CUBIC EQUATIONS. 583 



found b = 13. Hence the equation to be satisfied is 3a 2 + 13k 2 — 160. There 

 are two solutions, viz., a = 7 with k = 1, and a = 6 with k — 2. To decide 

 between them, we take the root — 2a, which is — 14 in the first case, and — 12 

 in the second case; and, substituting these numbers in the original equation, 

 x 3 — IQOx + 504 = 0, we find that — 14 satisfies the equation, while — 12 fails. 

 Therefore the root is 7 4- V 13. Hence we see that there may be & plurality of 

 solutions of the equation 3a 2 + bk 2 = — q ; but they are easily found, when the 

 coefficients of the given equation are of moderate magnitude. 



I think it may be of some interest to add, to these examples of my method, 

 Wood's own example, in illustration of Cardan's rule, taken from his Algebra, 

 6th edition, p. 172. He proposes to find the roots of x 3 +6x — 20 = where 

 ^=6, r=20. Since q is positive, two roots must be imaginary. Computing — 4<f 

 — 27r 2 = <P 2 we find that it equals — 11664. This number = 2 4 . 3 6 , multiplied 

 by — 1. Hence = 2 2 . 3 3/ /— 1, and, consequently, we discover the value of 

 b= — 1. 



The formula 3d 2 + bk 2 = —? becomes 3a 2 — k 2 = — 6\ or 3a 2 + Q = k 2 . It is 

 easy to satisfy this by putting a = =*= 1, # = 3. We find on trial that the negative 

 sign is required. Therefore the roots are — 1 + 3\/— 1, and — 1 — 3^—1, and 2. 

 To verify this, we may observe that the two former are the roots of the quad- 

 ratic x 2 + 2% + 10 = 0, which, being multiplied by x — 2 = 0, gives back the 

 proposed cubic. 



Part II. — Some properties of Cubic Equations whose Roots are whole numbers. 



Let x 3 + qx — r = be the proposed cubic, wanting the second term. I shall 

 suppose q to be negative, because otherwise the equation would have impossible 

 roots. Let the roots x, y, z be whole numbers, either positive or negative. 



Since 



x + y + z 



= 0, 



therefore z = — (x + y) 



Hence 







q, or xy + xz + yz = xy — (x + yf 



or, 







— q=x 2 + xy + y 2 . 



This may be shown in another way. Since x 3 + qx — r = 0, and y 3 + qy — r = 0. 

 . " . by subtraction x 3 — y 3 = — q (x — y), . • . — q = x 2 + xy + y 2 , as before. 

 Theorem 1. — " The coefficient q is either of the form 3n or 3n + 1." 

 Demonstration. — All numbers are of one of the three forms, 3n, 3^+1, or 

 3w + 2. Therefore x — y is of one of those forms. And therefore (x — yf is of 

 one of the two forms 3n or 3n + \. Add 3xy, which makes no difference, since it 

 is a multiple of 3. Therefore, x 2 + xy + y 2 is of the form 3n, or else of the form 

 3w + 1. Therefore the coefficient q is of one of those forms. 



