y= 9, 10, 12, 16, x = 72, 40, 24, 16. Thus the question is susceptible of 4 solutions. Let us verify the set y = 10, x = 40. The base of the rectangle contains 40 feet, and its altitude 10, and its surface is 400 square feet. On the other hand its perimeter is equal to 2 (40+ 10) or 100 feet; now 400 is quadruple of 100. Let us generalise this question, and propose to ourselves to determine a rectangle whose surface contains m times as many square yards as its perimeter contains yards. We have the equation x y = m (2 x + 2 y) = 2 m x + 2 my. Whence, deducing the value of x by performing the division, We remark, in the first place, that it is useless to take any notice of the negative divisors of 4 m2; since if y - 2 m is negative and numerically smaller than 2 m, y is then positive; but then when y 4 m2 if. is negative and numerically greater than 2 m, the corres2 m ponding value of x is negative. The contrary would take place if y 2 m was negative, and numerically greater than 4 m2; that is, y would be negative and a would be positive. Now it is only proposed to admit direct solutions of the question. This being premised, let d, d', d", d'", be the divisors of 4 m2, we have whence y-2 md, d', d", d'"'; y = 2m+d, 2 m + d', 2 m + d", 2 m + d'"', and consequently, by designating by q, q', I", q'", the entire quotient of 4 m2 by d, d', d", d"", x = 2m+q, 2 m + q', 2 m + g′′, 2 m + q"", The number of solutions at first appears equal to the number of the divisors of 4 m2; but as the equation is symmetrical* in x and y, *We call a symmetrical function of two or more quantities any expression which contains these quantities combined in the same manner; that is, such that when we change these quantities the one into the other, the expression does not change except in the order of the terms. the values of x, deduced from the equation first resolved with reference to y, will be the same as those of y taken in the inverse order. Thus, the whole number of distinct solutions is really equal to only half the number of divisors, if this number is even, and to half the number plus unity if it is odd. As a second application, let m = 3, whence x = 6 + seeking the divisors of 36, we have whence Then 36 y 6 = 36 9, 12, 18, 36; y= 7, 8, 9, 10, 12, 15, 18, 24, 42. = 36, 18, 12, 9, 6, 4, 3, 2, 1. x=42, 24, 18, 15, 12, 10, 9, 8, 7; which gives five distinct solutions. 144. Question 2. Having given the side a of a square, to find in whole numbers the sides of a rectangle whose perimeter shall be to that of the square in the same ratio as their surfaces. Let x and y be the sides of the rectangle; 2(x + y) and xy represent its perimeter and its surface. Moreover, 4 a and a2 express the perimeter and the surface of the square; so that we have the equation This may present two cases, when a is an even and when it is an odd number. (1.) If a is an even number and equal to 2 a', it becomes, by suppressing the factor 2, x y = a' (x + y), whence x = a + a'2 y · a" In order that this expression of x may be entire, it is evidently only necessary that 2 ya should be a divisor of a2. the By designating by d, d', d", these divisors, and by I, I′, I′′, quotients of the division of a2 by d, d', d", (the quantities d, d', d", q, q', q', are necessarily odd numbers, since by hypothesis a is an odd number), we shall have + a d' 2 2 = ¶, q', q", odd numbers, - atq, atq, atq", entire expressions. x= 2 In the first place, let a = 20; the equation is 2xy = 20 (x + y), or, dividing by 2, and resolving with reference to x, 100 x = 10 + 10' seeking the divisors of 100, we find whence 100 y y 10 1, 2, 4, 5, 10, 20, 25, 50, 100; y = 11, 12, 14, 15, 20, 30, 35, 60, 110; =100, 50, 25, 20, 10, 5, 4, 2, 1; then x = 110, 60, 35, 30, 20, 15, 14, 12, 11, which gives five different solutions. In the second place, let a = 15; we have the equation y, 8, 9, 10, 12, 15, 20, 30, 45, 120; 225 2y-15 Bour. Alg. =225, 75, 45, 25, 15, 9, 5, 3, 1; 26 then x = 120, 45, 30, 20, 15, 12, 10, 9, 8; in all, five different solutions. 145. Question 3. Having given the side a of a cube, in a whole number, it is required to find, also in whole numbers, the side of the base and the altitude of a rectangular parallelopiped, with a square base, such that their solidities may be to each other as their surfaces. Let x be the side of the base, and y the altitude of this parallelopiped, x2 y and 2x2 + 4 x y will represent the solidity and the surface of this solid; moreover, a3 and 6 a2 are the expressions for the solidity and the surface of the given cube; we have then the This being premised, let us designate by d, d', d", all the divisors of 2 a2, and let us put there hence results 3 ya d, d', d"; y=a‡d a‡d a+d" Moreover, let q, q', q', be the quotients of 2 a2 by d, d', d", we By taking account of entire expressions only in the two series of values of x and y, we shall obtain the sets of values in whole numbers which are capable of verifying the equation. For example, let a 8, the equation becomes seeking the divisors of 128, and making them equal to 3 y — 8, then 3 y 128 8 = 3 x 16 +128, 16 +【32, 16 + 8, 16 + 2, We propose for a new exercise the equations. (1.) xy=2x+2y+20 { (2.) 8 x y=6x+5y + 12; (3.) 4 x y=3x+2y-12 (x 26, 14, 10, 8, Sx = 0, 4, y = 6, 0. (4.) The general equation m x y = a x + by + c. Question 4. It is required to find, in whole numbers, the rectangular parallelopipeds with square bases, such that their solidities may contain five times as many cubic feet as their surfaces contain square feet. Side of the base, x=220, 120, 70, 60, 45, 40, 30, 28, 25, 24, 22, 21, altitude y = 11, 12, 14, 15, 18, 20, 30, 35, 50, 60, 110, 210. Twelve solutions. Of the Formation of Powers and the Extraction of Roots of any Degree whatever. Introduction.-As the resolution of equations of the second degree supposes the process for the extraction of the square root already known, so, the resolution of equations of the third and |