[485 
485] 
PROBLEMS AND SOLUTIONS. 
577 
3 form For instance, n = 4 ; partitions of (a, b, c, d) into two parts are (a, bed), (b, eda),, 
(c, dab), (d, abc), (ab, cd), (ac, db), (ad, be) ; no. is = 7. Partitions into three parts are 
(ab, c, d), (ac, b, d), (ad, b, c), (be, a, d), (b, d, ac), (cd, a, b) ; no. is = 6. Partition into 
4 parts is (a, b, c, d) ; no. is = 1. And we have 
1-1.7 + 2.6-6.1 = 13-13 = 0. 
iscriminant 
equation,) 
g equal to 
may take 
> 0 ; and we 
0, U' = 0 
e equation 
a) (S' - a') 
S'- a'= 0 
n UU' = 0 
at infinity 
= 0, q' = 0 
1 therefore 
the curve 
, the line 
point at 
Solution by the Proposer. 
Write n= aa + 6/3 + cy +..., where a, /3, 7... are positive integers all of them 
different, and a, /3, 7... are positive integers; and consider the partitions wherein we 
have a parts each of a things, b parts each of ¡3 things, &c. Writing as usual 
II (n) =1.2.3 ...n, the number of partitions of the form in question is 
Tin 
~Ua.llb...(Ua) a (U/3) b ... ’ 
whence, putting for shortness a + /3 + ... = p, the theorem may be written 
2(-r 
II (p — 1) Tin 
lia. 115 ... (IIa) № (II/3) 6 .. 
= 0, 
the summation extending to all the partitions n = aa + b/3 + ... , as explained above. 
Now if the n quantities x, y, z,... are the wth roots of unity, we have x + y + z ... =0, 
and therefore also (x + y + z...) n =0, and the general term of the left-hand is 
(na) a (n^) 6 ...^ №/36 '“^ 
where [a a /3 6 ...] denotes the symmetrical function 1 t x a y a ...(a factors) uPv&... (b factors)... 
of the roots x, y, z, u, y...of the equation 6 71 —1 = 0; where, as above, n= aa + b/3 + ... . 
Now by a formula not, I believe, generally known, but which is given on p. 175 of 
the translation of Hirsch’s Algebra (Hirsch’s Collection of Examples &c. on the Literal 
Calculus and Algebra, translated by the Rev. J. A. Ross, London, 1827), the value of 
the sum in question is = (— n > where p = a + b+ ..., (the sign +, given 
in the formula as quoted, is at once seen to be (—)^ _1 ); whence, substituting and 
omitting the factor n, we have 
2(-)* 
II (p — 1) Tin 
Tla. Tib ... (IIa) a (II/3) 6 
= 0, 
which is the required theorem. 
Observation. In Cauchy’s Exercices d’Analyse &c., t. 111., p. 173, is given a 
formula relating to the same mode of partition of the number n, viz. this is 
2 
Tin 
lia. 116 ... a a /3 b ... 
= Tin. 
,rts) = 0. 
C. VII. 
73