(16) Find two numbers whose sum is a square, the sum of their squares a square, and either added to the square of the other a square.

1

Let-
-x and x be the numbers; then their sum is a square, and

1

2

=

= -X

1 x

4

a square, and +x+2x=0=1+

in order to satisfy the other condition, we assume

nx 1

2

1

-x

a square; and,

are numbers that answer the conditions as follows:

2

2

+16'

and

28

(17) Find two such numbers, that if their product be added to the sum of their squares, the sum shall be a square.

Let 2r be their sum and 2y be their difference; then, the greater will be x+y and the less x-y; hence x2-y2= their product, and 2x2+2y2= the sum of their squares; then, by the question, 3x2+y2= □=nx-y and 2ny

2

x= ;; if n=2 and y=2,..x=8, which will answer the conditions.

n2-3

(18) To find two square numbers, such, that the difference of their cube roots shall be a square number.

Let x and y be the required numbers. Then x2-y2=; consequently, x and y may be any two numbers which are the hypotenuse and one leg of a right-angled triangle, and the least numbers of this description are 5 and 3, and the numbers themselves 15625=1252 and 729=27*.

(19) Find three numbers, such, that not only the sum of all three of them, but also the sum of every two, shall be a .

Put 4r, x2-4.x, and 2x+1 for the three numbers; then it only remains to render 6x+1=0.

if n=12, x=20, which will answer the conditions of the problem.

(20) Find two numbers, such, that the sum of their squares and the sum of their cubes shall be both squares.

Let b be the base, p the perpendicular, and h the hypotenuse of a rational right-angled triangle, r any multiplier of b, p, and h; then (br)+(px)=(hx), but (bx)+(px)= a rational square r2; hence (b3+p3).x=r2, or

;;
; now if r=b+p3, .. x=b+p3, and .. br=b(b+p3), px=p>

(b3+p3); now let b=3, p=4; then is x=91, bx=273, and pr=364. If b=6 and p=8, then r=728, br=4368, and pr=5824, and so on in general. (21) Find a number to which if 8 be added, the sum shall be a cube, and from which if 1 be subtracted, the remainder shall be a cube.

Let x be the number; b=2, c=1; then r+b3= a cube and r−03= a cube;

Assume x-c(a—c)3= a cube a3-3a2c+3ac-c3, and ..1=a -3a2c+3ac2; and, equating both values of x, we get

and, putting the right-hand member of this equation into numbers, we get

(22) To find three square numbers,such, that the sum of every two of them shall be a square number.

Let x, y, and z2 be the numbers sought.

Then x2+z2, y2+z3, and x2+y2 are the three numbers; i. c.,

which are evidently two squares; and therefore it remains to make

square number.

(m2-1)2. n2+(n2-1). m2, or (m+1). (m-1)'. n2+(n+1)'. (n−1)'. m2=

a square number.

Hence (m+1)2. (m −1)2. (m+2)2+m2. (m+3)o × (m+1)3,

or

or

is a square number.

(m-1). (m+2)2+m2. (m+3)3,
2m++8m3+6m2-4m+4,

5m2

Let the root of this quantity be assumed= -m+2.

(5m3 —m+2)2=2m2+8m3 + 6m2 —4m+4 ;

m=-24, and n—— -22.

Then

whence

To obtain the answer in whole numbers, let z=528;* then x=-6325, and y=-5796. Hence 528, -5796, -6325 are the roots of the squares, and

5282, 57962, 63252 are the squares required.

(23) To find three cube numbers, such, that if from every one of them a given number 1, be subtracted, the sum of the remainders shall be a square. Let 1+x, 2-x, and 2 represent the required roots. Then, per question, (1+x)3—1+(2−x)3—1+8—1=0;

(24) It is required to find three integral square numbers, such, that the difference of every two of them shall be a square number.

Let the roots of the required numbers be denoted by

Assume

then

and

but

sy, s-y, and r2+2.

r2x2=s2 + y2;

p? —x2—s2=y2 =☐

y=r1-2r2x2-2r2s2+x1+2x2s2+s1 ;

(r2+x2)2 — (s2 —y2)2= □

=(r2+x2)3 — (s2 —r2+x2+s2)2=r*+2r2x2+x1—s1+2r2s2 —2s2x2 — 254 — yok

* The least common multiple of the denominators, 48 and 44.

G G

Take then

a=340,

x=256 and y2—r2—s2 —x2—441-256-169=16,

••. (r2+x2)2=(441+256)2=(697)= one number,

and (x)=(s2+y)=(441-256)=(185)

and

the second number;

(s2—y2)3=(169—16)2=(153), which is the other number.

(25) To find three square numbers such, that their sum, being severally added to their three roots, shall make square numbers.

Let 2x, 6x, and 9r denote the three roots; .. by the question,

Assume y2+2y=(~2,2-1)"; and .. y2+2y+1=(-1)*+1—2—222+1

+1=

2

2z

4z2

2* — 222 +422 + 1 _2* +222+ 1 _ (+1)*; and, consequently, y+1

22+1

=

2z

4z2

=

422

-1=

2z

; hence, by substitution in the

But 422 is a square number;

But

.. (2-1)+12zX (z—1)2= □

=(-1)3× (2-1)2+12%. (≈—1)2=(≈—1)°× { (z−1)2+12z}.

(x-1) is,

.. (2-1)+12%==22+10x+1=0.

Again, by substitution in the third, we have

(-1)+182X(2-1), and .. (2-1). (2-1)+18z. (2-1)2=0.

(2-1) 2x {(2-1)+18},

.. (2-1)+18z=22+16x+1;
(2+16x+1)-(22+10x+1)=6z=3z × 2,
3x+2 3z

2

=

2

+1, the root of the greater.

(1) Required six numbers whose sum and product shall be equal.

Ans. 1, 2, 3, 4, 5, and (2) Required five square numbers whose sum shall be a square.

15

119

Ans. 1, 4, 9, 16, and .

(3) Divide the number 3 into four rational squares.

Ans.

16 1 9
49
and
25' 25' 25' 25

(4) Divide unity into three rational squares.

Ans.

9 4 and 49' 49' 49'

36

(5) Find two numbers whose sum is a cube, and difference a square.

Ans. 1512 and 216.

(6) Find two numbers whose product plus their sum or difference is each

(7) To find two numbers, such, that when each is multiplied into the cube of the other, the products will be squares.

(8) To find two square numbers whose difference is 40.

Ans. 2 and 8.

Ans. 49 and 9.

(9) To find two square numbers, such, that their sum added to their product may be a square number.

(10) It is required to find two whole numbers, such, that their difference, the difference of their squares, and the difference of their cubes shall be squares. Ans. 10 and 6.

(11) Find two numbers, such, that the sum of their squares shall be both a square and a cube.

Ans. 75 and 100.

(12) Find two numbers whose sum shall be a cube, but their product and quotient squares.

Ans. 25 and 100.