And consequently the integral values of the two unknown quantities are x=16-3, or 13 | 48 12-3, or 91 3, or 98-3, or 5 | 6-3, or 3=4-3, or 1 48 48 - | y=16-2, or 1 =12-2, or 2 | -2, or 4 8 Which are all the answers in whole positive numbers that the question admits of. 2. Given x + xy = 2x + 3y +29, to find the values of x and y in whole positive numbers. Here, by the second form, a=2, b=3, & c= = 29. 26 X3 Where it is plain, that x must be such a number, that, when diminished by 3, shall be a divisor of 26. But the several divisors of 26 are 1, 2, 13, and 26, the only ones of which, that will render the expression positive, are 1 and 2. Therefore x=1+3=4=2+3=5, Which are all the answers, in whole numbers, that the question admits of. 3. Given x2=y2+ 20y, to find the values of x and y in whole positive numbers. Here, by the third form, a=20, and b=0, Where it is plain, that n must be some even number which is a divisor of 50. But the only number of this kind, that will give positive results, is 2. 50 Therefore y=+1-10=16, and x=10+16-2=24. 2 4. Given x2=5y-12y+64, to find the values of x and y in whole positive numbers. Here, by the 4th form, a=5, b= − 12, & c=8. Where it is plain, that n must be less than √5, and greater than ; which numbers are only 1 and 2. And x=8+1x7=15=8+ 2 x 44=96. 5. It is required to find two numbers, such, that their product, added to their sum, shall be 79. 6. Given x + xy = 4x + 3y + 27, to find the several values of x and y in whole numbers. 7. Given x=y2+ 100y+ 1000, to find the two least values of x and y in whole numbers. Ans. x 70 and y = 30 8. Given x 50y+100y+100, to find the values of x and y in whole numbers. Ꮖ Ans. x 190 and y=40 OF THE DIOPHANTINE ANALYSIS. (G) This branch of Algebra, which is so called from its inventor, Diophantus, a Greek mathematician of Alexandria in Egypt, who flourished in or about the third century after Christ, relates chiefly to the finding of square, cube, and other similar numbers, or to the rendering certain compound expressions free from surds; the method of doing which is by making such substitutions for the unknown quantity, as will reduce the resulting equation to a simple one, and then finding the value of that quantity in terms of the rest. It is to be observed, however, that questions of this kind do not always admit of answers in rational numbers, and that, when they are resolvable in this way, no rule can be given that will apply in all the cases that may occur; but, as far as respects a particular class of these problems, relating to squares, they may generally be determined by means of some of the rules derived from the following formula; observing, that the possibility or impossibility of the different cases of it depend upon the nature of the coefficients a, b, c. (x) PROBLEM I, To render surd quantities of the form (a+ bx + cx) rational; or, to find such values of r as will make a + bx + cx a square. (x) The subject here treated of, which forms one of the most curious and abstruse branches of the Indeterminate Analysis, has been amply investigated and exemplified, both by Diophantus and several modern writers on Algebra, whose works abound with inquiries of this nature; but as some of the most valuable of these performances have now become extremely scarce, and others are too voluminous to be brought into general use, it is hoped that the following methodical abstract of this part of the science, which comprehends most of the methods hitherto known, for resolving problems of this kind, will be found acceptable to such readers, as may wish, by means of a ready compendium, to acquire some knowledge of this interesting branch of the Analytic Art. RULE. 1 1. In the case where c=0, put the remaining part √(a + bx)=n, or a + bx= n2; and we shall have n2 -a x= ; where n, both in this and the following b cases, may be any number, either integral or fractional, that will render the value of a positive. 2. In the case where a=0, put √(bx + cx2)=nx, or bx + cx2 = n2x2, then, by dividing by x, and transposing the terms, we shall have n'x- cx=b; and 3. When a is a square number, put it =d, and make √(d2 + bx + cx2) =d+nx; then d2 + bx + cx2 = d2 + 2dnx + n2x2, or b+cx=2dn + n2x; and conse quently x= 2dn-b 2dn Or, if b=0, x=-n2° C = 4. When c is a square number, put it e2, and make √(a + bx + e2x2) = n + ex ; then a + bx+ e2x2 = n2 + 2¢nx + e2x2, or a + bx=n2 + 2enx ; and conse quently x= a- -n2 2en-b Or, if b=0, x= a-n3 5. When neither a nor c are square numbers, yet if the formula can be resolved into two simple factors, (which it always can when b2-4c is a square, but not otherwise,) the irrationality of it may be taken away, by putting (a + bx + cx) = ~{(d+ex)(f + gx)} = n(d+ex); in which case we shall have (d+ex) (ƒ + gx) = n2(d+ex)3, or ƒ +gx= n2(d (d+ex); and consequently x= d=0, x= and if ƒ =0, x= f en2-g g- en2* Or, if 6. When neither of the above rules will apply, if the formula can be resolved into two parts, one of which is a square, and the other the product of any two simple factors, the irrationality of it may be destroyed, by putting √(a + bx + cx2) = √ { (d + ex)2 + (ƒ + gx)(h+kx) } = (d+ex) +n(f + gx); in which case we shall have (d+ex)2 + (f + gx)(h+ kx) (d+ex)2 + 2n(d + ex}{ƒ+gx) +n2(ƒf +gx)2, or h+ kx = 2n(d+ex) +n2(f+gx); and consequently a= An h Or, if d=0, x= k-n(2e+gn)* (2d+fn) - h k-n(2e + gu) Or, if the part, in this case, which is found to be a square, be a known quantity, put √(a + bx + cx2) = ~ { d2 + (e +ƒx) (g+hx) } = d + n(e+fx); then we shall have d+(e+fx) (g+hx) = d° + 2dn(e + fx) + (e+fx), or g+hx = 2dn + (e + fx); and consequently, by transposing and uniting the different e + 2dn-g terms, x= h-f 7. These being all the cases of the general formula that are resolvable by any direct rule, it only remains to observe, that, in other instances of a different kind, if we can, by trials, find any one simple value of the unknown quantity that satisfies the conditions of the question, an expression may be derived from this that will furnish as many other values of it as we please. 2 2 == Thus, let p be a value of x so found, and make a + bp + cp2=q'; then, by putting a=p+y, we shall have a + bx + cx2= a + b(p + y) + c(p + y)2 = a + bp + cp2+by+2cpy+cy, or a + bx + cx2=q° + (b+2cp)y+cy; from which latter form the value y, and consequently that of x, may be found as in Case 3. of |