and it is required to find how many solutions this equation admits of in positive integers. and since one of the quantities,, is a whole number, ..the number of solutions required is 190. 360. To find the number of solutions in positive integers of the equation ax + by + cz = d, each term being positive*. Transposing one of the terms, cz, the equation becomes where z may have any integral value greater than 0 and less than zero values of the unknown quantities being supposed to be excluded. Hence, by Art. 359, the number of solutions will be the number of integers which lie between where Р and q are those quantities which satisfy the equation ap - bq = 1. (d − c z) p, (d- cz)q, a and b Thus a certain number of solutions is determined for each value of %, and the sum of these numbers will be the number required. OBS. The application of this rule is attended with considerable difficulty (see Barlow's Theory of Numbers, Art. 162, or Peacock's Algebra, Art. 506); but the following Example will shew the student how he may determine the number of solutions in some cases without much trouble. Ex. Required the number of integral solutions of - by = C, * If any term be negative, as by, then the equation being of the form axthe number of solutions is at once declared infinite. (Art. 359. Cor. 3). Hence it appears, (1) that t must be an odd number; (2) that it must be less than 23; also since x + y + z cannot be less than 3, (zero values being excluded), t cannot be greater than 20, that is, its greatest value is 19; and therefore all the different values of t are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19; and the number of solutions is 10. 361. The theory for the solution of indeterminate equations of more than one dimension is too difficult to be admitted into an elementary work, like the present. The reader is referred for farther information to Barlow's Theory of Numbers, Chaps. III. and IV. But there are two classes of such equations, which admit of easy solution: 1st. Such as do not involve the square of either of the unknown quantities; and 2nd. Such as involve the square of one of them only. Ex. 1. Required the integral solutions of 3xy-4y+3x=14. hence y +1 must be either 10, or a divisor of 10; that is, it must be either 1, 2, 5, or 10; and therefore the values of y can be only 0, or 1, or 4, or 9; of which the first and last make a negative. .. x = 3, 2 are the solutions required. y = 1, 4 Required the integral solutions of 2xy-3x2+ y = 1. Ex. 2. Hence, 7 being a prime number, 2x + 1 can be no other number but 7, or 1; or .. 2x=6, or x = 3, and .. y=4;, 2x=0, i. e. x = 0, and . y=1. Which are the only solutions in positive integers. 362. In the solution of different kinds of unlimited problems different expedients must be made use of, which expedients, and their application, are chiefly to be learned by practice. Ex. 1. To find a "perfect number ", that is, one which is equal to the sum of all the numbers which divide it without remainder. Suppose y" to be a "perfect number "; its divisors are 1, y, y2... y", x, xy, xy3 x y1-1; ... + y" + x + xy + xy2 - 1) x x and, that a may y 20, that is, be a whole number, let y+1-2y" =0, or y = 2; then x = 2"+1 - 1. Also, let n be so assumed that 2"+1 - 1 may have no divisor but unity, which was supposed in taking the divisors of y" x; then y" x, or 2" × (2"+11) is a "perfect number." Thus, if n = 1, the number is 2 × 3 or 6, which is equal to 1+ 2+ 3, the sum of its divisors: If n = 2, the number is 22 × (23 - 1) X = 4 × 7 = 28. Ex. 2. To find two square numbers, whose sum is a square. Let x and y be the two square numbers; Assume x2 + y2 (nx − y)2 = n2x2 – 2n xy + y2, And if n and y be assumed at pleasure, such a value of x is obtained, that x2 + y2 is a square number. = But if it be required to find integers of this description, let y = n2 - 1, then a 2n, and n being taken at pleasure, integral values of x and y, and consequently of 2 and y', will be found. Thus, if n = 2, then y = 3, and ≈ = 4, and the two squares are 9 and 16, whose sum is 25, a square number. Ex. 3. To find two square numbers, whose difference is a square. Let x and y be the two squares; 363. To explain the different systems of notation. DEF. In the common system of notation each figure of any number* increases its value in a tenfold proportion in proceeding from right to left. Thus 3256 may be expressed by The figures 3, 2, 5, 6, by which the number is formed, are called its digits, and the number 10, according to whose powers their values proceed, is called the radix of the scale. It is purely conventional that 10 should be the radix; and therefore there may be any number of different scales, each of which has its own radix. When the radix is 2, the scale is called Binary; when 3, Ternary; when 10, Denary, or Decimal; when 12, Duodenary, or Duodecimal; &c. * In this Section and in the following one by number a whole number is always meant. If 3256 expressed a number in a scale whose radix is 7, that number might be expressed thus, 6+5×7+2×72 + 3 × 73. And generally, if the digits of a number be a。, a1, а,, α„, &c., reckoning from right to left, and the radix r, the number will be properly represented by Or, if there be n digits, the number will be (reversing the order of the terms) OBS. In any scale of notation every digit is necessarily less than r, and the number of them, including O, is equal to r. Also in any number the highest power of r is less by 1 than the number of digits. 364. To express a given number in any proposed scale. N = a + a ̧r + а«r2 + a ̧μ3 + &c.; and if N be divided by r, the remainder is a.. If the quotient be again divided by r, the remainder is a and so on, until there is no further quotient. is av Therefore all the digits a, a, a, a,, &c. are found by these repeated divisions; and consequently the number in the proposed scale. Ex. To express 1820, written according to the denary scale, in a scale whose radix is 6. .. the number required is 12232. This may be easily verified. Thus if the result be correct, 2+3×6+2 × 62 + 2 × 63 + 1×64 must amount to 1820; which, upon trial, it is found to do. |