588 
PEOEESSOE SYLYESTEE ON THE EEAL 
and reasoning as in the preceding case for n= 4 (with the sole difference, that if &N 
vanishes by virtue of de—cf vanishing, we should have P=0, N = 0, and the criterion- 
series — +00, which at once indicates the existence of four imaginary roots), we see 
that there remains only to consider the case where the criterion-series takes the form 
OH — f* 0. It is scarcely necessary to observe that all the criteria can never vanish 
simultaneously; for that would indicate the equality of all the roots in the transformed, 
and therefore in the given equation, whose own criteria, contrary to hypothesis, would 
also be all zero. The zero values of the two extreme criteria indicates that the three first 
and the three last literal parts of the coefficients are in geometrical progression, from 
which it will immediately be seen that the equation to be considered may be thrown (by 
substituting in lieu of x and y suitable multiples of x and y, which will not affect the 
characters of the criteria) into the convenient form 
x 5 + 5s x 4 y + 1 OeVy 2 + 10 qVy 3 + 5 v\xy 4 +/=0, 
with the two conditions s 4 — zrf positive, rf — m 3 positive. 
The form of the criterion-series, apocopated from either end, shows that two of the 
roots must be imaginary ; and consequently, in order to establish the existence of two 
imaginary pairs of roots, it is only necessary to show that the discriminant of the above 
equation, subject to the above conditions, must remain always positive. That discrimi- 
nant I proceed to determine; but as a guide to the form under which it is to be 
expressed, the following observation is important. Let us take the more general form 
ax 5 + bx 4 y + cx 3 y 2 + dx 2 y 3 + exy 4 - \-fy 5 = 0 , 
where 
a=l, b=Xs, c—(jij s 2 , d=(Aif, e=\?i, f— 1, 
X, ft being any numerical quantities. 
The discriminant will evidently be a symmetrical function of e and e. 
Let a p b q c r d s e t be the literal part of any term in the discriminant. By the law of weight 
we must have 
q-\-2r-\-3s~\-4t=5 X 4=20. 
But in the equation before us, a p b q c r d s e t (to a numerical factor pres ) is s ?+2 V s+< , and 
(g+2r)—(2s+t)=(g+2r+3s+4t)—5(s+t) 
=5(4— s+t). 
Hence the difference between the indices of e and y in each term is a multiple of 5, 
and consequently, since the discriminant is a symmetrical function in e and q, it will be 
a rational integral function of s 5 H~^ 5 and zr\. Moreover, as no such term as c 4 d 4 can figure 
in the discriminant, which, as we know, must in all cases contain one or the other of the 
two final and of the two initial coefficients, we see that no term can be of higher than 
the 14th degree in s, n, nor yet so high, for the only terms that could be of that degree 
would be bc 3 d 3 e ; but making a and /‘each zero in the original form, it becomes obvious 
