61 
[721 
722] 61 
s of the great 
;overy of the 
on, which, on 
; whereas the 
:r the radical 
functions l/r 
mid, from the 
mod. p), [see 
3 [Rosenhain] 
ons —1 = 0 
an example— 
its difference 
x- + x + 1 =0. 
u 2 + u — 1 = 0 ; 
mbined with 
2{5 + V(5)}]; 
ner to obtain 
to be given 
be consistent 
hat a prime 
denote F (a), 
722. 
A PROBLEM IN PARTITIONS. 
[From the Messenger of Mathematics, vol. vn. (1878), pp. 187, 188.] 
Take for instance 6 letters; a partition into 3’s, such as abc . def contains the 6 
duads ab, ac, be, de, df ef. A partition into 2’s such as ab.cd .ef contains the 3 
duads ab, cd, ef. Hence if there are a partitions into 3’s, and /3 partitions into 2’s, 
and these contain all the duads each once and only once, 6ct + 3/3 = 15, or 2a+ /3=5. 
The solutions of this last equation are (a = 0, /3=5), (a = 1, /3 = 3), (a = 2, /3 = 1), and 
it is at once seen that the first two sets give solutions of the partition problem, but that 
the third set gives no solution; thus we have 
a = 0, /3 = 5 a = 1, /3 = 3 
and 
; the values 
ith this last 
)ur equations 
nt with each 
ab . cd . ef abc . def 
ac .be .df ad .be .cf 
ad . bf. ce ae. bf. cd 
ae . bd. cf of. bd . ce. 
af. be . de 
lis condition 
Similarly for any other number of letters, for instance 15; if we have a partitions 
into 5’s and /3 partitions into 3’s, then, if these contain all the duads, 4a + 2/3 = 14, 
or what is the same 2a + /3 = 7; if a = 0, /3 = 7, the partition problem can be solved (this 
is in fact the problem of the 15 school-girls): but can it be solved for any other values 
(and if so which values) of a, /3 ? Or again for 30 letters; if we have a partitions into 
5’s, /3 partitions into 3’s and y partitions into 2’s; then, if these contain all the duads, 
4a + 2/3 + y = 29; and the question is for what values of a, ¡3, y, does the partition- 
problem admit of solution.