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not always admit of answers in rational numbers, and that when they are resolvable in this way, no rule can be given, that will apply in all the cases that may occur ; but as far as respects a particular class of these problems relating to squares, they may generally be determined by means of some of the rules derived from the following formula

PROBLEM 1.- To find such values of x as will make v (axa + bx + c) rational, or ax+ bx +c= a square.*

Rule 1.-When the first term of the formula is wanting, or a == 0,

0,"put the side of the square sought == n; then bx: +c=

na. And consequently, by transposing c, and dividing by the

na coefficient h, we shall have x =

where n may be any

b number taken at pleasure.

2. When the last term is wanting, or c= 0, put the side of the square sought nx, or, for the sake of greater generality ; then, in this case, we shall have ax? + bx

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тах

n

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And consequently, by multiplying by na, and dividing by

bna X, there will arise an-x + bn= mʻx, and x

where

ma

an?

m and

and nig

both in this and the following cases, may be any whole numbers whatever, that will give positive answers.t

3. When the coefficient a, of the first term, is a square number, put it da, and assume the side of the square sought

2dm ma -- ; then, 12 202 + bx to

x +

n And consequently, by cancelling dạx®, and multiplying by

т

do t

d2x2 +

2

n

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* The coefficients a, b, of the unknown quantities, as well as the absolute term c, are here supposed to be all integers; for if they were fractions, they could be readily reduced to a common square denominator; which, being afterwards rejected, will not alter the nature of the question ; since any square number, when multiplied or divided by a square number, is still a square.

† The unknown quantity x, in this case, can always be found in integers when b is positive; and, in Case 4, next following, its integral value can always be determined, whether h be positive or negative, See Vol. II. of Bonnycastle's Treatise on Algebra, Art. (H.)

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na

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N

an 2din

4. When the last term c is a square number, put it = f*, and assume the side of the square sought

te; then mox2

2em axa + ba +

十 x + e*. And consequently, by cancelling e*, and dividing by x, we shall have ax+b= in a 2em

bna 2emn + +

ma ana 5. When the given formula, or general expression,

axa + bx toc can be divided into two factors of the form fx +g and hx + k, which it always can when b2 - 4ac is a square, let there

2722 be taken (fx + 3) x (hx + k) (fx + g)'; then, by re

gm? - kn? duction, we shall have x ==

; where it may be

hna fm observed, that if the square root of b2 4ac, when rational, , the two factors abovementioned, will be bd

bto ax +

and x + 2

2a And, consequently, hy substituting them in the place of the former, we shall have,

ama (b ----) — na (b+

2a (no - am) 6. When the formula, last mentioned, can be separated into two parts, one of which is a square, and the other the product of two factors, its solution may be obtained by putting the sum of the square and the product so formed, equal

2 N°

be put

m

to the square of the sun of its roots, and

times one of the

n

* These factors are found by putting the given formula a.x2+bx to 0, and then determining its roots; which, by the rule for quadratics, are 1

b 1 + V!(b2 - 4ac), and x=

V (02.--4ac). 2a 2a

2a 2a Whence, if 62 4ac be a square, of which the root is 8, we shall have a, b 8

b x +

and a +

十一十 for the divisors of ax2 + bxt, or 2a 20?

2a

2 za 6 - 8

b+d ax + and x +

for its two factors, as in the above rule. 2a

9

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factors, and then finding the values of x as in the former instances.

7. These being all the cases of the general formula that are resolvable by any direct rule, it only remains to observe, that, either in these, or other instances of a different kind, if we can find, by trials, any one simple value of the unknown quantity which satisfies the condition of the question, an expression may be derived from this that will furnish as many other values of it as we please.

Thus, let p, in the given formula axa + bx + c, be a value of x so found, and make ap2 + bp +c= 92.

Then, by putting & = y + p, we shall have axa + bx +c= a(y + p)2 + b(y + p)+c=ayo +(zap+b) y tapa + bp + €, or axa + bx + c = ay+ (2ap+b) y +99.

From which latter expression, the values of y, and consequently those of x, may be found as in Case 4.

Or, because c= q* bp - apa, if this value be substituted for c, in the original formula ax2 + bx + c, it will become a (x2 - pa) + b (x - ) + , or

q2 + (x p) x (ax + ap+b). = a square; which last expression can be resolved by Case 6. It may here, also, be farther observed, that by putting the

ya

6 given formula, ax? + bx+6=

and taking a shall have, by substituting this value for æ in the former of these expressions, and then multiplying by 4a, and transposing the terms ay + (b*x

+ (box — 4ac) = 22; or putting, for the sake of greater simplicity, 62 - 4ac = b', this last expression may then be exhibited under the form aya + b =, where it is obvious, that if ay: +(62 4ac), or its equal aya + b', it can be made a square, axa + bx + c will also be a square.

And as the proposed formula can always be reduced to one of this kind, which consists only of two terms, the possibility or impossibility of resolving the question, in this state of it, can be more easily perceived.*

; we

4

2a

* It may here be observed, that an infinite number of expressions, of the kind ay2-+(62— ac), or ay2 +6=22,where mentioned, are wholly irresolvable; among which we may reckon

292 +3, 542 £6, 712 £5, &c. none of which can ever become squares, whatever number, either whole or fractional, be substituted for y; although there are a variety of instances in which the value of y may be found, even in integers, so as to render the formula ay2 + b

For a further detail of which circumstances, as well as for other particulars relating to this part of the subject, see the second volume of Eyler's Algebra, or the second volume of Bonnycastle's Algebra.

22.

EXAMPLES.

22

= 6;

5

5

1. It is required to find a number, such that if it be multiplied by 5 and then added to 19, the result shall be a square. Let x = the required number: then, as in Case 1, 5x +

19 19 = no, or x =

; where it is evident that n may be

5 any number whatever greater than 19. Whence, if n be taken = 5, 6, 7, respectively, we shall 25 - 19 36 - 19

49 - 19 have x =

11, or

33, or

5 the latter of which is the least value of x, in whole numbers, that will answer the conditions of the question; and consequently 5x + 19 = 5 X6 + 19 = 30 + 19 = 49, a square number as was required.

2. It is required to find an integral number, such that it shall be both a triangular number and a square.

It is here to be observed, that all triangular numbers are of the form ; and therefore the question is reduced to 2

2x2 + 2x the making or its equal

a square. 2

4 Where, since the divisor 4 is a square number, it is the same as if it were required to make 2x4 + 2x a square.

m22

agreeably to the

x2 + x

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22 + x

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2

Let, therefore, 2x2 + 2x = (

n

n?

ma

method laid down in Case 2.

Then, by dividing by x, and multiplying the result by n*, the equation will become 2nRx + 2na = m*x, or (m2

m*x, or (2no)

2n? 2n; and consequently, x=

where, if n be 2n2

2 + 64+8 taken =2, and m= = 3, we shall have x=8, and

2

2 72

36, which is the least integral triangular number 2 that is at the same time a square.

3. It is required to find the least integral number, such that if 4 times its square be added to 29, the result shall be a square.

Here it is evident, that this is the same as to make 42c2 529 a square.

And, as the first term in the expression is a square, let

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ma

ma

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conse

2

n

n

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m2

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n

12

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N

Case 3. 4m

4 m Then, 2 + 29, or

29

and nea

2972 -- ma quently, x =

; where, if m and n be each taken 4mn

29-1 =1, we shall have x =

:7, and 422 +29=4 X 49 +

4 29 = 225=(15)", which is a square number, as was required.

4. It is required to find such a value of x as will make 17 x2 5x + 1 a square.

Here the last term 1 being a square, let there be taken, according to Case 4,

ma

2 m 7x2 5x + 1 - x - 1)?

u tl. Then, by rejecting the l on each side of the equation,

m? 2mn and dividing by a, we shall have 7x 5

and 2mn

512 consequently, x=

; where, if m and n be each ma 7.no

2-5 3 taken 1, the result will give a =

2, or by tak1.-7 6

48 – 45 ing n=3, and m=8, we shall have a

= 3, which

64 63 makes 7 X 32 - 5 X 3+1 = 49 = 71, as required.

5. It is required to find such a value of x as will make 8àcé + 14x + 6 a square.

Here, by comparing this expression with the general formula, ax2 + bx + c, we shall have a = 8,6 = 14, and c = 6. And, as neither ä nor ċ, in the present instance, are

4ac = 196 192 = 4 is a square, the given expression can be resolved, by Case 5, into the two following factors, 8x + 6 and x + 1.

ma Let, therefore, 8x2 + 14x +:6=(8x + 6) (x + 1) + 1), agreeably to the rule there laid down. Then there will arise, by dividing each side by x + 1,

ma 8x +6

(x + 1). And consequently, by multiplication and reduction, we shall

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squares, but 12

=(x = (*

na

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