And if p is the root of the greatest integral square contained in mAB, and g the integer to be added to p (according to the directions given for using (1) and (2)), so as to mA + B then from (1) and (2) we shall get (mA + B)3 — (mA — B); 2 2 = (p + q)2 — c2 = mAB, (5). Hence, putting p2 + r=mAB, = r = 2pq + q2 — c we get c√2pq + q3 — r, (7), which gives c when p, q, and r are known. If 1 or 2 is put for q, it is clear from (6) that when has Р its least value, must have nearly its greatest value; conse Р quently, in finding the factors A and B of N, it will be necessary to pay particular attention to those values of m which give the greatest values of Р Supposing (by approximation) that the least value of mA - B is expressed by 4, then we shall get B=mA± 2 A and mAB = m2A2 ± mA2, or √mAB = A√m2±m; and since p is not much less than mAB, we shall (approximately) have p = A√m2±m, which (if m is much greater than unity) may be reduced to p = mA nearly. nearly, which, if q equals 1 or 2, may be reduced to Р Р = = A 4m 29 If A is very small in comparison to B, and 1 or 2 is put for q, then, since m must be a large number, and that A less than it is clear from (8) that the greatest values of ? 4m will generally be greater than unity, P Reciprocally, if, in finding the factors A and B of N, m is a large number, and the values of that correspond to m2, m, and m+2 are greater than unity, or if two of them that are successive are greater than unity, while the remaining value does not differ much from it, then it is clear that A is small in comparison to B, and since the value of c that corresponds to the greatest of the values of r will geneA rally be less than neglecting the sign, it is also manifest 2 that A and B can (generally) be found immediately from (1) and (2). To illustrate what has been done, take the following EXAMPLES, 1. To resolve 997331, of the form 4n + 3, into its factors. After a few trials, we find that by representing m -- 2, m, and m + 2 by 59, 61, and 63, we shall get and for the corresponding values of and since p 10377 7926 7799 — 7670 = 129 and 7926 — 7799 127, it follows from = (4) (if 59, 61, and 63 are the correct values of m2, m, and m+2) that A ought not to differ much from 129 or 127. Indeed, since 997331 × 61 is of the form 4n+ 3, we shall get a=p+q=7799 + 1; consequently, from p = 7799, g= 1, r = 12790, and c=2pq + qr, we get c± 53; hence, from ac we have 7853 and 7747 = 127 × 61 for the factors of 997331 × 61, and of course 127 and 7853 are the factors of 997331. 2. To resolve 5428681, of the form 4n +1, into its factors. Putting 55, 57, and 59 for m2, m, and m + 2, we get 13614 26717 25363 and for the corresponding values of ; 17279' 17590' 17896 consequently, if 55, 57, and 59 are the correct values of m-2, m, and m + 2, since 17590-17279 = 311 and 17896 P - 17590306, we ought to have A nearly equal to 311 or 306. Because 5428681 × 57 is of the form 4n + 1, we put a = P+ q = 17590 +1, and thence from p=17590, q = 1, r = 26717, and c = ± √2pq + q2 −r, we get c = ± 92. Hence, from ac we get 17683 and 17499 = 307 × 57 for the factors of 5428681 × 57, and of course 307 and 17683 are the factors of 5428681. 3. To resolve 262657, of the form 4n + 1, into its factors. Multiplying the number by 1, 3, 5, 7, 9, 11, etc., succes513 1202 2260 2574 1544 2626 2432 3599 sively, we get 512' 887' 1145' 1355' 1537' 1699' 1847' 1984' 400 4194 2693 4262 2581, and so on, for the correspond2113' 2233' 2348' 2457' 2562' Because so many of the first values of are greater than 2 Ρ unity, it clearly follows from what has been done, that the factor A must be extremely small in comparison to the remaining factor, B, of the given number; indeed, it is easily shown from what has been done, that the given number is a prime, or has no other factors than unity and itself. 4. To resolve 333667, of the form 4n + 3, into its factors. 738 Multiplying by 1, 3, 5, 7, etc., successively, we get 577' 1001 1654 885 3179 3012, and so on, for the correspond1000' 1291' 1528' 1732' 1915' ing values of 2. Hence, it is easy to infer (as in the last Р question) that the given number is a prime. Remarks. This and the preceding example have been taken from p. 288 and p. 287 of Legendre's Théorie Des Nombres, where the numbers are shown to be primes. SECTION XXII. PERMUTATIONS AND COMBINATIONS. (1.) THE different arrangements that may be made of quantities or things, when any number of them are taken at a time, as to their order, are called their Permutations. Thus, if we take the letters a, b, c, etc., singly; since each letter can have one position, there will of course be as many positions or permutations as there are letters; and if we take the letters a and b together, since they can be written in the orders ab and ba, there will be two permutations. Supposing the letters a, b, c, d, e,.... t, to represent n quantities or things; then if m is a whole number not greater than n, we propose to find the number of permutations that can be made of the n letters, by taking m of them at a time. If we find the permutations that may be made of the n-1 letters b, c, d, e, . . . . t, when m 1 of them are taken at a time, and write a before each of them, we shall have all the permutations that can be made when a stands first. Similarly, the n-1 letters a, c, d, e, . . . . t, when m - 1 of them are taken at a time, and b written before each of them will give the same number of permutations (as before) in which b stands first; and so on, for the letters c, d, e, . . t.. Hence, adding the number of permutations in which a, b, c, . . . . t, severally stand first, it is clear that the sum equals n times the number of permutations that may be made of the n 1 letters a, b, c, ...s, when taken m - 1 at a time. If we take n 1 and m 1 for n and m, it follows, from what has been shown, that the number of permutations of the n 1 letters a, b, c, . .... 8, when taken m - 1 at a time, equals n 1 times the number of permutations that may be made of the n 2 letters a, b, c, d, e, . . . . r, taken m — 2 at a time, and the number of permutations of the n-2 letters taken m2 at a time, equals n-2 times the number of permutations of the n-3 letters a, b, c, d, e, . . . . q, taken (m m-3 at a time, and so on, to the number of permutations that may be made of n (m − 1) = n — m + 1 letters a, b, c, etc., taken m - (m − 1) = 1 at a time, which clearly make nm + 1 permutations. Hence, it follows that the number of permutations that can be made of the n letters a, b, c, d, e, . . . . t, when taken m at a time, is truly expressed by the product n(n − 1)(n − 2) (n − 3) × .... × (n − m + 1), (1); consequently, if m = n or if all the letters are taken at a time, the number of permutations is expressed by the product 1.2.3.4.5 x .... X (n − 1)n, (2). EXAMPLES. 1. To find the number of permutations of the 5 letters a, b, c, d, e, taken 3 at a time, and when they are all taken at a time. Because n = 5, m3, and n-m+1=3, (1) gives 5.4.360 the number of permutations when 3 letters are taken at a time; and (2) gives 1.2.3.4.5 120 the = number of permutations when the letters are all taken at a time. 2. To find the number of permutations that may be made of 7 different things taken 5 at a time, and when they are all taken at a time. Ans. 7.6.5 210, and 1.2.3.4.3.6.7 5040. 3. To find the number of different permutations that may be made of the letters in the word good, when the letters are all taken together. If the letters were all different from each other, we should have 1.2.3.4 = 24 for the answer; but since the letter o is repeated, and that interchanging o and o in any permutation does not affect it, it is clear that we must divide the preceding result by 1.2 the number of permutations between two different things, when both are taken together; consequently, = 12 the required number of per mutations. 1.2.3.4 4. To find the number of different permutations among the letters in aaabbed, when the letters are all taken together. |