Question of solving the Equation of the Fifth Degree. 543 
In this manner we find, that, under the circumstances sup- 
posed in the enunciation of the theorem, the function 
J (x)—ge 
vanishes, for every value of x which makes the polynome 
x°+D2+E vanish; and since these values have been sup- 
posed unequal, we must have, therefore, 
SJ (@)—qe = (#°+De2+E). (x), ....... (46.) 
the function ($ x) being rational, like f(x), and not containing 
x2°+D2+E as a divisor; which was the thing to be proved. 
Corollary. It is evident that, under the circumstances 
above supposed, the coefficients B! D! E! of (4.) will be ex- 
pressed as follows: 
BS OP Ca ey Ore eee. (47.) 
that is, the equation of the 5th degree in y will be of the form 
0 = 7+Q*Dy+Q°E. ...... vows (48.) 
At the same time the relation between y and «x will reduce 
itself, by (2.) and (7.), to the form 
y = Ucrt(2?+Dx+E). 9 (x), ...... (49.) 
Q still denoting Q+g. If, then, we were to establish this 
. additional supposition 
Dh ei BERT. Sb altiducets 0 GD 
in order to complete the reduction of (4.) to De Moivre’s 
solvible form, we should have 
that is, 
GY ess Pyitracalita oes, otste sewee (52) 
the equation of the fifth degree in y would become 
Gt, rh leahann ch tnighen ah prarmetbanl sks (53.) 
and the relation between y and x would become 
¥y =(P +DrtE).¢ G3 «2-0 (54.) 
and thus, although the equation in y would indeed be easily 
solvible, yet it would entirely fail to give any the least assistance 
towards resolving the proposed equation of the fifth degree 
in 2. 
Observatory, Dublin, May 15, 1836. 
