in any case of this kind, can be resolved into factors, such that one of them shall be a square, it will be sufficient to make the remaining factor a square, in order to render the whole expression so; since a square, multiplied or divided by a square, is still a square.* EXAMPLES. 2 n 2x2 + 1. It is required to find such a value of a as will make 1123 + 3x2 a square. Let the given expression 113 + 3x2 = nx?: agreeably to Case 1. Then, by dividing by x*, we shall have 11x + 3 = na; 3 and consequently, x= where n may be any number, 11 positive or negative, that is greater than 73. Taking, therefore, n=2, 3, 4, 5, &c., respectively, we 1 6 13 shall have, in this case, x == or 2, the last of 11' 11'11' which is the least integral answer that the question admits of. 2. It is required to find such values of x as will make 23 2x2 + 2x + 1 a square. Here the last term 1, being a square, let 1 + 2x 23 -- (1 + x) = 1 + 2x + a*, agreeably to the first part of Case 2. Then, since the first two terms, on both sides of the equation, destroy each other, we shall have 3 - 2x2 = 2*, or 203 = 3x, and consequently x= 3; which, by substitution, makes 1 + 2x 2009 + 23 =1+6 - 18 +27=16 a square, as required. Again, by putting x=y +3, according to Case 3, we shall * The method of determining the factors of which any formula is composed, when it can be done, is to put the given expression then find the roots r, qo', &c., of the equation so formed; each of which will give a factor 2 go', and these are generally easily discovered, as we here seek only the rational roots, which are always divisors of the absolute term, or of that which does not contain x. Thus the formula x3 2+1 is resolvable into the factors (1-2) X(1+x) X(1-2), or(1 - 2)2X(1+x); and by putting 1+x= n2, we have * = n2—1; where, if 'n be taken equal to any number whatever, 23 :C2 — +1 will be a square; though, by any other mode of solution, it would be difficult to find even two or three values of x. It may here also be observed, there are but few questions in this problem that can be determined in whole numbers. Several of them, likewise, admit only of one answer, and others are totally irresolvable, either in integers or fractions. Thus, if it were required to make 23 4 1 a square, the only positive value of that renders this possible, is 2; and the making of 3.22 — 1 a square, is impossible. S 0, and 22 have 1 + 2x - 2x + x3 = 1+ 2 (y + 3) - 2 (y + 3) + ( (y + 3) = 16+ 17y.+ 7y + y. And consequently, by making 16 +177 + 7y + y3 = 17 289 (4+ä yle=16 + 174 + y', agreeably to the first part of 8 64 Case 2, by cancelling 16 + 17y, there will arise 7y2 + y = 289 289 644°, or y +7= 64 289 289 448 159 Whence y -7 and x=3 64 64 159 192 159 33 for another value of x. 64 64 64 Which number, being substituted in the original formula, 429025 655 makes 1 + 2x - 2x2 + ? a square, as 262144 512 before. 64 2 (655) 9 3. It is required to find such values of x as will make 3.2ce3 5x + 6x + 4 a square. Here, 4 being a square, let 4 + 6x — 5x2 + 3x = (2+2x)? 4 + 6x + 2x2, as in the first part of Case 2. Then, since the first two terms on each side of the equation destroy each other, we shall have 303 5x2 2012, or 5+ 29 3x - 5=, and consequently, in this case, x 3 12 29 29 45 Whence (2 + # x1) = (2+3)=(g)2 a square, as was required. Or, by the second method of the same Case, let 4 + 6 0 29 87 5x2 + 3x3 = (2 + 2x 1642)2 =4 + 6x - 5x2 16 841 24; then, as the first three terms on each side of this 256 841 87 equation destroy each other, we shall have 256 16 841 87 3x, or 3, or 841X -- 1392 768; and con256 16 1332 + 768 2160 sequently, x= which is another value 841 841? of X, that, being substituted in the original formula, will make it a square. act 203 4. It is required to find such values of was will make 23 + 3 a square. Here, it is evident that the expression is a square when 1. Let, therefore, X =1+ y, and we shall have 3 + 2 =4 + 3y + 3y2 + y . And, as the first part of this is a square, make, according to the first part of Case 2, 4 + 3y + 3y2 + 38 =(2 + 2y) 4 + 3y + 4y. Then, because the first two terms on each side of the equation destroy each other, we shall have y + 3y =ty, or y +3=tr 9 39 Whence y 3 and x=1 16 16 16 - 39 23 ; which is a second value of x. 3 39 4 64 y, according to the second part of 128 Case 2. Then, as the first three terms on each side of the equation 1521 117 destroy each other, we shall have 4096 1521 117 = 1. 4096 128 352 352 1873 Whence, also, y y which 1521' 1521 1521 is a third value of x. And by proceeding in the same way with either of these new values of x as with the first, other values of it may be obtained; but the resalting fraction will become continually more complicated in each operation. PROBLEM 3.-To find such values of x as will make v (axe + bx3 + cæa + dx + e) rational, or ax* + bx3 + cxa + doc te= a square. The resolution of expressions of this kind, in which the indeterminate, or unknown quantity, rises to the fourth power, is the utmost limit of the researches that have hitherto been made on the formula affected by the sign of the square root; and in this problem, as well as in that last given, there are only a few particular cases that admit of answers in rational numbers; the rest being either impossible, or such as afford y + 128yo = y, or yt and a = 1+ x3 + one or two simple solutions; which may generally be found as follows:* RULE 1.-When the last term e, of the given formula, is a square, put it = f, and make fa + dx + cx2 + x3 + axt d 4cfd2 d4cfo - da (f+ a + -20%) =f? + dx + cx + xx3 at 2f 863 8f4 (4cf2 — da) 64f8 Then, by expunging the first three terms, which are common to each side of the equation, there will remain bxc3 + ((4cfo – d) (40f-da) a24 x4. And consequently, by 8f4 641 dividing by 23, and reducing the result, we shall have x = 64bf – 8df” (4cf2 - do) (4cf" - 2)2 -- 64afe which form fails when the coefficients c and d, or b and d, are each = 0. 2. When the coefficient a, of the first term of the formula, is a square, put it = g, and make gc4 + bx + cx+ dx + e b b(4cg” - b^)* = (g* ++ -)=-g4x4 + 6x3 +ca? + 893 8g4 (4cg2 – 62) + 64g b(4cg? - 12) (4cg2 – 62) Then, då te= x + ; and conse894 64g: (4cgo - 62) - 64ege quently, a which form likewise 64dg6 – 8bgʻ (4cgo – bo) fails under similar circumstances with the former. 3. When the first and last terms of the formula are both squares, put a == g?, and e =-f%, and make fa + dx + cx2 + d d2 + )2 + ac + gx22 2fg 2f" 4f2 dg d2 dg bx3 -) 2 + (2fg + f f And consequently, x = f? (2fg - c) + dd f(bf - dg) Or, because g enters the given formula only in its second 2g * As an instance of what is above said, it may be observed, that the only value of 2 that renders the formula 2x4 322+2 a square, is 1; and the formula x4--22+1 can never be a square, except when x=+1, or -1. 4 X, power, it may be taken either negatively or positively; and 4d2 — f2 (2fg + c) consequently, we shall also have x== f(bf + dg) So that this mode of solution furnishes two different answers. Also, if there be taken for another supposition fa + dæ b bf + 0x2 + bx3 + g2x4 = (f+ x + gx2)2=f+ x + (2 fg 2g 8 62 + + 482) 12 + bx®+gozo, hence, by cancelling, dx + ca? = 5fx b2 g (bg - bf) + (2f8 +407) x*; and consequently, *= 482 + g2 (2fg -c) And because f enters the given formula only in the second power, it may be taken either negatively or positively; and g (bg + bf) consequently, we shall also have x = 112 - g2 (2fg + c) So that this solution likewise furnishes two values for which are each different from the former. But these forms all fail under similar circumstances with those of the second Case. 4. When neither the first nor the last terms are squares, the formula cannot be resolved in any other way than by first endeavouring to discover, by trials, some simple value of the unknown quantity that will answer the conditions of the question, and then finding other values of it according to the methods pointed out in the last two problems. Thus, let р be a value of x so found, and make ap4 + bp3 to cp2 + dp te= 92. Then, by putting x = y +p, we shall have ap4 + bp3 + cp2 + dp te=a(y + p)' + b(y + p)' + c(y + p) + d(y + p) te=ayt + (ap+b) y +(6ap2 + 3b + c) y2 + (4ap3 + 3bp +2cp + d)y + ape + bp3 + cps + dp te, or aw* +683 + cæa to da te ayt + (ap + b)y3 + (6apa + 3b + c) ya + (4aps + 3bps + acp + d) y + 02." From which last formula, the value of y, and consequently that of x, may be found by Case 1. EXAMPLES. 1. It is required to find such a value of ac as will make 1 2:0 + 332 4203 + 5x4 a square. Here, the first term 1, being a square, let 1 2x + 3x24x3 + 5w4 = (1 - 2 + x-2)2 =1 -- 2x + 30c2 2003 + ", agree ably to the method in Case 1. Then we shall have 5x4 4003 = 24 - 2x3. |