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 ax2 + bx + c = □, and dx2 + ex +ƒ=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 □, cx + dy = 0, and ex+fy = 0, let the first of them ax + by u2, the second cx + dy = v2, and the third ex+fy == w3. Then, by first eliminating x in each of these equations, and afterwards y in the two resulting equations, we shall have (af - be)v2 - (cf — de) u2 = (ad — bc) w2; or, putting v=uz, and reducing the terms, the result will give af - be ad- bc z2 cf — de where the the simple equality righthand member being a square, it only remains to find a value of z that will make the lefthand member a square; which, when possible, may be done by Problem 1. Hence, having z, we have, as above, v=uz; and the first d bz2 u2 and y az3 с ad bc 2 u2, two equations will give x= where u may be any whole or fractional number whatever. But if the three formulæ, 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. EXAMPLES. 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 — w2, and x + 192 = 22. Then, by eliminating a, and equating the result, we shall have w2 128 z2 192, or w2 + 64 22. And as the quantity on the righthand side of the equation is now a square, it only remains to make w2 + 64 a square. For which purpose, put its root =wn; then w2 + 64 w2 + 2nw + n2, or 2nw † n2 = 64; and consequently, w= 64 — n2 ; where, taking n, which is arbitrary, 2, we shall 2n have w= 1 152 64 4 60 4 4 15; and consequently x=w2- 128 128225 128 97, the answer. 2. It is required to find a number x, such that x2+x and x shall be both squares. Here, according to Case 2, of the last Problem, let x = 1 1 or, by reduction, (1+ y) y2 y2 (1+ y), and (1 ૫) 0. 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 口; For this purpose, therefore, let 1+y then 1 1; z3, or y = z2 y = 2 22; 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 simple number that will answer the condition required; which, it is evident from inspection, will be the case when z 1. Let, therefore, z = = 1 w, agreeably to Problem 1, Case 7, and we shall have 1 2 — 22 2 — (1 (1 − w)2 = 1 + Or, putting 1 nw for the root of the former of these expressions, there will arise, by squaring, 1+2w w2 1 2nw + n2w2. Whence, expunging the 1 on each side, we shall have 2 W where, in order to render the value of x positive, n may be taken equal to any proper fraction whatever. (m2 + n2)2 4mn (n2 m2)' where m and n may now be taken equal to any integral numbers whatever, provided n be made greater than m. and if n = 3 and m 2, x = 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 x, x+y, and x+2y, be the three numbers sought; and put 2x+y= u2, 2x + 2y = v2, and 2x + 3y = w2, agreeably to Case 3. Then, by eliminating x and y from each of these equations, we shall have v2 — u2 w2 - v2, or 2v2 u2 w2. And if we now put v=uz, there will arise 2u2z2. w2 u2 = w; or, by dividing by u2, 222-1= ; where the righthand u2 member being a square, it only remains to make 22a square, which it evidently is when z = = 1. 1 a 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)2 — 1 = 1 − 4p+2p2. And consequently, if this last expression be put = (1 -np), we shall have, by squaring, 1-4p+2p2-12np + n2p2, or 4+2p 2n+n2p; whence 2n 2n+2 4 2n 4 n2 p and z = 1 Or if, for the sake of greater generality, n in the last expression, we shall have m2 ·2mn + 2n2 n And since, by the two first equations, y u2 = (22 — 1) u2, and x= 1⁄2 (u2 — y) evident that 2 must be some number greater than 1, and less than √2 Or, taking u = 2 × 31, ∞ = 482, and y = 2880, we have 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 a2 = given square number, and a2 ̄and a2 x2 its two parts. Then, since a2 is a square, it only remains to make a2 x2 a square. For which purpose let its root have a2 x2 =n2x2 2anxa2, or whence, by reduction, x = nx a, and we shall 2anx; n2x2 the root of the first part, 2an 2 n2 + 1' where a and ʼn may be any numbers taken at pleasure, pro vided 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+b2 be the given numbers, and x2, y2, the two required numbers, whose sum, x2+ y2, is to be equal to a2 + b2. 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, x = a + mz, and y a+mz, = b nz, and we shall have a2 + 2amz + m2 22 + b2 — 2bnz + n2z2 — a2 + b2. 2 Or, by transposition and rejecting the terms which are common to each side of the equation, m22 + n2z2 2bnz *To this we may add the following useful property: If s and 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, s2r2, and s2r2. will be the perpendicular, base, and hypothenuse of a right-angled triangle. From which expressions, two square numbers may be found, whose sum or difference shall be square numbers; for (2rs)2+(s2r^2)2. ——— (s2r2)2, and (s2 + r2)2 — (2rs)2 = (s2 — r2)2, or (s2 † 1′2)2 — (s2 — 7:2)2 = + (2rs)2; where's and r may be any numbers whatever, provided r be greater than s. 2amz, or m2z + n2z = 2bn 2am; whence 2bn - 2am 2bmn + a(n2 — m2) 2amn+b(m2-n2) m2 + n2 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 z, 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. x Then, putting the side of the less square, and x + b= side of the greater, we shall have (x + b)2 x2 = x2 + 2bx + b2 — x2 = d = (ab), or 2bx + b2 = d = (ab). Whence, dividing each side of this equation, by b, we shall a -b have x= = the side of the less square sought, and 2 If, for instance, d 60, take a × b 30 X 2, and we shall have x= 30-2 14, and x+2 30+2 2 = 256 19660, the given difference. =16, or 162-142 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, x2-y2, shall be an integral square. Here the factors of x2 y2, being x + y and x shall have (x + y) × (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 m, -y ms2. Then, by the question, we shall have (x+y) × (x − y) or its equal x2 y2 = m2r2s2; which is evidently a square, whatever may be the value of m, r. s.. But by addition and subtraction, the above equations give, when properly reduced, where, as above said, m, r, and s, may be assumed at pleasure. Thus, if we take m2, we shall have x = = r2+sa, and y = r2 = s2, which expressions will obviously give integral |