420 solution of (a, b, c, d)=(a 2 , b 2 , c\ d 2 ). [855 
and it hence appears that, writing 7, e, 6 to denote respectively an imaginary cube 
root, fifth root, and seventh root of unity, then the values of (a, b, c, d) are 
1, 
1 , 
1, 
i ; 
7> 
7- 
7 2 > 
7 2 ; 
1, 
1, 
7> 
7 2 ; 
e, 
e 2 , 
e 3 , 
e 4 ; 
67, 
e y, 
e 4 7 , 
e 3 7 2 ; 
e 2 7, 
e 4 7 2 , 
e 3 7 , 
e 7 2 ; 
1, 
e, 
e\ 
1, 
0 3 , 
6\ 
0 5 ‘, 
viz. for each of these systems we have the required relation 
(a, b, c, d) = (a 2 , b 2 , c 2 , d 2 ). 
It may be noticed that out of the eight equations we have the following three 
which are irreducible :— 
x 4 + a? + x 2 + x + 1 =0, 
x A + £(— 1 + i V15) x 3 — 2x 2 + -|(— 1 — i \J15)x + 1=0, 
ic 4 + ^ (— 1 — i V15) x? — 2x 2 + ^ (— 1 + i \Z15)ic+ 1 = 0. 
Each of these is an Abelian equation, viz. the roots are of the form 
a, 6 (a), 6 2 (a), 6 s (a), (= a, a 2 , a 4 , a 8 ), 
where 6 4 (a) = a, not identically but in virtue of the value of a, viz. we have 0 4 (a) = a 16 = a, 
in virtue of a 15 =l: (in the first equation a 5 = 1, and therefore a 15 =l; in each of 
the other two, a 15 is the lowest power which is =1). 
In the first equation, we have evidently 
as the irreducible factor of x 5 — 1. 
The second and third equations combined together give 
(x 4 — \x 3 — 2x 2 — \x + l) 2 + -Ui {a? — x) 2 = 0 ; 
that is, 
&' 8 — od + x 5 — x 4 + x 3 — x + 1 = 0, 
where the left-hand side is the irreducible factor of x 15 — 1.