Table of the least Values of x and y in the Equation x?-ay=1, for all Values of a from 2, to 99. (E. v. 2; p. 89.) 9100 66 12 2 20 2574 5 59 69 4 226153980 3 19 10 7 649 15 4 33 60 61 62 63 65 66 67 17 68 2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27 28 29 30 31 32 33 34 35 37 38 39 40 41 42 43 44 45 46 47 69 70 71 72 73 74 75 76 77 78 66249 31 1766319049 63 8 129 65 48842 33 7775 251 3480 17 26 55 28 197 500001 19 1574 1151 12151 2143295 39 49 62809633 99 6 2 2 170 9 55 197 24 5 51 26 127 9801 11 13 3482 199 161 24335 48 7 99 50 649 79 80 82 83 1 16 8 5967 4 936 30 413 .2 267000 430 3 6 3 21 53000 2 165 120 1260 221064 4 5 6377352 10 84 85 86 87 88 89 90 91 92 93 94 95 48 96 97 50 51 52 98 99 (63.) Solution of the equation me – ay: = +6, in which b < Va. If this equation is possible, b will be found amongst the denominators of the complete quotients (a,, az, az, &c. Art. 31.) of the converging fractions which express the value of Va. Given one solution of the above equation to determine the general values of x and y. Suppose that we have found, m, n, p, q, such that m“ – ané= +b, and på – aq'= +1, then x=mpang, y=np+ma, in which the general values of x and y obtained in the preceding Art. must be substituted for p and q respectively. The equation på - aqi=1, or p’ - aq=-1 must be employed, according as the known solution and the given equation have the same or contrary signs. To determine the general values of x and y in the above equation, b being > Va. [1] Suppose b to be composed of factors bı, bą, &c. each <Va: the solution of the given equation, when possible, may be deduced from those of the equations m,' – an,'= +by me - an, = + bą, &c. for (m, - an,%) (mze - ang) (mz+ang) &c. x - ay®. Having found one integral value of x and y, the general values may be found as before by means of the equation př - aqi= +1. [2] Suppose b not to be composed of factors < Va: it will in this case be necessary to find the values of t and u in the equation ť - au’= +bx?, m? - an? +b; calling this the known solution, the general values may be found as before. In this case however the values may be fractional. (B. 181—4.) (64.) Solution of the equation a wo + bay + cy= +e. In this equation, x and y, as well as y and e, may be considered prime to each other. This may be reduced to a similar equation of a simpler form, by assuming X = ,X, +by C=ab,? + bb, b, + cb. If b> a, and >c, the equation may be transformed into another in which 6 $ a, or c. Suppose a < c; assume w=X1 - my, b m being the nearest integer to i assume also b,= am- b, and C=am' — bm +c, the equation becomes a 0,2 – 6,819 +cıyo = +e, in which b, $ a. If b,>c, the same process must be repeated. In the successive transformed equations, the value of bé - 4ac is always the same. If he - 4ac < 0, a and c are the least numbers contained in the transformed equation, in which 6 $a, or c. (Leg. 53–8; B. 100_2.) a [1] Suppose b— 4ac <0, and= -f. Multiplying the equation by 4a, and assuming x=2ax + by, we obtain z + fy=4ae. y<? ( If any integral value of y< 2 which satisfies this f equation, also renders by+x2a, a solution may be obtained; if both these conditions are not fulfilled, the equation is impossible in integers. [2] Suppose 6? —- 4ac=0. In this case the first first side of the equation becomes a perfect square, and is therefore possible, if e is a square. [3] Suppose bo - 4ac>0, and =k?. In this case the proposed equation may be decomposed into two simple equations ayx + bay=eng and aq x + bey=ég. [4] Suppose b? — 4ac=4f, f not being a square. First; let e be < Vf: expand a root of the equation in a continued fraction. (p. 32.) If amongst the quantities 201, 202, &c. the denominators of the complete quotients, we find the number e, the numerator and denominator of the corresponding converging fraction, if substituted for x and y respectively, will satisfy the given equation. As often as e occurs, a different solution may be obtained; if it does not occur at all, there is no integral solution of the proposed equation. Secondly; suppose e > Vf: assume cy=ae; the equation is reduced to a,x? + 6,01. +cy=+1, which must be solved for each value of n. The given equation is impossible in integers, unless n may be so assumed, that a, may be an integer. (Leg. 75–83.) Solution of the equation ax + bay + cy'= +1. b$ a, or c, and a < c; multiply the equation by a, and assume x=ax + by, the equation is reduced to + eye = +a, which may be solved by preceding methods. (E. Add. 66—71.) FORMS OF CUBES. (65.) Every cube * 4n, or 4n+1; #7n, or 7n +1; #9n, or 9n+1. 03, 19, 23, &c. (a– 1)*. 2013 yo 63 202 + y =2x", and generally M 23 + naye =%, in which m is a number of an impossible form to modulus a, and n prime to a, are impossible in integers. No triangular number < 1 is a cube. (B. 60-70; Leg. 328-32.) |