Where the fractional part of the last result being the same as that of the first, we shall have 13 } Quotients............6, 2, 2, 12, 2, 2, 12, &c.) Converging fractions...,, 2, 32, 397, 128, 2040, &c. And as the last figure of the first period of these quotients, 2, 2 12, occupies an odd place, and has the fraction 32 standing under it, we shall have x = 32, and y 5, which numbers, being substituted in the proposed equation, give 322-415-1024-1025 And if m be put = 1, in the formulæ expressing the general values, we shall have = 1. y (32+5/41)+(32-5/41)3 x = =131168, and (32+5/41)-(32-5/41) 241 2 20485; which numbers, when substituted as above, also give 1311682-41204852-1. And when m= 2, 3, 4, &c. other values may be found. Also, if the proposed equation had been 2-41y2+1, we should have had, from the above process, X- =2049, and y=320; these being the numerator and denominator of the fraction standing under the last figure of the second period, which here occupies an even place. for 3. Given -13y=1, to find the values of x and y. Here, according to example 2 of the last article, we shall have, 13, the following quotients and converging fractions: S3, 1, 1, 1, 1, 6 where the last figure 6, of the first period, being in an odd place, we must take the fraction that stands under the 6 in the next period; which gives x= 649, and y: 180, for the least values of these quantities that satisfy the conditions of the question. And if the proposed equation had been 2-13---1, we should have had, for their least values, in this case, x 18, and y=5. And = greatly deceived if, after having failed in trying a few moderately large numbers, we should thence conclude that the question was impossible. These circumstances induced Euler to form a table of the values of x and y, necessary for the solution of the equation ay 1, for all the values of a from 1 to 100; which is given in Vol. II. of his Algebra, together with another of double the extent, by Lagrange. Legendre, in his Essai before mentioned, has also extended the same table to upwards of 1000, and has shown its application to the solution of the general case x2-ay-b. from these other values may be found, as in the examples before given. y. 4. Given 22 - 14y=1, to find the least values of x and Ans. x= 15, and y = = 4. 5. Given x2-22y2 = 1, to find the least values of x and y. Ans. x = 197, and y = 42. 6. Given 22-29y2 8. Given x2- 113y2 9. Given x2 1, to find the least values of x and y. Ans. x= 66249, y = 9100. y. == 1, to find the values of x and y Ans. x= 1204353, and y - 113296. — 7y2 = 1, to find the first values of x and y. Ans. x= 3, or 45, y= 1, or 17. 167. To determine the values of x and y in the equation x-ay-b, when the absolute term b is less than a. RULE. Find the partial quotients arising from extracting the square root of a, after the manner before used, to the end, if necessary, of the first period, and set their corresponding converging fractions under them. Then, if the absolute term b be found in the denominator of any one of the complete quotients obtained from this development, the converging fraction standing under the partial quotient answering to that term, will give the solution of x2-ay—b, or x-ay--b, according as the fraction occupies an even or an odd place in the period. But if the term b be not found in an odd place, the latter equation is impossible; and if it be not found in the denominator of any of the complete quotients, the proposed equation is impossible under either sign.* Secondly, when the equation is possible, if the numerator and denominator of the fraction abovementioned be denoted by m and n respectively, and the values of p and q be found, by the last Article, in the equation p-aq+1, or p2-aq-1, according as the substitution of m and n for x and y, gives a result with the same or a contrary sign to that proposed in the equation, we shall have, for new values of these quantities, xpmaqn, or yqm pn. And by substituting these values of x and y for m and n *The quotients arising from this mode of extracting the square root of any nonquadrate number being, in general, symmetrical, the absolute term b may be found several times in the same period; in which case we shall have as many solutions of the equation x2-ay2=+b, or a2—ay2——b, as there are recurrences of this kind; and the same will likewise take place in all the other pe in the same formulæ, and so on, other values of them may be obtained at pleasure.* 1. Given x2-23y-2, to find the values of x and y. where, the absolute term 2 being found in the denominator of the second complete quotient, we shall have { Partial quotients......4, 1, 3, &c. and consequently as the Converging fractions..,,, &c. fraction answers to the partial quotient 3, in the last term, we shall have x=m=5, and x-n=1; which numbers being substituted in the proposed equation, give 52-23×1=25-23-2. Also if there be now taken the auxiliary equation p-23q2-1, we shall have, by continuing the above process for 23, and taking the fraction which stands under the last quotient figure of the first period, p 24, and q-5; whence, according to the formulæ given in the latter part of the rule, x=pmagn ating pr} that is, {5, or 235, y=1, or 49, And by putting these values for m and n in the same formulæ, other values of x and y may be found; and so on. Given 2-13y=3, to find the values of x and y. whence, the absolute term 3 being found in the denominator of the last of these complete quotients, we shall have {Converging fractions....3, 1, 1, &c. And consequently, as the fraction answers to the partial quotient 1 in that term, *The general values of x and y in this Article may be otherwise obtained by determining those of p and q, according to the rules before given, and then substituting them for those letters in the above formulæ ; but the method here followed is more convenient in practice. we shall have x = m = 4, and y=n=1; which numbers being substituted in the proposed equation, give 42-13 x 1 = 16 — 13 3; and, as the result has here a different sign from that in the question, we must take the equation p2-ag? —— 1, which being resolved by the method beforementioned, gives p-18, and q-5; whence, by substituting these values, together with those of m and n, in the common formulæ, we shall have | y=qm±pn 137, Į y=2, 38, (=pman} that is, 22, or 397 for the values sought ; 1371333; which give, by substitution, means of these values, others may be readily found. and, by We might here also proceed to the solution of the general case of these equations -ay-b, where a and b are supposed to be any given integral numbers; but the artifices and train of reasoning required for this purpose would extend the present article to too great a length. We must therefore refer those who are desirous of farther information on this interesting subject, to_Lagrange's Additions, at the end of Euler's Algebra, or to the Essai sur le Theorie des Nombres of Legendre, second edition. where they will find almost every branch of the Indeterminate Analysis treated with great perspicuity and elegance. Of Continued Fractions. 168. A continued fraction is that which has for its denominator a whole number and a fraction; and so on, continually, or till the series terminates, by being broken of, after a certain number of terms.' These expressions arise in various ways, and are of great use in finding the approximate values of fractions and ratios, that are expressed in large numbers, as well as in the resolution of certain unlimited problems of the first and second degrees; in the latter of which the answers, in whole numbers, cannot be easily obtained by any other method. Thus, in order to represent the irreducible fraction, or ratio,, by a continued fraction, let b be contained in a p times, with a remainder c; also, let c be contained in b, q times, with a remainder d; din c, r times, with a remainder e; and so so on, as in the following operation: Art. 48 p. 33. a Then we shall have, by the common method for the division of a c b C d c e numbers =+;,=9+=+, &c. where the integers p, q, r, &c. are called partial quotients: and each of these, with its remainder, or depending fraction, complete quotients. And since, by taking the reciprocals of the second, third, &c. of be substituted for their equals in the former, there will arise d=p+ 1 d &c. Whence, by extending the number of terms, and generalizing the formulæ, we shall have P+ 1 according as the nu merator is greater or case, will less than the denominator; which expressions, in this consist of a greater or less number of terms, according as the frac a tion is more or less complicated; but they will always terminate a when is rational. Hence, from what has been here shown, it is Ъ obvious that any given irreducible fraction may be converted into a continued fraction, as follows; RULE. Divide the greater of the two terms of the fraction by the less, and the last divisor continually by the last remainder, till nothing remains, as in finding their greatest common measure; then the successive quotients, thus found, will be the denominators of the several terms of the continued fraction, and their numerators are always unity, or 1. 1. Thus, if it be required to reduce 1 to a continued fraction, 1051)2431(2 where, since the numerator of the proposed fraction is greater than the denominator, the first quotient figure 2, will be an integer, and the rest of 1)9(9 the quotients, taken in order, are 3, 329)1051(3 64)329(5 5,79. Whence, taking these for the denominators of the continued fractions sought, and 1 for each of their numerators, we shall have, according to the above rule, |