this, as well as in the preceding problems, there are many expressions in which the unknown quantity admits of only one value, and, in a great many instances, the value is impossible. We now proceed to show how to find other values from having one value already given. 185. Suppose it is already known that the expression √(ax+bx2+cx+dx+e) becomes rational when xr, and that we have ar+b+cr2+dr+e=s2. Assume y+r= x, and we have ay+4ary+6ar2y2+4ar3y+art = ax1 the terms in the bx3 last line representcx2ing the sums of the by3+3bry2+3br1y+br3 = = = dx e= e ay+ b'y3+ c'y2+d'x + s2 = 0 Hence the expression is reduced to a form in which the preceding case will apply, and therefore the value of y, and thence that of x, may be determined. As formula of the form ax1 + b often occur in the solution of Diophantine Problems, it demands some further notice here. Supposing p to be a value of x, that renders ax+b a square, so that ap+b=q, to find a general expression for other values: Assume x=pty; then because ap1+b= q, by substitution a+b becomes q+4apy+6apy +4apy3 +ax = 0; which formula falls under case 3, and by comparing each term with the general expression, we have qe, 4ap-d, 6apc, 4apb, and aa; whence we have d 4ap3 2ap3 n = = 2e 2q ap2(3q2 2ap1) y= 2 = __ap2 (q2+26) q 3 b-2mn 4apq-4a2p3q2 (q2+2b)___4pq°—(q2—b) (q2+26)4pq m (q2—b) (q2+2b)2——qʻ (dividing by a, and substituting q3—b for ap1) 8bq2-4q1 39-46; hence we have x=p+y=p × This simple value of x shows that the most complex expressions may often be reduced, by skilful management; the reduc tion was first given by Euler. This formula may often be applied to difficult questions, where expressions of the form ax + b often occur. 1. Find such values of x, that the expression 3x+2x3-5x2+ 7x-3 may be a square. It appears, upon trial, that if 1 be substituted for x, that the expression will become a square, viz. 4. Put therefore xy+1, and we have 3x2+2x3—5x2+7x—3—3y2+14y3+19y2+15y+4, which must be made a square; therefore, according to the last case, denote this square by (y+5y+2)2={f} by + 418 by3 + 19y2+15x+4, we shall then have 3y+14y2=3y+1853, or 3y+14=$3$ty; whence y=13424, and, consequently, 2=25471. 6047 Problem IV. To find such values of x as will render rational the expression (ax+bx2+cx+d.) In this problem there are likewise only three cases in which a direct solution can be obtained; these are as follows. CASE I. When both first and last terms are cubes, or when the expression is of the form ⇓ (a3x3+bx2+cx+d3.) Put a3ï3+bx2+cx+d3=(ax+d)3—a3x3+3a2dx2+3ad2x+ď3, and we have bx2+cx=3a2dx2+3ad2x, or bx+c=3a2dx+3aď2; whence x = 3ad-c 1. Find a value of x such, that the expression +9x+4x+8 may be a cube. Put x+9x+4x+8=(x+2)3—x3+6x2+12x+8, = and we shall 23. then have 9x+4x 6x2+12x, whence x = CASE II. When the first term only is a cube, or when the pression is of the form (a3x3+bx2+cx+d). Put a3x3+bx2+cx+d=(ax+m)2=—a3x3+3a2mx2 + 3am3x +m3, b and make 3amb, or m= and we shall then have cx+d = 3a2 1. Find a value of x that will make the expression 8x-4x+2x-12 a cube. Put 8x-4x+2x-12=(2x-})3—8x3 —4x2+}x+2, and we get 2x-12-3x+; whence x-225. 36 CASE III. When the last term only is a cube, or when the expression is of the form (ax+bx2+cx+d3). Put ax+bx2+cx+d3=(mx+d)3—m3x3-+3m2 dx2+3md2x + d3, C and make c=3md2, or m=- and there results ax3-bx2= m3x3 3d2 +3m2dx, or ax+b=m3x+3m2d; whence x 34** 1. Required such a value of x that will make the expression 2x+3x-4x+8 a cube. Put 2x+3x2-4x+8=(—}x+2)3 = — 212x2 + 3x2 - 4x+8, and we have 2x+3x=-24x3+Zx2 - zres or 2x+3=- +whence —— 63. These two last cases are evidently applicable to those forms belonging to case 1st, and, therefore, when the first and last terms are both cubes, three solutions may be obtained, one from each case; it must however be observed, that they all fail when b and c are both 0. Having now given all the cases in which a direct solution of the problem can be obtained, it remains to show, as in the preceding problems, how, from having a particular solution, others may be derived from it. (6.) This expression when xr, and that then ar3+br2+cr+d=s3. -x, and we have (ax + bx2 + bx+ d) becomes rational ay3 + 3ary? 3arly by+2bry cy ar3 = ax3 = a cube. Assume y +r The expression is therefore reduced to a form which is resolva ble by last case. ay3 + b'y + c'y + s3 1. It is required to find such values for x, that the expression 2x3-4x2+6x+4 may be a cube. It appears upon trial, that x=1 is a satisfactory value; put then x-y+1, and the expression becomes 2y+2y+4y+8, which put equal to (x+2)3= »y3+}y2 +4y+8, and there results 2y+2y=24y3+3y2, or 2y+2=y +; whence y=-39, and, consequently, x=1}. On Double and Triple Equalities. 195. In the preceding problems, the object of our investigations has been to find rational values for expressions under a surd form, and our inquiries have been directed to each expression separately. Questions, however, often occur in the diophantine analysis, that requires us to find values for the unknown quantity, or quantities, that shall not only render a single expression, a square, cube, &c. but that shall also, at the same time, fulfil similar conditions in one or more other expressions, containing the same unknown quantity or quantities. In the case where two expressions are concerned, it is called a double equality, and when there are three expressions, a triple equality, &c. The following methods of resolving these equalities will be of service to the student in ordinary cases; but in those where the methods here given are and found to be insufficient, he must be guided by his own penetration and ingenuity, since no general method of proceeding, that shall be suitable to every case that may occur, can be given.. Problem I. To resolve the double equality ax + b = 0, cx+d. Put ax+b=p2, and cx + d=q2, then equating the p2 b two values of x, which the equations furnish, we have a q'-d C ́, or cp-cbaq-ad; therefore cpcaq-cad + cb, and, consequently, g must be such a value that the expression caq? cad+cb may become a square, which value may be ascertained by one or the other of the preceding methods, and thence the value of x may be determined. Problem II. To resolve the double equality, or } Put x= x= y then if each equality be multi plied by y, there will result the double equality a+by, and c+ dy , which belongs to the preceding problem. Or put ax2 + bx = p2x2, then ax+b=px, and consequently, x= b p2-a? a p -a or multiplying by the square (pa), it becomes co-abd+dbp ; whence p may be determined, and thence x. Problem III. Resolve the double equality c az2+ bx+= Here it will be necessary first to resolve the dx2+ex+f= ᄆ equality by Problem 1, Art. ( 1 ), and to substitute the value x so. deduced in the second equality dr2+ex+f=□, which will in consequence, rise to the fourth power, and therefore its solution will belong to Prob. 3, p. 397. Problem IV. To resolve the triple equality Put cx+dy-u then, by expunging y ex+fx=s2 from the two first equations, we have x = punging x from the same equations, we hove y ad-bc ; there fore, by substituting for x and y, in the third equation, their re af-be cf-de spective values here exhibited, we shall have ・u2 سلم ad-bc adbc: or putting u tz, and dividing the expression by the Having then found which the values of z may be determined. the values of z, we shall have, from the above values of x and y, observing to write tz for u, the following results: viz. x— and y = = az2 -C value whatever. any d- bz ad-bc where t may be 189. The above are the most general methods hitherto discov ered for the resolution of double and triple equalities; we may now therefore proceed to show the practical application of the foregoing parts of the present chapter to the solution of diophantine questions; but, as has been already said, the student must expect to meet with cases in which the mode of proceeding must be left, in a great measure, for his own judgment and penetration to suggest. Indeed, the subject on which we are now treating has exercised the ingenuity of some of the most eminent mathematicians of Europe; but Euler and Lagrange have been the most successful in combating the difficulties with which it is attended. The performances of the former are contained in the second volume of his Algebra, which with the additions of Lagrange, forms the most complete body of information on the diophantine analysis extant, and it to this work, chiefly, that the attention of the student is directed. In the following solutions it will frequently be observed that much depends upon the nature and relation of the assumptions made at the commencement, as a little artifice and ingenuity here will often enable us readily to satisfy one or two conditions of the question, when those that remain may be obtained by one or other of the known methods already given. 1. Find three numbers such, that if to the square of each the product of the other two be added, the results shall be all squares. If x, y, and z denote the numbers, then x2+yz=□, y2+xz= 0,22+xy=0. Assume x=mz, y = nz, then the expressions become m2+n□, n2+m=¤, 1+mn=0; now the first and second expressions are evidently squares, if m+n = 1, ··· -n; by substitution, 1+mn=1+‡n—n2—■—(1—cn)2; 2c+1 8c+1 c2+14(c2+1)' m= whence n = c(c-8) 4(c2+1) z, y=nz c(c-8) 4(c2+1) z; hence if z be ta and m=n= ken=4(c+1), then x=c(c-8), y=8c+1; where c may be any number greater than 8: taking c-9, we get z=328, y=73, x=9. Let y, x y and 4x denote the numbers. Then it only remains to make 16x2+xy—y'—■—(4x—ry)2=— 16x2—8rxy+ry'. |