28x + 29.

Therefore

4x^2 + 28x + 34 = a square = (2x – 6)^2, say.

Hence

x = 1/26,

and (2861/676, 7645/676, 5084/169) is a solution.

4. Given one number, to find three others such that any one of them, or the product of any two, minus the given number gives a square.

Given number 6.

Take two consecutive squares x^2, x^2+2x+1.

Adding 6 to each, we assume for the first number x^2+6, and for the second x^2+2x+7.

For the third we take twice the sum of the first and second minus 1, or 4x^2+4x+25.

Therefore third minus 6 =

4x^2 + 4x + 19 = square = (2x – 6)^2, say.

Therefore

x = 17/28,

and (4993/784, 6729/784, 5665/196) is a solution.

5. To find three squares such that the product of any two added to the sum of those two, or to the remaining square, gives a square.

“We have it in the Porisms” that, if the squares on any two consecutive numbers be taken, and a third number be also taken which exceeds twice the sum of the squares by 2, we have three numbers such that the product of any two added to those two or to the remaining number gives a square.

Assume as the first square x^2+2x+1, and as the second x^2+4x+4, so that third number = 4x^2+12x+12.

Therefore

x^2 + 3x + 3 = a square = (x – 3)^2, say, and

x = 2/3.

Therefore (25/9, 64/9, 196/9) is a solution.

6. To find three numbers such that each minus 2 gives a square, and the product of any two minus the sum of those two, or minus the remaining number, gives a square.

Add 2 to each of three numbers found as in the Porism quoted in the preceding problem.

Let the numbers so obtained be x2+2, x^2+2x+3, 4x^2+4x+6.

All the conditions are now satisfied, except one, which gives

4x^2 + 4x + 6 – 2 = a square.

Divide by 4, and

x^2 + x + 1 = a square = (x – 2)^2, say.

Therefore

x = 3/5,

and (59/25, 114/25, 246/25) is a solution.

Lemma I to the following problem.

To find two numbers such that their product added to the squares of both gives a square.

Suppose first number x, second any number (m), say 1.

Therefore

x.1 + x^2 + 1 = x^2 + x + 1 = a square = (x – 2)^2, say.

Thus

x = 3/5, and

(3/5, 1) is a solution, or (3, 5).

Lemma II to the following problem.

To find three right-angled triangles (rational) which have equal areas.

We must first find two numbers such that their product + the sum of their squares = a square, e.g. 3, 5, as in the preceding problem.

Now form right-angled triangles from the pairs of numbers (7, 3), (7, 5), (7, 3+5).

[i.e. the right-angled triangles (7^2+3^2, 7^2-3^2, 2.7.3), etc.]

The triangles are (40, 42, 58), (24, 70, 74), (15, 112, 113), the area of each being 840.

7. To find three numbers such that the square of any one +/- the sum of the three gives a square.

Since, in a right-angled triangle, (hypotenuse)^2 +/- twice product of perps. = a square, we make the three required numbers hypotenuses and the sum of the three four times the area.

Therefore we must find three right-angled triangles having the same area, e.g., as in the preceding problem, (40, 42, 58), (24, 70, 74), (15, 112, 113).

Reverting to the substantive problem, we put for the numbers 58x, 74x, 113x; their sum

245x = four times the area of any one of the triangles = 3360x^2.

Therefore

x = 7/96, and

(203/48, 259/48, 791/96) is a solution.

Lemma to the following problem.

Given three squares, it is possible to find three numbers such that the products of the three pairs shall be respectively equal to those squares.

Squares, 4, 9, 16.

First number x, so that the others are 4/x, 9/x; and 36/x^2 = 16.

Therefore

x = 3/2,

and the numbers are (3/2, 8/3, 6).

We observe that x = 6/4, where 6 is the product of 2 and 3, and 4 is the side of 16.

Hence the following rule. Take the product of two sides (2, 3), divide by the side of the third square 4 [the result is the first number]; divide 4, 9 respectively by the result, and we have the second and third numbers.

8. To find three numbers such that the product of any two +/- the sum of the three gives a square.

As in Lemma II to the 7^th problem, we find three right-angled triangles with equal areas; the squares of their hypotenuses are 3364, 5476, 12769.

Now find, as in the last Lemma, three numbers such that the products of the three pairs are equal to these squares respectively, which we take because each +/- 4.(area) or 3360 gives a square; the three numbers then are

4292x/113, 3277x/37, 4181x/29.

It remains that the sum of the three = 3360x^2.

Therefore

32824806x/121249 = 3360x^2,

x = 32824806/407396640 = 781543/9699920,

and the numbers are 781543/255380, 781543/109520, 781543/67280.

9. To divide unity into two parts such that, if the same given number be added to either part, the result will be a square.

Necessary condition. The given number must not be odd and the double of it + 1 must not be divisible by any prime number which, when increased by 1, is divisible by 4 [i.e. any prime number of the form 4n – 1].

Given number 6. Therefore 13 must be divided into two squares each of which > 6. If then we divide 13 into two squares the difference of which < 1, we solve the problem.

Take half of 13, or 13/2, and we have to add to 13/2 a small fraction which will make it a square, or, multiplying by 4, we have to make 1/x^2 + 26 a square, i.e.

26x^2 + 1 = a square = (5x + 1)^2, say, whence

x = 10.

That is, in order to make 26 a square, we must add 1/100, or to make 13/2 a square, we must add 1/400, and

1/400 + 13/2 = (51/20)^2.

Therefore we must divide 13 into two squares such that their sides may be as nearly as possible equal to 51/20.

Now 13 = 2^2 + 3^2. Therefore we seek two numbers such that 3 minus the first = 51/20, so that the first = 9/20, and 2 plus the second = 51/20, so that the second = 11/20.

We write accordingly (11x + 2)^2, (3 – 9x)^2 for the required squares [substituting x for 1/20].

The sum

= 202x^2 – 10x + 13 = 13.

Therefore

x = 5/101, and

the sides are 257/101, 258/101.

Subtracting 6 from the squares of each, we have, as the parts of unity, 4843/10201, 5358/10201.

10. To divide unity into two parts such that, if we add different given numbers to each, the results will be squares.

Let the numbers be 2, 6 and let them and the unit be represented in the figure, where DA=2, AB=1, BE=6, and G is a point in AB so chosen that DG, GE are both squares. Now DE=9. Therefore we have to divide 9 into two squares such that one of them lies between 2 and 3.

Let the latter square be x^2, so that the other is 9-x^2, where 3>x^2>2.

Take two squares, one > 2, the other < 3 [the former being the smaller], say 289/144, 361/144.

Therefore, if we can make x^2 lie between these, we shall solve the problem.

We must have x>17/12 and <19/12.

Hence, in making 9-x^2 a square, we must find x>17/12 and <19/12.

Put

9 – x^2 = (3 – mx)^2, say, whence

x = 6m/(m^2 + 1).

Therefore

17/12 < 6m/(m^2 + 1) < 19/12.

The first inequality gives 72m > 17m^2 + 17; and 36^2 – 17.17 = 1007, the square root of which is not greater than 31; therefore m <= (31+36)/17, i.e. m<=67/17.

Similarly from the inequality 19m^2 + 19 > 72m we find m<=66/19.

Let m=7/2. Therefore

9 – x^2 = (3 – 7x/2)^2, and

x = 84/53.

Therefore x^2 = 7056/2809, and the segments of 1 are (1438/2809, 1371/2809).

11. To divide unity into three parts such that, if we add the same number to each of the parts, the results are all squares.

Necessary condition. The given number must not be 2 or any multiple of 8 increased by 2.

Given number 3. Thus 10 is to be divided into three squares such that each > 3.

Take 1/3 of 10, or 10/3, and find x so that 1/(9x^2) + 10/3 may be a square, or 30x^2 + 1 = a square = (5x + 1)^2, say.

Therefore x = 2, x^2 = 4, 1/x^2 = ¼, and

1/36 + 10/3 = 121/36 = a square.

Therefore we have to divide 10 into three squares each of which is as near as possible to 121/36.

Now 10 = 3^2 + 1^2 = the sum of the three squares 9, 16/25, 9/25.

Comparing the sides 3, 4/5, 3/5 with 11/6, or (multiplying by 30) 90, 24, 18 with 55, we must make each side approach 55.

[Since then 55/30 = 3 – 35/30 = 4/5 + 31/30 = 3/5 + 37/30], we put for the sides of the required numbers

3 – 35x, 31x + 4/5, 37x + 3/5.

The sum of the squares = 3555x^2 – 116x + 10 = 10.

Therefore x = 116/3555, and this solves the problem.

12. To divide unity into three parts such that, if three different given numbers be added to the parts respectively, the results are all squares.

Given numbers 2, 3, 4. Then I have to divide 10 into three squares such that the first > 2, the second > 3, and the third > 4.

Let us add ½ of unity to each, and we have to find three squares such that their sum is 10, while the first lies between 2, 5/2, the second between 3, 7/2, and the third between 4, 9/2.

It is necessary, first, to divide 10 (the sum of the two squares) into two squares one of which lies between 2, 5/2; then if we subtract 2 from the latter square, we have one of the required parts of unity.

Next divide the other square into two squares, one of which lies between 3, 7/2; subtracting 3 from the latter square, we have the second of the required parts of unity. Similarly we can find the third part.

13. To divide a given number into three parts such that the sum of any two of the parts gives a square.

Given number 10.

Since the sum of each pair of parts is a square less than 10, while the sum of the three pairs is twice the sum of the three parts or 20, we have to divide 20 into three squares each of which is < 10.

But 20 is the sum of two squares, 16 and 4;

and, if we put 4 for one of the required squares, we have to divide 16 into two squares, each of which is < 10, or, in other words, into two squares, one of which lies between 6 and 10. This we learnt how to do [IX. 10].

We have, when this is done, three squares such that each is < 10, while their sum is 20; and by subtracting each of these squares from 10 we obtain the parts of 10 required.

14. To divide a given number into four parts such that the sum of any three gives a square.

Given number 10.

Three times the sum of the parts = the sum of four squares.

Therefore 30 has to be divided into four squares, each of which is < 10.

(1) If we use the method of approximation, we have to make each square approximate to 15/2; then, when the squares are found, we subtract each from 10, and so find the required parts.

(2) Or, observing that 30 = 16 + 9 + 4 + 1, we take 4, 9 for two of the squares, and then divide 17 into two squares, each of which < 10.

If then we divide 17 into two squares, one of which lies between 17/2 and 10, as we have learnt how to do [IX. 10], the squares will satisfy the conditions.

We shall then have divided 30 into four squares, each of which is less than 10, two of them being 4, 9 and the other two the parts of 17 just found.

Subtracting each of the four squares from 10, we have the required parts of 10, two of which are 1 and 6.

15. To find three numbers such that the cube of their sum added to any one of them gives a cube.

Let the sum be x and the numbers 7x^3, 26x^3, 63x^3.

Therefore 96x^3 = x, or 96x^2 = 1.

But 96 is not a square; we must therefore replace it by a square in order to solve the problem.

Now 96 is the sum of three numbers, each of which is 1 less than a cube; therefore we have to find three numbers such that each of them is a cube less than 1, and the sum of the three is a square.

Let the sides of the cubes be m+1, 2-m, 2, whence the numbers are m^3 + 3m^2 + 3m, 7 - 12m + 6m^2 – m^3, 7; their sum =

9m^2 – 9m + 14 = a square = (3m – 4)^2, say;

therefore

m = 2/15,

and the numbers are 1538/3375, 18577/3375, 7.

Reverting to the original problem, we put x for the sum of the numbers, and for the numbers respectively 1538x^3/3375, 18577x^3/3375, 7x^3, whence 43740x^3/3375 = x, that is (if we divide out by 15 and by x),

2916x^2 = 225, and

x = 5/18.

The numbers are therefore found.

16. To find three numbers such that the cube of their sum minus any one of them gives a cube.

Let the sum be x and the numbers 7x^3/8, 26x^3/27, 63x^3/64.

Therefore 4877x^3/1728 = x,

and, if 4877/1728 were the ratio of a square to a square, the problem would be solved.

But 4877/1728 = 3 – (sum of three cubes).

Therefore we must find three cubes, each of which < 1, and such that (3 – their sum) = a square.

If, a fortiori, the sum of the three cubes is made < 1, the square will be > 2. Let it be 9/4.

We have therefore to find three cubes the sum of which = ¾ or 162/216;

that is we have to divide 162 into three cubes.

But 162 = 125 + 64 – 27; and “we have it in the Porisms” that the difference of two cubes can be transformed into the sum of two cubes.

Having thus found the three cubes, we start again, and x = 9x^2/4, so that x = 2/3.

The three numbers are thus determined.

17. To find three numbers such that each of them minus the cube of their sum gives a cube.

Let the sum be x and the numbers 2x^3, 9x^3, 28x^3.

Therefore 39x^2 = 1; and we must replace 39 by a square which is the sum of three cubes + 3;

therefore we must find three cubes such that their sum + 3 is a square.

Let their sides be m, 3 – m, and any number, say 1.

Therefore 9m^2 + 31 – 27m = a square = (3m – 7)^2, say, so that m = 6/5, and the sides of the cubes are 6/5, 9/5, 1.

Starting again, we put x for the sum, and for the numbers

341x^3/125, 854x^3/125, 250x^3/125,

whence 1445x^2 = 125, x^2 = 25/289, and x = 5/17.

The required numbers are thus found.

18. To find three numbers such that their sum is a square and the cube of their sum added to any one of them gives a square.

Let the sum be x^2 and the numbers 3x^6, 8x^6, 15x^6.

It follows that 26x^4 = 1; and if 26 were a fourth power, the problem would be solved.

To replace it by a fourth power, we have to find three numbers such that each increased by 1 gives a square, while the sum of the three gives a fourth power.

Let these numbers be m^4 – 2m^2, m^2 + 2m, m^2 – 2m [the sum being m^4]; these are indeterminate numbers satisfying the conditions.

Putting any number, say 3, for m, we have as the required auxiliary numbers 63, 15, 3.

Starting again, we put x^2 for the sum and 3x^4, 15x^4, 63x^4 for the required numbers,

and we have 81x^4 = x^2, so that x= 1/3.

The numbers are thus found (1/243, 5/243, 7/81).

18. To find three numbers such that their sum is a square and the cube of their sum added to any one of them gives a square.

Let the sum be x^2 and the numbers 3x^4, 8x^4 and 15x^4.

It follows that 26x^4 = 1; and, if 26 were a fourth power, the problem would be solved.

To replace it by a fourth power, we have to find three numbers such that each increased by 1 gives a square, while the sum of the three gives a fourth power.

Let these numbers be m^4-2m^2, m^2+2m, m^2-2m [the sum being m^4]; these are indeterminate numbers satisfying the conditions.

Putting any number, say 3, for m, we have as the required auxiliary numbers 63, 15, 3.

Starting again, we put x^2 for the sum and 3x^6, 15x^6, 63x^6 for the required numbers, and we have 81x^6 = x^2, so that x=1/3.

The numbers are thus found (1/243, 5/243, 7/81).

19. To find three numbers such that their sum is a square and the cube of their sum minus any one of them gives a square.

19a. To find three numbers such that their sum is a square and any one of them minus the cube of their sum gives a square.

19b. To find three numbers such that their sum is a given number and the cube of their sum plus any one of them gives a square.

19c. To find three numbers such that their sum is a given number and the cube of their sum minus any one of them gives a square.

The given sum is 2, the cube of which is 8.

We have to subtract each of the numbers from 8 and thereby make a square.

Therefore we have to 22 into three squares, each of which is greater than 6; after which, by subtracting each of the squares from 8, we find the required numbers.

But we have already shown [cf. IX 11] how to divide 22 into three squares, each between 6 and 8.

20. To divide a given fraction into three parts such that any one of them minus the cube of their sum gives a square.

Given fraction ¼.

Therefore each part = 1/64 + a sqaure.

Therefore the sum of the three = ¼ = the sum of three squares + 3/64.

Hence we have to divide 13/64 into three squares, "which is easy".

21. To find three squares such that their continued product added to any one of them gives a square.

Let the "solid content" = x^2.

We now want three squares, each of which increased by 1gives a square.

They can be got from right-angled triangles by dividing the square of one of the sides about the right-angle by the square of the other.

Let the squares then be 9x^2/16, 25x^2/144, 64x^2/225, the continued product = 14400x^6 = x^2, by hypothesis.

Therefore 120x^2/720 = 1; and, if 120/720 were a square, the problem would be solved.

As it is not, we must find three right-angled triangles such that, if b's are their bases, and p's are their perpendiculars, p1p2p3b1b2b3 = a square; and if we assume one triangle arbitrarily (3, 4, 5), we have to make 12p1p2b1b2 a square.

"This is easy", and the three triangles are (3,4,5), (9,40,41), (8,15,17) or similar to them.

Starting again, we put for the squares 9x^2/16, 225x^2/64, 81x^2/1600.

Equating the product of these to x^2, we find x to be rational, namely 16/9, and the squares are 16/9, 100/9, 4/25.

22. To find three squares such that their continued product minus any one of them gives a square.

Let the solid content be x^2, and let the numbers be obtained from right-angled triangles, being 16x^2/25, 25x^2/169, 64x^2/289.

Therefore the continued product 25600x^6/1221025 = x^2.

If then 25600/1221025 were a fourth power, i.e. if 4.5.8/5.13.17 were a square, the problem would be solved.

We have therefore to find three right-angled triangles with hypotenuses h1,h2,h3 respectively, and with p1,p2,p3 as one of the perpendiculars in each respectively, such that h1h2h3p1p2p3 = a square.

Assuming one of the triangles to be (3,4,5), so that e.g. h3p3 = 5.4 = 20, we must have 5h1p1h2p2 = a square.

This is satisfied if h1p1 = 5h2p2.

With a view to this we have first (cf. the last proposition) to find two right-angled triangles such that, if x1,y1 are the two perpendiculars in one and x2,y2 the two perpendiculars in the other, x1y1 = 5x2y2. From such a pair of triangles we can form two more right-angled triangles such that the product of the hypotenuse and one perpendicular in one is five times the product of the hypotenuse and one perpendicular in the other.

Since the triangles found satisfying the relation x1y1 = 5x2y2 are (5,12,13) and (3,4,5) respectively, we have in fact to find two new right-angles triangles from them, namely the triangles (h1,p1,b1) and (h2,p2,b2), such that h1p1 = 30 and h2p2 = 6, the numbers 30 and 6 being the areas of the two triangles mentioned.

These triangles are (13/2, 60/13, 119/26) and (5/2, 12/5, 7/10) respectively.

Starting again, we take for the numbers 16x^2/25, 576x^2/625, 14400x^2/28561. [12/5 divided by 5/2 gives 24/25, and 60/13 divided by 13/2 gives 120/169]

The product = x^2; therefore, taking the square root, we have 4.24.120x^2/5.25.169 = 1, so that x = 65/48, and the required squares are found.

23. To find three squares such that each minus the product of the three gives a square.

Let the "solid content" be x^2, and let the squares be formed from right-angled triangles, as before. If we take the same triangles as those found in the last problem and put for the three squares 25x^2/16, 625x^2/576, 28561x^2/14400, each of these minus the continued product (x^2) gives a square. It remains that their product = x^2; this gives x = 48/65, and the problem is solved.

24. To find three squares such that the product of any two increased by 1 gives a square.

Product of first and second + 1 = a square, and the third is a square; therefore "solid content" + each = a square. The problem therefore reduces to IX. 21 above.

25. To find three squares such that the product of any two minus 1 gives a square.

This reduces, similarly, to IX. 22 above.

26. To find three squares such that, if we subtract the product of any two of them from unity, the result is a square.

This again reduces to an earlier problem, IX. 23.

27. Given a number, to find three squares such that the sum of any two added to the given number makes a square.

Given number 15.

Let one of the required squares be 9;

I have then to find two other squares such that each + 24 = a square, and their sum + 15 = a square.

To find two squares, each of which + 24 = a square, take two pairs of numbers which have 24 for their product.

Let one pair of factors be 4/x, 6x, and let the side of one square be half their difference or 2/x – 3x.

Let the other pair of factors be 3/x, 8x, and let the side of the other square be half their difference or 3/(2x) – 4x.

Therefore each of the squares + 24 gives a square. It remains that their sum + 15 = a square; therefore (3/(2x) – 4x)^2 + (2/x – 3x)^2 + 15 = a square, or 25/(4x^2) + 25x^2 – 9 = a square = 25x^2, say.

Therefore x = 5/6, and the problem is solved.

28. Given a number, to find three squares such that the sum of any two minus the given number makes a square.

Given number 13.

Let one of the squares be 25;

I have then to find two other squares such that each + 12 = a square, and (sum of both) – 13 = a square.

Divide 12 into factors in two ways, and let the factors be (3x, 4/x) and (4x, 3/x).

Take as the sides of the squares half the differences of the factors, i.e. let the squares be (3x/2 – 2/x)^2, (2x – 3/(2x))^2.

Each of these + 12 gives a square.

It remains that the sum of the squares – 13 = a square, or 25/(4x^2) + 25x^2/4 – 25 = a square = 25/(4x^2), say.

Therefore x = 2, and the problem is solved.

29. To find three squares such that the sum of their squares is a square.

Let the squares be x^2, 4, 9 respectively.

Therefore x^4 + 97 = a square = (x^2 – 10)^2, say; whence x^2 = 3/20.

If the ratio of 3 to 20 were the ratio of a square to a square, the problem would be solved; but it is not.

Therefore I have to find two squares (p^2, q^2, say) and a number (m, say) such that m^2 – p^4 – q^4 has to 2m the ratio of a square to a square.

Let p^2 = z^2, q^2 = 4 and m = z^2 + 4.

Therefore m^2 – p^4 – q^4 = (z^2 + 4)^2 – z^4 – 16 = 8z^2.

Hence 8z^2/(2z^2 + 8), or 4z^2/(z^2 + 4), must be the ratio of a square to a square.

Put z^2 + 4 = (z+1)^2, say; therefore z=3/2, and the squares are p^2 = 9/4, q^2 = 4, while m = 25/4;

or, if we take 4 times each, p^2 = 9, q^2 = 16, m = 25.

Starting again, we put for the squares x^2, 9, 16; then the sum of the squares = x^2 + 337 = (x^2 – 25)^2, and x = 12/5.

The required squares are 144/25, 9, 16.

30. A man buys a certain number of measures of wine, some at 8 drachmas, some at 5 drachmas each.

He pays for them a square number of drachmas; and if we add 60 to this number, the result is a square, the side of which is equal to the whole number of measures. Find how many he bought at each price.

Let x = the whole number of measures; therefore x^2 – 60 was the price paid, which is a square = (x – m)^2, say.

Now 1/5 of the price of the 5-drachma measures + 1/8 of the price of the 8-drachma measures = x; so that x^2 – 60, the total price, has to be divided into two parts such that 1/5 of one + 1/8 of the other = x.

We cannot have a real solution of this unless x > (x^2 – 60)/8 and x < (x^2 – 60)/5.

Therefore 5x < x^2 – 60 < 8x.

(1) Since x^2 > 5x + 60, x is not less than 11.

(2) Since x^2 < 8x + 60, x is not greater than 12.

Therefore 11 < x < 12.

Now (from above) x = (m^2 + 60)/2m; therefore 22m < m^2 + 60 < 24m.

Thus (1) 22m < m^2 + 60, m is not less than 19.

(2) 24m > m^2 + 60, m is less than 21.

Hence we put m = 20, and (x^2 – 60 = (x – 20)^2, so that x = 23/2, x^2 = 529/4, and x^2 – 60 = 289/4.

Thus we have to divide 289/4 into two parts such that 1/5 of one part plus 1/8 of the other = 23/2.

Let the first part be 5z.

Therefore (second part)/8 = 23/2 – z, or second part = 92 – 8z; therefore 5z + 92 – 8z = 289/4, and z = 79/12.

Therefore the number of 5-drachma measures = 79/12.

Therefore the number of 8-drachma measures = 59/12.