1. To find three numbers such that, if the square of any one of them be subtracted from all three, the remainder is a square.

Take two squares x^2, 4x^2; the sum is 5x^2.

If then we take 5x^2 as the sum of the three numbers, and x, 2x as two of them, we satisfy two conditions.

Next divide 5, which is the sum of two squares, into two other squares 4/25, 121/25 [II. 9], and assume 2x/5 for the third number.

Therefore

x + 2x + 2x/5 = 5x^2

x = 17/25.

The numbers are 17/25, 34/25, 34/125.

2. To find three numbers such that the square of the sum of all three added to any one of them gives a square.

Let the square of the sum of all three be x^2, and the numbers 3x^2, 8x^2, 15x^2. Hence

26x^2 = x

x = 1/26.

The numbers are 3/676, 8/676, 15/676.

3. To find three numbers such that the square of the sum of all three minus any one of them gives a square.

Sum of all three 4x, its square 16x^2, the numbers 7x^2, 12x^2, 15x^2. Then

34x^2 = 4x

x = 2/17.

The numbers are 28/289, 48/289, 60/289.

4. To find three numbers such that, if the square of their sum be subtracted from any one of them, the remainder is a square.

Sum x, numbers 2x^2, 5x^2, 10x^2. Then

17x^2 = x

x = 1/17.

The numbers are 2/289, 5/289, 10/289.

5. To find three numbers such that their sum is a square and the sum of any pair exceeds the third by a square.

Let the sum of the three be (x+1)^2; let first + second = third + 1, so that third = (x^2)/2+x; let second + third = first + x^2, so that first = x+1/2.

Therefore second = (x^2)/2 + ½.

It remains that first + third = second + a square.

Therefore

2x = square = 16, say

x = 8.

The numbers are 17/2, 65/2, 40.

6. To find three numbers such that their sum is a square and the sum of any pair is a square.

Let the sum of all three be x^2+2x+1, sum of first and second x^2, and therefore the third 2x+1; let sum of second and third be (x-1)^2.

Therefore the first = 4x, and the second = x^2-4x.

But first + third = square, that is,

6x + 1 = square = 121, say.

Therefore

x =20,

and the numbers are 80, 320, 41.

7. To find three numbers in A.P. such that the sum of any pair gives a square.

First find three square numbers in A.P. and such that half their sum is greater than any one of them. Let x^2, (x+1)^2 be the first and second of these; therefore the third is

x^2 + 4x + 2 = (x – 8)^2, say.

Therefore

x = 31/10;

and we may take as the numbers 961, 1681, 2401.

We have now to find three numbers such that the sums of pairs are the numbers above.

The sum of the three = 5043/2,

and the three numbers are 241/2, 1681/2, 3121/2.

8. Given one number, to find three others such that the sum of any pair of them added to the given number gives a square, and also the sum of the three added to the given number gives a square.

Given number 3.

Suppose first required number + second = x^2 + 4x + 1,

second + third = x^2 + 6x + 6,

sum of all three = x^2 + 8x + 13.

Therefore third = 4x+12, second = x^2+2x-6, first = 2x+7.

Also, first + third + 3 = a square, that is

6x + 22 = square = 100, suppose.

Hence

x = 13,

and the numbers are 33, 189, 64.

9. Given one number, to find three others such that the sum of any pair of them minus the given number gives a square, and also the sum of the three minus the given number gives a square.

Given number 3.

Suppose first required number + second = x^2 + 3,

second + third = x^2 + 2x + 4,

sum of all three = x^2 + 4x + 7.

Therefore third = 4x+4, second = x^2–2x, first = 2x+3.

Lastly, first + third – 3 = a square, that is

6x + 4 = a square = 64, say.

Therefore

x = 10,

and the numbers are 23, 80, 44.

10. To find three numbers such that the product of any pair of them added to a given number gives a square.

Let the given number be 12. Take a square (say 25) and subtract 12. Take the difference (13) for the product of the first and second numbers, and let these numbers be 13x, 1/x respectively.

Again subtract 12 from another square, say 16, and let the difference (4) be the product of the second and third numbers. Therefore the third number = 4x.

The third condition gives 52x^2 + 12 = a square; now 52 = 4.13, and 13 is not a square; but, if it were a square, the equation could easily be solved.

Thus we must find two numbers to replace 13 and 4 such that their product is a square, while either + 12 is also a square.

“This is easy, and as we said, it makes the equation easy to solve” [II. 34]. The squares, 4, ¼ satisfy the condition.

Retracing our steps, we now put 4x, 1/x and x/4 for the numbers, and we have to solve the equation

x^2 + 12 = square = (x + 3)^2, say.

Therefore

x = ½,

and the required numbers are 2, 2, 1/8.

11. To find three numbers such that the product of any pair minus a given number gives a square.

Given number 10.

Put product of first and second = a square + 10 = 4 + 10, say, and let first = 14x, second = 1/x.

Let product of second and third = a square + 10 = 19, say; therefore third = 19x.

By the third condition, 266x^2 – 10 must be a square; but 266 is not a square.

Therefore, as in the preceding problem, we must find two squares each of which exceeds a square by 10.

The squares 121/4, 49/4 satisfy these conditions.

Putting now 121x/4, 1/x, 49x/4 for the numbers , we have, by the third condition, (5929x^2)/16 – 10 = square.

Therefore

5929x^2 – 160 = square = (77x – 2)^2, say.

Therefore

x = 41/77.

The numbers are 4961/308, 77/41, 2009/308.

12. To find three numbers such that the product of any two added to the third gives a square.

Take a square and subtract part of it for the third number; let x^2+6x+9 be one of the sums, and 9 the third number.

Therefore product of first and second = x^2+6x; let first = x, so that second = x+6.

By the two remaining conditions 10x+54, 10x+6 are both squares.

Therefore we have to find two squares differing by 48; “this is easy and can be done in an infinite number of ways.”

The squares 16, 64 satisfy the condition. Equating these squares to the respective expressions, we obtain

x = 1,

and the numbers are 1, 7, 9.

13. To find three numbers such that the product of any two minus the third gives a square.

First x, second x+4; therefore product = x^2+4x, and we assume third = 4x.

Therefore, by the other conditions, 4x^2+15x, 4x^2-x-4 are both squares.

The difference = 16x+4 = 4(4x+1), and we put

((1/2)(4x+5))^2 = 4x^2+15x.

Therefore

x = 5/4,

and the numbers are 5/4, 21/4, 5.

14. To find three numbers such that the product of any two added to the square of the third gives a square.

First x, second 4x+4, third 1. Two conditions are thus satisfied.

The third condition gives

x + (4x + 4)^2 = a square = (4x – 5)^2, say.

Therefore

x = 9/73,

and the numbers are 9, 328, 73.

15. To find three numbers such that the product of any two added to the sum of those two gives a square.

[Lemma.] The product of the squares of any two consecutive numbers added to the sum of the said squares gives a square.

Let 4, 9 be two of the required numbers, x the third.

Therefore 10x+9, 5x+4 are both squares.

The difference = 5x + 5 = 5(x + 1).

Equating the square of half the sum of the factors to 10x+9, we have

((1/2)(x + 6))^2 = 10x + 9.

Therefore

x = 28,

and 4, 9, 28 is a solution.

16. To find three numbers such that the product of any two minus the sum of those two gives a square.

Put x for the first, and any number for the second; we then fall into the same difficulty as in the last problem. We have to find two numbers such that (a) their product minus their sum = a square, and (b) when each is diminished by 1, the remainders have the ratio of squares.

Now 4y+1, y+1 satisfy the latter condition.

The former (a) requires that

4y^2 – 1 = square = (2y – 2)^2, say,

which gives

y = 5/8.

Assume then 13/8, 28/8, x for the numbers.

Therefore 5x/2–7/2, 5x/8–13/8 are both squares, or if we multiply by 4, 16 respectively, 10x–14, 10x–26 are both squares.

The difference is 12 = 2.6, and the usual method gives

x = 3.

The numbers are 13/8, 7/2, 3.

17. To find two numbers such that their product added to both or to either gives a square.

Assume x, 4x-1 for the numbers, since

x(4x – 1) + x = 4x^2, a square.

Therefore also, 4x^2+3x-1, 4x^2+4x-1 are both squares.

The difference is x = 4x.(1/4), and we find

x = 65/224.

The numbers are 65/224, 9/56.

18. To find two numbers such that their product minus either, or minus the sum of both, gives a square.

Assume x+1, 4x for the numbers, since

4x(x+1) – 4x = a square.

Therefore also, 4x^2+3x-1, 4x^2-x-1 are both squares.

The difference is 4x = 4x.1, and we find

x = 5/4.

The numbers are 9/4, 5.

19. To find four numbers such that the square of their sum plus or minus any one singly gives a square.

Since, in any right-angled triangle, (sq. on hypotenuse) +/- (twice product of perps.) = a square, we must seek four right-angled triangles [in rational numbers] having the same hypotenuse, or we must find a square which is divisible into two squares in four different ways; and “we saw how to divide a square into two squares in an infinite numbers of ways” [II. 8].

Take right-angled triangles in the smallest numbers, (3, 4, 5) and (5, 12, 13); and multiply the sides of the first by the hypotenuse of the second and vice versa.

This gives the triangles (39, 52, 65) and (25, 60, 65); thus 65^2 is split up into two squares in two ways.

Again, 65 is “naturally” divided into two squares in two ways, namely into 7^2 + 4^2 and 8^2 + 1^2, “which is due to the fact that 65 is the product of 13 and 5, each of which numbers is the sum of two squares.”

Form now a right-angled triangle from 7, 4. The sides are (7^2–4^2, 2.7.4, 7^2+4^2) or (33, 56, 65).

Similarly, forming a right-angled triangle from 8, 1, we obtain (2.8.1, 8^2-1^2, 8^2+1^2) or (16, 63, 65).

Thus 65^2 is split into two squares in four ways.

Assume now as the sum of the numbers 65x and

as first number 2.39.52x^2 = 4056x^2,

as second number 2.25.60x^2 = 3000x^2,

as third number 2.33.56x^2 = 3696x^2,

as fourth number 2.16.63x^2 = 2016x^2,

the coefficients of x^2 being four times the areas of the four right-angled triangles respectively.

The sum

12768x^2 = 65x

x = 65/12768.

The numbers are 17136600/163021824, 12675000/163021824, 15615600/163021824, 8517600/163021824 or, 714025/6792576, 528125/6792576, 46475/485184, 4225/80864.

20. To divide a given number into two parts and to find a square which, when either of the parts is subtracted from it, gives a square.

Given number 10, required square x^2 + 2x + 1.

Put for one of the parts 2x+1, and for the other 4x.

The conditions are therefore satisfied if

6x + 1 = 10.

x = 3/2.

The parts are (4, 6) and the square is 25/4.

21. To divide a given number into two parts and to find a square which, when added to either of the parts, gives a square.

Given number 20, required square x^2 + 2x + 1.

If to the square there be added either 2x+3 or 4x+8, the result is a square.

Take 2x+3, 4x+8 as the parts of 20, and

6x + 11 = 20,

whence,

x = 3/2.

Therefore the parts are (6, 14), and the square 25/4.