(63.) Solution of the equation x2 — ay2 = ±b, in which b<Va. If this equation is possible, b will be found amongst the denominators of the complete quotients (a,, a, a,, &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. then Suppose that we have found, m, n, p, q, such that m2 — an2 = ±b, and p2 — aq2= ±1, x=mpang, y=np±mq, in which the general values of x and y obtained in the preceding Art. must be substituted for p and q respectively. 2 = The equation p2- aq=1, or p2-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 b1, b, &c. each <Va: the solution of the given equation, when possible, may be deduced from those of the equations for (m2 — an ̧2) (m2 — an ̧2) (m ̧2 + an ̧2) &c. ‡x2 — ay2. Having found one integral value of x and y, the general values may be found as before by means of the equation p2 - aq2= ±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 t2 — au2 = ±b≈2, 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 ax2 + bxy+cy2 = ±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 ɑ1, ɑ2, and b1, b2, being so assumed that a ̧ b2 — ab1= ±1. 2 If b> a, and >c, the equation may be transformed into another in which ba, or c. Suppose a < c; assume x=x1 — my, b m being the nearest integer to ; assume also a b1=am—b, and c1 = am2-bm+c, 2 = Þa. the equation becomes ax-b1xy+cy2+e, in which b1a. 1 If b1>c1 the same process must be repeated. 1 1 In the successive transformed equations, the value of b2-4ac is always the same. If b2-4ac < 0, a and c are the least numbers contained in the transformed equation, in which ba, or c. (Leg. 53-8; B. 100—2.) equation, also renders by + 2a, a solution may be obtained; if both these conditions are not fulfilled, the equation is impossible in integers. [2] Suppose b2-4ac=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 b-4ac>0, and k2. = In this case the proposed equation may be decomposed into two simple equations [4] Suppose b2-4ac4f, f not being a square. First; let e be < Vƒ: expand a root of the equation in a continued fraction. (p. 32.) If amongst the quantities 2a, 2a, &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 2 the equation is reduced to a112+b11ÿ1+c1ÿ12 = ±1, which must be solved for each value of n. The given equation is impossible in integers, unless n may be so assumed, that a1 may be an integer. (Leg. 75-83.) Solution of the equation ax + bxy+cy2= ±1. Let it be transformed so as to fulfil the conditions ba, or c, and a < c; multiply the equation by a, and assume = ax + by, the equation is reduced to x2+ey2= ±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. All cubes are of the same form to modulus a, as the cubes 03, 13, 23, &c. (a−1)3. 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.) |