588 MR TALBOT ON SOME MATHEMATICAL RESEARCHES. 



Investigation of the Theory of the above Method. 



Since x + y + z — 0, one root must have an opposite sign to the other two. 



To fix the ideas, let two roots x, y, be positive, then the third or negative root 

 z is the largest of the three, since it equals the sum of the other two. Let the 

 equation be as before, x 3 + q x — r = 0. We have found that — q — x 1 + xy + y 2 , 

 whence — 4^ = 4 x 2 + 4xy + 4y 2 . Subtract from this 3 s 2 = 3 (x + y) 2 — Sx 2 

 + 6xy + 3 y 2 , and we obtain — 4^ — 3 z 2 — x 2 — 2 xy + y 2 = (x — y) 2 = D 2 (putting 

 D for the difference of the roots). Hence we find, as before (though more directly), 

 that if z be the greatest of the three roots V— 4 q — 3 z 2 = D, which is the* differ- 

 ence of the two smaller and positive roots. But since — 4 q is a positive 

 quantity, let us put it = + Q. In the following lines, I think I shall be clearer, 

 if I neglect the negative sign of s for the present, and speak only of its magnitude ; 

 restoring the negative sign at the end of the inquiry. We have seen that 3 z 2 

 always falls short of Q, by the quantity D 2 . Now, let us inquire what are the 

 conditions necessary, in order that 3 (s + l) 2 may not fall short of Q, but may 

 exceed it? Evidently that circumstance requires that the increment 3(2s + 1) 



D 2 

 should exceed D 2 , or that 2z + 1 should exceed -h-; that is, that 2 (x + y) + 1 



D 2 

 ^> — . Accordingly, this is the definition of " approximate roots," which I 



have proposed in the preceding pages. Hence it follows then, that whenever the 

 roots are approximate, although z is less than — q- (as it always is), yet z + 1 is 



greater than -A that is, z is greater than — ==- — 1. Let —Q be equal to the 



whole number A + the decimal portion d, so that d is some positive quantity 

 less than unity. Hence, we have proved that z is less than A + d, but greater 

 than A — 1 + dj therefore, a fortiori, it is greater than A — 1. But it must be 

 remembered that z is a whole number. How can it be greater than A — 1, yet 

 less than A + d? Evidently in one way only; it must be equal to A. But A is 



the square root of -^-, minus the decimal portion d ; and, therefore, restoring 



the value of Q= — 4§s we have z = J — *# rejecting the decimals, which is 



what we undertook to prove. But since z is the negative root of the equation, we 

 must, of course, give the negative sign to this square root. 



For simplicity, I have hitherto supposed that the cubic equation wants its 

 second term, but if that is not the case, it makes no material difference, because, 

 an equation with integer roots, having its second term, can, with a little trouble, 

 be converted into one equally with integer roots, but wanting its second term. 



The method of solution which I have pointed out is sufficiently simple. I was 



