Page images

af - be

cf - de


[ocr errors]

from which last formulæ, the value of y, when the question is possible, and consequently that of X, may be determined as in Case 1. But if it were required to make the two general expressions

axa + bx+c = , and dx=' + ex +f=0, the solution could only be obtained in a few particular cases, as the resulting equality would rise to the fourth power. In the case of a triple equality, where it is required to make

ax + by = 0, cx + dy = 0, and ex +fy= 0, let the first of them ax +- by =u», the second cx + dy =va, and the third ex +fy = w*.

Then, by first eliminating & in each of these equations, and afterwards y in the two resulting equations, we shall have (af - be)u2 – (cf - de) vo = (ad -- bc) w2; or, putting v=uz, and reducing the terms, the result will give

wa the simple equality


where the ad bc ad be righthand member being a square, it only remains to find a value of 2 that will make the lefthand member a square ; which, when possible, may be done by Problem 1. Hence, having %, we have, as above, v=uz; and the first

d bza

az? two equations will give x=

22 and

Y ad - be

ad - be where u may be any whole or fractional number whatever.

But if the three formula, here proposed, contained only one variable quantity, the simple equality, to which it would be necessary to reduce them, would rise, as in the last case, to the fourth power, and be equally limited with respect to its solution.

4. In other cases of this kind, all that can be done is to find successively by the former rules, several answers, when one is known; and if neither this nor any of the abovementioned modes of solution are found to succeed, the problem under consideration can only be determined by adopting some artifice of substitution that will fulfil one or more of the required conditions, and then resolving the remaining formulæ, when they are possible, by the methods already delivered for that purpose; but as no general precepts can be given, for obtaining the solution in this way, the proper mode of proceeding, in such cases, must chiefly depend upon the skill and sagacity of the learner.

[ocr errors]
[ocr errors]


1. It is required to find a number x, such that x + 128 and + 192 shall be both squares.

Here, according to Case 1, let x + 128 = wʻ, and x + 192


have w=


[ocr errors]


Then, by eliminating r, and equating the result, we shall have wa · 128 = 22- 192, or wa + 64 22.

And as the quantity on the righthand side of the equation is now a square, it only remains to make wa + 64 a square.

For which purpose, put its root = w+n; then wa + 64 = 22 + 2nw +na, or 2nw na 64; and consequently, w 64 - 12

; where, taking n, which is arbitrary, = 2, we shall 2n 64 4 60

15; and consequently x=w2 -- 128 4

4 152 128=225


97, the answer. 2. It is required to find a number X, such that ** + x and

2 shall be both squares. Here, according to Case 2, of the last Problem, let a 1

1 1

1 1 ; then we shall have to make


squares ; y


y 1

1 or, by reduction, +(1+y)= 0, and -2 (1 - y) 口。

Or, since a square number, when divided by a square number, is still a square, it is the same as to make 1+y== , and 1

- Y For this purpose, therefore, let 1 +y= 7*, or y=24 - 1; then 1


2 z; which is also. to be made a square. But as neither the first nor the last terms of this formula are squares, we must, in order to succeed, find some siniple number that will answer the condition required; which, it is evident from inspection, will be the case when %:= -1.

Let, therefore, z=1 -- w, agreeably to Problem 1, Case 7, and we shall have 1 y= 2 - 2 = 2 - (1 - w)= 1 + 2w - W2;. or y = wa Or, putting 1 nw for the root of the former of these ex. pressions, there will arise, by squaring, 1 + 2w W2 = 1 2nw to now

Whence, expunging the 1 on each side, and dividing by w, we shall have 2

2n + naw; and consequently 2n2 + 2

1 (m+1)*
na + 1


2w 4n where, in order to render the value of x-positive, n may be taken equal to any proper fraction whatever.





[ocr errors]


and a




[ocr errors]
[ocr errors]

Or if, for the sake of greater generality,

be substituted for n, we shall have

(ma + na)

4mn (na -- mo) where m and n may now be taken equal to any integral numbers whatever, provided n be made greater than m.

25 If, for instance, n 2 and m=1, we shall have x


169 and if n=3 and m = 2, X = and so on, for any other

120 number.

3. It is required to find three whole numbers in arithmetical progression, such, that the sum of every two of them should be a square.

Let *, * + y, and x + 2y, be the three numbers sought; and put 2x+y=u, 2x + 2y=va, and 2x + 3y=w*, agreeably to Case 3.

Then, by eliminating x and y from each of these equations, we shall have v2 - = 22 -- 02or 2v2 U" = w2.

And if we now put v=uz, there will arise 2u<22 22 wa; or, by dividing by u2, 222 1 where the righthand member being a square, it only remains to make 222 - 1 a square, which it evidently is when = -1.

But as this value would be found not to answer the conditions of the question, let z=1-p; then 2z2 – 1= 2 (1 - p)" -- 1=1 - 4p + 2po.

And consequently, if this last expression be put = (1 np), we shall have, by squaring, 1 --- 4p + 2pa = 1 - 2np + napa, or -4+2p

2n + np; whence 2n 4

2n 4 722 2n + 2 P

1 na 2

na 2 na - 2


[ocr errors]

and ?=



Z 2

Or if, for the sake of greater generality,

be substituted for n in the last expression, we shall have

ma 2mn + 2na

ma - 2n2 And since, by the two first equations, y=v

- ua u222 - u* = (22 – 1) u, and x = } (u~ -- y) = 1 (2 – 2*) u», it is evident that 2 must be some number greater than 1, and less than V 2

and y

[ocr errors]


[ocr errors]





If, therefore, m= 9 and n = 5, we shall have
81 – 90 + 50



81 - 50
31' 312

2 Or, taking u= 2 X 31, x = 482, and y=2880, we have x = 482, x+y=3362, and x + 2y = 6242, which are the numbers required.

4. It is required to divide a given square number into two such parts that each of them shall be a square.* Let aa given square number, and and a

2% its two parts. Then, since - is a square, it only remains to make c2

22 a square. For which purpose let its root = nx a, and we shall have a - 22 n 22 2anx + a, or - ox2 now?

2апх; whence, by reduction, w=

the root of the first part,

na +1' 2ana

the root of the second. na +1 2ап

ana Therefore


are the parts required; 1

na + 1 where a and n may be any numbers taken at pleasure, provided n be greater than 1.

5. It is required to divide a given number, consisting of two known square numbers, into two other square numbers.

Let a2 + 32 be the given numbers, and xa, ya, the two required numbers, whose sum, 2 + y2, is to be equal to a + 62.

Then it is evident, that if x be either greater or less than a, y will be accordingly less or greater than b. Let, therefore,

=a + m2, and y=b nz, and we shall have aa + 2amz to moz2 + 12 ~ 2nzt naz= a + 62.

Or, by transposition and rejecting the terms which are common to each side of the equation, mʻza + no 2 2bnz

and nx

[ocr errors]
[ocr errors]


[ocr errors]


* To this we may add the following useful property :

If s and 7 be any two unequal numbers, of which r is the greater, it can then be readily shown, from the nature of the problem, that

2rs, s2 r2, and s2 +12. will be the perpendicular, base, and hypothenuse of a right-angled triangle.

From which expressions, two square numbers may be found, whose suon or difference shall be square numbers; for (ars)2+($2 ----2)2. (52 + 2)2, and (52+72)2-- (Prs)2 =(82----2), or (52+-2)2 – (s - 1:2) (2rs)2; where's and may be any numbers whatever, provided ý be greater than s.


[ocr errors]


have . ==


[ocr errors]

have a =

2amz, or maz + néz = 2bn

2am; whence
2bn - 2am 2bmn + a(n? - m?) 2amn+b(m2-n2

m2 +- 122
m2 + n?

m2 +n% where m and n may be any numbers, taken at pleasure, provided their assumed values be such as will render the values of x, y, and 2, in the above expressions, all positive.

6. It is required to find two square numbers, such that their difference shall be equal to a given number.

Let d the given difference; which resolve into two factors a, b, making a the greater and b the less. Then, putting x =

the side of the less square, and at b = side of the greater, we shall have (x + 5)2 -- X2=

x2 = x + 2bx + b2 — =d=(ab), or 26x + b2 =d=(ab). Whence, dividing each side of this equation, by b, we shall

- B

the side of the less square sought, and 2

atb +5

the side of the greater. 2

2 If, for instance, d= 60, take a xb= 30 x 2, and we shall 30-2.

30 +2 =14, and x + 2

16, or 162_142 2

2 = 256 - 196 = 60, the given difference.

7. As an instance of the great use of resolving formula of this kind into factors, let it be proposed, in addition to what has been before said, to find two numbers, x and y, such that the difference of their squares, 22 -y?, shall be an integral square. Here the factors of x2

y?, being x + y and a shall have (a + y) x (x - y) x2 - y2. And since this product is to be a square, it will evidently become so, by making each of its factors a square, or the same multiple of a square. Let there be taken, therefore, for this purpose,

X + y=mra, a Y Then, by the question, we shall have (x + y) x (x - y) or its equal x2 72 = m2r252 ; which is evidently a square, whatever may be the value of rn, 1, s..

But by addition and subtraction, the above equations give, when properly reduced,

m (32 + ) m (72 - )

gy where, as above said, m, r, and s, may be assumed at pleasure. Thus, if we take in=2, we shall have a y=p2 = s, which expressions will obviously give integral

y, we


[ocr errors]



[ocr errors]


« PreviousContinue »