-0, we shall

Or, by rejecting r, there will remain

put = s.

29

Then r29s+20; where, by taking s =

the number required.

3. To find a number, which, being divided by 6, shall leave the remainder 2, and when divided by 13, shall leave the remainder 3.

Ans. 68.

4. It is required to find a number, which, being divided by 7, shall leave 5 for a remainder, and if divided by 9, the remainder shall be 2. Ans. 47, 110, &c.

5. It is required to find the least whole number, which, being divided by 39, shall leave the remainder 16, and when divided by 56, the remainder shall be 27. Ans. 1147.. 6. It is required to find the least whole number, which, being divided by 7, 8, and 9, respectively, shall leave the remainders 5, 7, and 8.

Ans. 215.

7. It is required to find the least whole number, which, being divided by each of the nine digits, 1, 2, 3, 4, 5, 6, 7, 8, 9, shall leave no remainders. Ans. 2520.

8. A person receiving a box of oranges observed, that when he told them out by 2, 3, 4, 5, and 6 at a time, he had none remaining; but when he told them out by 7 at a time, there remained 5; how many oranges were there in the box?

Ans. 180.

OF THE DIOPHANTINE ANALYSIS..

THIS branch of Algebra, which is so called from its inventor, Diophantus, a Greek mathematician of Alexandria in Egypt, who flourished in or about the third century after Christ, relates chiefly to the finding of square and cube numbers, or to the rendering certain compound expressions free from surds: the method of doing which is by making such substitutions for the unknown quantity, as will reduce the resulting equation to a simple one, and then finding the value of that quantity in terms of the rest.

It is to be observed, however, that questions of this kind do

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 √(ax2 + bx + c) rational, or ax2 + bx + c = a square.*

RULE 1.-When the first term of the formula is wanting, or α== = 0, put the side of the square sought = n; then bx + c = n2.

And consequently, by transposing c, and dividing by the

coefficient b, we shall have x =

number taken at pleasure.

n2
b

с

where n may be

any

2. When the last term is wanting, or c= O, put the side of the square sought =nx, or, for the sake of greater gene

; then, in this case, we shall have ax2 + bx:

==

And consequently, by multiplying by n2, and dividing by x, there will arise an3x + bn2 = m23x, and x =

m2

bn2

and,

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

3. When the coefficient a, of the first term, is a square number, put it d2, and assume the side of the square sought

2dm

n

m2

And consequently, by cancelling d2x2, and multiplying by

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 b be positive or negative. See Vol. II. of Bonnycastle's Treatise on Algebra, Art. (H.)

n2, we shall have bn2x + cn2 = 2dmnx + m2, and x= m2

bn2 - 2dmn

4. When the last term c is a square number, put it = e2,

and assume the side of the square sought

mx

= +e; then

n

+ xe. And consequently, by

n

cancelling e, and dividing by x, we shall have ax + b =

bn2 2emn

5. When the given formula, or general expression,

ax2 + bx + c

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

(fx +g); then, by re

observed, that if the square root of b2 - 4ac, when rational, be put, the two factors abovementioned, will be

And, consequently, by substituting them in the place of the former, we shall have,

am2 (bs) — n2 (b + s) x= 2a (nam3)

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

to the square of the sum of its roots,

*These factors are found by putting the given formula ax2+b+c= 0, and then determining its roots; which, by the rule for quadratics, are

4ac be a square, of which the root is d, we shall have a,

+
2a 2a'

b

and x+

2a 2a'

b

ax +

for the divisors of ax2 + bx+c, or

for its two factors, as in the above rule.

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 ax2 + bx + c, be a value of x so found, and make ap2 + bp + c = q2.

Then, by putting x = y+p, we shall have ax2 + bx + c = a (y+p)2+b (y + p) + c = ay2 +(2ap + b) y + ap2+bp+c, or ax2 + bx + c = ay2 + (2ap + b) y + q2.

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

Or, because c = q2 bp ap3, if this value be substituted for c, in the original formula ax2 + bx + c, it will become a (x3 — p3) + b (x − p) + q2, or

q2 + (x − p) × (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

y2 4'

Z - b

and taking x ==

2a

; we

given formula, ax2 + bx + c = shall have, by substituting this value for x in the former of these expressions, and then multiplying by 4a, and transposing the terms ay2 + (b3x — 4ac) = 22; or putting, for the sake of greater simplicity, b2-4acb', this last expression may then be exhibited under the form ay2 + b' = z3, where it is obvious, that if ay2 + (b2 - 4ac), or its equal ay+b', it can be made a square, ax2 + 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.*

* It may here be observed, that an infinite number of expressions, of the kind ay2+(b2— ac), or ay2+b′ = 22, here mentioned, are wholly irresolvable; among which we may reckon

2y23, 5y2 ± 6, 7y2±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

22.

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 Euler's Algebra, or the second volume of Bonnycastle's Algebra.

EXAMPLES.

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 the required number: then, as in Case 1, 5x +

any number whatever greater than ✓ 19. Whence, if n be taken

=

= 5, 6, 7, respectively, we shall 36-19

5

49 19

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 × 6 + 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 ; and therefore the question is reduced to

of the form

x2 + x

2

Where, since the divisor 4 is a square number, it is the same as if it were required to make 2x2 + 2x a square.

Then, by dividing by x, and multiplying the result by n2, the equation will become 2n2x + 2n2 = m3x, or (m2 — 2n2)

x = 2n2; and consequently, x =

2n2

where, if n be

m2 2n2

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

36, which is the least integral triangular number

that is at the same time à 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 4x2 +29 a square.

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