A similar mode of solution may be applied to continued surds or expressions of the form √a+√a+ √a, &c., the value of which, though apparently infinite, is always determinable by a certain equation, and in some cases in a real integral or fractional quantity; for, putting x=√a+√a+ √a, &c., we shall have, by squaring both numbers, x2=a+√a+ √a+, &c., the latter part of which is evidently equal to the original surd; whence r2=a+r, or r2-r=a.. x=¦±√ita, where, if a=2, the expression becomes 2+√2+ √2+, &c., =2 or -1. 433. The process for developing any consists in making successively x=a+; 1 quantity, x, in a continued fraction, x'=b+x" 1 1 &c., a be ing the greatest whole number contained in x, b the greatest whole number contained in x', c the greatest whole number contained in x'', &c. The numbers a, b, c, &c., being found, it is evident that if x', x', &c., are replaced by their values, the required development is obtained, viz., Let it be required to convert 19 into a continued fraction. By proceeding in this way we shall obtain the following: 5 * Multiplying both numerator and denominator by 19+4. VII being the same as x1, it is evident that, omitting the 4, the greatest integral part of √19, the quotients 2, 1, 3, 1, 2, 8, already found, will always return again in the same order to infinity. Should it be required to convert the square root of 19 into a series of converging fractions without first reducing it to the continued form, they may be obtained, after the method before employed, from the integral parts of the above results only. (3) √31. 1 7 8 23 100 523 623 1769 Convergents, 3' 22' 25'72' 313' 1637' 1950' 5537° Ans. The quotients are 5, 1, 1, 3, 5, 3, &c. 5 16 37 127 1307 &c. (5) √45. And the convergents, 1' 3' 7' 24' 247' Ans. Quotients, 6, 1, 2, 2, 2, 1, 12, &c. 6 7 20 47 114 161 2046 Convergents, 1' 1' 3'7' 17' 24' 305* 434. The converse of the proposition stated in Art. 432 is true, viz., that the root of an equation of the second degree may be expressed in functions of the coefficients of the equation by continued fractions. The general form of the equation of the second degree may be written ax2-bx-c=0.... (1) in which b and c may be essentially negative. This may be put under the form a2 being the greatest square contained in c, we have +i+, &c. The above result may be obtained in a more simple manner; thus, put which shows that the square root of any number which is the sum of a square, and of another number, is a continued fraction. 435. Continued fractions furnish a method of resolving in whole numbers the indeterminate equation In this equation a, b, c are whole numbers, and the first two are supposed to have no common factor. Let us conceive that we have developed the rela a tion into a continued fraction, and that we have calculated all the conЂ vergents; the last will be equal to this relation itself. Let us subtract from a' it the next to the last, which I represent by. The numerator of the difference will be ab'-ba', and by the property of Art. 430 we have then equation (1) is satisfied by taking x=±b'e, y=±a'c. This solution being known, we know (Art. 161) that all the others are given by the formulas =±b'c—bt, y=Fa'c+at, t designating any whole number whatever. We take the upper or lower sign according as we have + or in the equality (2), or, what is the same thing, a according as the convergent is of an even or uneven rank. Let there be the equation b EXAMPLE. 261x-82y=117. The equation, then, is satisfied by making r=11x117=1287 and y=35 X117=4095; then, finally, the general values of x and y are x=1287+82t, y=4095+2611. If we divide 1287 by 82, and 4095 by 261, we find 1287=82 × 15+57 and 4095 261 × 15+180. Then, observing that t is any whole number whatever, we can write more simply x=57+82t, y=180+261t. 436. The following theorem will be found useful in the resolution of indeterminate equations of the second degree. Let p-Aq=D be an indeterminate equation, in which D<√Ā. I will be found among assert, that if this equation is resolvable, the fraction the fractions which converge toward √Ā. From the above equation we derive P D and, therefore, Dq ; then d q2 q(p + q √ A) Po Let be the converging fraction which precedes and which is of such a lo nature that the sign of d will be the same with that of D; it will remain to be proved that we have In the second member, instead of p, I put its value, q√A; the in9 equality to be proved can then be written thus: (9+%)( √×D)+(2−%) √ño >0. But this inequality is manifest, since we have √A>D, q>%, and since the part (q—9%) √A, which is at least equal to √A, by itself surpasses δ which is less than unity. then, will always be found in the fractions which converge toward VA, so that it will only be necessary to develop √A in a continued fraction, and to calculate the converging fractions which result, in order to have all the solutions in entire numbers of the equation -AyD, D being <VA. METHOD OF RESOLVING IN RATIONAL NUMBERS INDETERMINATE EQUATIONS OF THE SECOND DEGREE. 437. Let the proposed general equation be ax2+bxy+cy2+dx+ey+f=0, in which x and y are the indeterminates, and a, b, c, d, e, f the given entire numbers, positive or negative. We first derive from this equation the following: 2ax+by+d=√ [(by+d)2—4a(cy2+ey+f)]. If we make, to abridge, the radical t, b2-4ac-A, bd-2ae=g, d2-4af=h, we shall have the two equations 2ax+by+d=t, If we multiply the last of these equations by A, and make, again, Ay+g =v, g2—Ah=B, we shall have the transformed equation v-At-B. Reciprocally, if we can find values of v and t which satisfy the equation |