582 MR TALBOT ON SOME MATHEMATICAL RESEARCHES. 



has two integer roots it must have three, else the coefficients would not be whole 

 numbers. Supposing, then, that a cubic equation has one integer root, and no 

 more, then the two other roots must be of the form a ± k*/b, where a, b, k, are 

 integers, and \/b is a surd in its lowest terms. I have shown how the roots can 

 be found when k = 1, and I am now inquiring whether they can be found when 

 k has other values, because that would amount to a general solution of this class 

 (which I have called the second class) of cubics. The example which I have just 

 given shows that it can be readily effected in certain cases, but how far is the 

 method general? 



Example 2. Let x z — 975# — 9972 = 0. In this equation |- comes out 



= 3 . 17 . 181 \/3. Whence b = 3 and the roots are seen to be of the form 



a ± k\Z"6. Hence, k being a factor of 2, must equal either 3, 17, or 181, unless 



it be the product of two of them, as 3, 17 = 51. But we have also 3 a 2 + bk 2 

 — — q, or 3 a 2 + Sk 2 — 975 ; whence a 2 + k 2 . = 325. Hence it is plain that k 

 cannot be so large a number as 181, or even as 51 ; therefore it must be either 

 3 or 17. If we try 3, we get a 2 = 325 — 9 = 316, which is not a square. We 

 must therefore have k = 17. This gives on trial a 2 = 325 — 289 = 36; whence 

 a = 6. The roots of the equation are therefore 6 + 17 \/3, 6 — 17 \/3, and — 12. 



It will be observed that in this instance the problem of solving the cubic is 

 converted into that of finding two squares such that their sum a 2 + k 2 may 

 = 325. Of course they are easily found. The general case gives 3a 2 + bk 2 

 = — q, where b and q are known integers. This is easily solved by trial when 

 the coefficient q is not too large. This, therefore, is an indeterminate problem 

 of the second degree. If its solution is regarded as within the domain of 

 ordinary arithmetic, the solution of cubic equations of the second class must be 

 considered as effected. At any rate, the problem is transformed into a very dif- 

 ferent one. It will be observed that unless we had found the value of \/b from 

 the properties of the function (p, we could not have effected the last-mentioned 

 transformation ; for b would have remained unknown. We come to the con- 

 clusion that this method reduces the solution of the general cubic (of the second 

 class) to the solution of the indeterminate problem 3a 2 + bk 2 = — q, where b and 

 q are given. 



Example 3. Resuming the former example 6 in p. 7 of this memoir # 3 — 1456 

 x — 456 = 0, in which we found b = 373, the equation to be solved is 3a 2 + 373k 2 

 = 1456. But here every value of k, even k — 2, is too large. The only remaining 



factor of |- is unity. Therefore we must have k = 1. The value k = 1 gives 



3a 2 = 1456 — 373 = 1083, which gives a = 19, as we found before. 



Example 4. Take the former example 7, z 3 — I60x + 504 = 0, in which we 



