And f(m, r) = 1 +f(r, r − 1) + f(r + 1, r) + ... + f(m − 1, r − 1) = a whole number; or the product of r consecutive numbers is divisible by [r' Ex. 1. If n be any whole number, then will n (no − 1) (no — 4) be divisible by 120. n (n2 − 1) (n2 — 4) = n (n − 1) (n + 1) (n − 2) (n + 2) which is the product of 5 consecutive numbers, and is therefore divisible by 1.2.3.4.5, or 120. Ex. 2. If n be any even number, n3 + 20n is divisible by 48. Now (m-1)m (m + 1) being the product of three consecutive numbers is divisible by 1.2.3 or 6; therefore n3 +20n is divisible by 48. 372. Every number which is a perfect square is of one of the forms 5m or 5m ± 1. For every number is of one of the forms 5m, 5m+1, 5m+2, 5m+3, 5m+4; all of which are included in the forms 5m, 5m±1, 5m±2, since 5m+3 = 5(m + 1) − 2 = 5 m′ − 2, and 5m+4=5(m + 1) − 1 = 5m′ — 1. But (5m)2= 5(5m3) = 5m', 5m + 1; (5m+ 2)2 = 25m2 + 20m + 4 form 5m-1; which is of the form 5m; = = 5 (5m2±2m) + 1, which is of the form 5 (5m2± 4m + 1) −1, which is of the .. every square is of one of the forms 5m, 5m+1, 5m – 1. 373. Every "prime number" greater than 2 is of one of the forms 4m ± 1. For every number is of one of the forms 4m, 4m + 1, 4m + 2, 4m+3; but neither 4m, nor 4m + 2 can represent prime numbers, since each is divisible by 2; therefore all prime numbers greater than 2 are represented by 4m+1 and 4m + 3. But 4m+3=4 (m + 1) − 1 = 4m′ −1; therefore the two forms are 4m ± 1. * This proof is not so simple as that found in Barlow's Theory of Numbers, Art. 12, and usually adopted; but the latter is liable to serious objection; for it assumes that, if the n(n-1) (n-2) &c. numerator of be divisible by each factor of the denominator separately, 3 &c. 1 2 it is divisible also by their continued product: which requires proof, since a number may obviously be divisible by each of two or more numbers, and yet not be divisible by their product. COR. Since m may be odd or even, that is, of the form 2n, or 2n+1, all prime numbers are represented by 8n±1, or 8 n±3. 374. Every prime number greater than 3 is of one of the forms 6m+1. For every number is of one of the forms 6m, 6m+1, 6m+2, 6m+3, 6m+4, 6m+5, of which the 1st, 3d, 4th, and 5th obviously cannot represent prime numbers; and therefore, all prime numbers greater than 3 are represented by 6m+1 and 6m+5. But 6m+5=6 (m + 1) − 1 Therefore, 6m+1 will include all prime numbers greater = 6m' - 1. than 3. COR. Since m may be odd or even, that is, of the form 2n, or 2n+1, all prime numbers greater than 3, will be included in 12n± 1, or 12n± 5. 375. No Algebraical formula can represent prime numbers only. Let p +9x+rx2 + &c. be a general algebraical formula; and let it be a prime number when x = m; therefore (P) the prime number in that case is p+qm + rm2 + &c. Now let x = m + n P ; then p+q (m + n P) + r (m + n P)2 + &c. is the number; and is equal to 66 p+qm+rm2 + &c. + MP, (M signifying some multiple of,”") = P + PM, which is divisible by P, and therefore not a prime; consequently the formula does not represent prime numbers only. 376. The number of primes is indefinitely great. For if not, let there be a fixed number of them, and let Р be the greatest: then ...... p is divisible by each of them, not one of them. and 1.2.3.5.7.11 p+1 If this latter number, then, be divisible by a prime number, it must be one greater than p; if not, it is itself a prime, (since every number is either a prime, or capable of being resolved into factors which are prime) and is greater than p. Therefore, in either case, there is a prime greater than p; that is, we may not assume any prime to be the greatest; or, the number of primes is indefinitely great. 377. To determine whether a proposed number be a prime or not. It is obvious that this may be done by dividing the proposed number by every number less than itself, beginning with 2, until we have either proved it to be divisible by some one of them without remainder, or that it is a prime, from not being divisible by any one of them. But there is no necessity to proceed so far, as may thus be shewn. If the proposed num ber (p) be not a prime, then p= ab, _the product of two other numbers. If then a > √p, b < √p; and if a <√p, b> p. Hence in both cases p is divisible by a number less than the square root of itself. Or it may be that the proposed number is an exact square, in which case it is divisible by its square root. If, therefore, a proposed number be not divisible by some number not greater than the square root of itself, it must be a prime. 378. To find the number of divisors of a given number. Let a, b, c, &c. represent the prime factors of which the given number is composed; and let a be repeated p times; b, q times; c, r times; &c.; so that the number a bic. &c.; then it is evident that it is divi = sible by each of the quantities, and also by the product of any two or more of them, that is, by every term of the continued product of which are all the different divisors (including 1) of ao b1. Similarly, if this result be multiplied by (1+c+c2+...+c), the product will consist of all the different divisors of a b'c'; and so on, if there be more factors of the given number. Now the number of these divisors is obviously p + 1 in a2; in arb the number is p + 1 taken as many times as there are terms in the second series, that is, (p + 1) (q + 1); in abc it is (p + 1) (q + 1) taken r+ 1 times, or (p + 1) (q + 1) (r + 1) ; and generally the number of divisors in a2bc". &c. = (p + 1) (q + 1) (r + 1). &c., including 1 and the number itself. Ex. Find the number of divisors of 2160. Here 2160 = 2 × 1080 = 22 × 540 = 23 × 270 = 21 × 135 = 2a × 3 × 45 = 21 × 3a × 15 = 24 × 33 × 5. Therefore the number of divisors = (4 + 1) (3 + 1) (1 + 1) or 40. COR. 1. The number of divisors will always be even unless each of the quantities p, q, r, &c. be even, that is, unless the proposed number be a perfect square. COR. 2. It is also obvious from what has been said above that the sum of all the different divisors of a b'c. &c. is the sum of all the terms in the continued product of (1 + a + a2 + + a3) (1 + b + b2 + + b2) (1 + c + c2 + ... + c') &c. ... . &c. ... 379. If m be any prime number, and N a number not divisible by m, then Nm-1-1 is divisible by m. (FERMAT'S THEOREM). For, it is easily seen by the Binomial Theorem, that in the expansion of (a + b + c + &c.)" m is a factor of every term except am, b", cm, &c. ; and that the coefficients are always whole numbers. But m, being a prime number, will not be divisible by any factor in the denominator of a coefficient, and will therefore remain as a factor of each term, when the coefficients are reduced. Therefore we may assume, (m being a prime number), (a+b+c+ &c.)" = am + b + cm + &c. + mP. Let, then, a = = c = &c. = 1, and the number of them be N; and But, by supposition, N is not divisible by m, .. Nm-1-1 is a multiple of m, or is divisible by m. 380. To prove that, for any positive integral value of n, Also by the Exponential Theorem, (" - 1)" = (1 + x + + ... − 1)" = (x + 1 ~.~2 + ...)" 1. 2 .... 1.2. 3 ..p −1+1=pQ, or is divisible by p; which is Wilson's Theorem, for determining whether any proposed number be a prime or not. 381. VANISHING FRACTIONS. To find the value of a fraction when the numerator and denominator are evanescent. Since the value of a fraction depends, not upon the absolute, but the relative, magnitude of the numerator and denominator, if in their evanescent state they have a finite ratio, the value of the fraction will be finite. To determine this value, substitute for the variable quantity its magnitude, when the numerator and denominator vanish, increased by another variable quantity; then, after reduction, suppose this latter to decrease without limit, and the value of the proposed fraction will be known. and when h = 0, or x = a, its value is 2a. |