tions, answering to the several quotients in the first period, no one should be found that satisfies the conditions of the question, it may be safely concluded, that neither the equation x2 - Ay2 = B, nor x2- Ay2=-B, can be resolved in whole numbers. Thus, as an example of this method, let it be required to find the least integral values of x and y in the equation x2-19y=-3. Here, since 19, when converted into a continued fraction, has been shown to give, in its first period, the following quotients and converging fractions. 1421 1 4 61 we shall find, by trial, that both and 14 326 which are the fifth and seventh terms of the period, will satisfy the conditions of the question. For, if 61 be substituted for x, and 14 for y, we shall have 19y=612-19 x 142-3721-3724 3. And, if 1421 be put for x, and 326 for y, we shall also have x2-19y=2019241-2019244-3, as before. Again, let it be required to find the least integral values of x and y in the equation x2 — 55y2=5. Here, by extracting √55, according to the method employed in a former example, there will arise Where, the fractional part of the last term being equal to that of the first, we shall have, for the several quotients of the first period, And, consequently, by following the second method, pointed out in the first part of this article, Whence, the several converging fractions, corresponding with the terms of this period, will be And by trial, it will be readily found, that the And if it were proposed to find the least values of x and y in the equation x2-55y2 = 1, 89 the penultimate fraction, 12 in the above series, will be found to satisfy the conditions of the question. For x2-55y=89a — 55 × 122 = 7921-7920 =1, as was to be shown (p). QUESTIONS FOR PRACTICE. 1. It is required to find the least fraction, having only two figures in each of its terms, that shall approach the nearest possible in value to 1103 887* 2. It is required to find the ratios next less and 450 next greater than that can be each expressed 109' by fractions having only two figures in their denominators. 289 128 Ans. & 70 -(g) 31 31 (p) Here, as in the former part of this problem, if m and n be made to represent the several successive values of x and y in the equation x2- Ay2 = ± 1, and p and q be the least values of r and y in the equation x2- Ay2= ± B, as above found, we shall have, in general terms, x= · pm ± Aqn, and y = pn ‡ qm, from which the various solutions, that any question of this kind is susceptible of, may be readily determined. (9) This question arises out of the method usually employed for adjusting the civil mode of reckoning time to that shown by the sun; the ratio 450 being that of 24h. to 5h. 48m. 48 sec., the time by which the solar year exceeds that in common use; 3. Given 9x+13y=200, to find all the values of x and y in whole positive numbers. Ans. 19, 2; y=5, 14. 4. Given 256x-87y=1, to find the least possible values of x and y in whole numbers. У Ans. x=52, y=153 5. In how many different ways is it possible to pay 100l. with seven shilling pieces and dollars at 4s. 6d. each? Ans. 6 different ways. 6. Given 11x+14y+19%=960, to find the three values of x, y, and %, in whole numbers, that approach the nearest to equality. Ans. x=23, y=22, z=21 7. It is required to find the least values of x and y in the indeterminate equation 7x+19y= 1921. Ans. x=3, y=100 8. It is required to find the number of solutions, in positive integers, of the equation 7x+9y+23x= 9999. Ans. 34365 9. Given x2-61y=-1, to find the least values of x and y in whole positive numbers. Ans. x=29718, and y=3805 10. Given x2-85y2 = 1, to find the least values of r and y in whole positive numbers. Ans. x=285769, y=30996 11. Given x-61y5, to find the least values of x and y in whole positive numbers. 12. Given x2- 61y=5, to find the least values of x and y in whole positive numbers. Ans. x=453, y=58 and, when investigated at length, will show the proper number of intercalary days that answer to a given number of years, OF THE DECOMPOSITION OF RATIONAL FRACTIONS INTO SIMPLE FRACTIONS. (1) THIS problem being of considerable utility in some of the higher branches of analysis, but more particularly in the integral calculus, where it is of frequent occurrence, it will be here proper to show some of the most simple methods of effecting the decomposition required; which, in many of the most useful cases, may be done, without the assistance of fluxions, as follows: RULE I. N D' 1. Let the fraction to be decomposed, have, at least, one dimension less in its numerator than in the denominator; to which state it can be always reduced, when necessary, by division; and will then have the form 2 Find all the m simple factors into which the denominator D can be reduced, either by trial, or by putting the sum of its terms =0, and then determining the roots r, r', r", &c. of the equation, so formed, in the usual way; which being done, the original equation may now be represented under the form 3. Let A, A', A", be the numerators of the several simple fractions into which the given fraction is re |