529] 
A “ smith’s PRIZE ” PAPER ; SOLUTIONS. 
415 
let this be (a, b). Then <£ — (f) 1 , changing its sign for the interchange in question, must 
vanish for a = b, that is, <£ — <£ z must contain the factor (a — b); let it contain it 
in the power (a — b) s ; then the quotient <f> — fa -r- (a — b) s , not vanishing for a = b, 
cannot change its sign by the interchange in question, and as by supposition (f>~4>i 
does change its sign, it appears that the exponent s must be odd. But cf> — fa, con 
taining the factor (a — b) s , must contain every other like factor (a — c) s or (c — d) s ; in 
fact, writing <f) — fa = K (a — b) s , then if the interchange (a, c) alters K into K 1} we have 
K (a — b) s = ± Kj (a — c) s . and K consequently contains the factor (a — c) s ; it does not 
contain any higher power (a — c) 9+s ', for if it did, by reversing the process it would 
appear that <f> — <f>i contained (contrary to supposition) the factor (a — b) s+s '. Similarly 
(£-</> i contains the factor (c — d) s , but no higher power (c — d) s+s '. Hence 4> — (f>i con 
tains the product of all the factors (a — b) s , that is, it contains V s , and writing 
<f) — <f) 1 = 2V s M, the quotient M does not contain any such factor as (ci — b)-, it therefore 
does not change its sign for any interchange whatever (a, b); and in consequence it 
remains unaltered for any such interchange, that is, if is a symmetrical function. 
Observing that any even power of V is a symmetrical function, we may without loss 
of generality include F s_1 in the symmetrical function M, and write therefore 
cf) —</>!= 2 VM, 
which is the subsidiary theorem in question. 
any prime exponent p, the function • - may be broken up into two factors each of the 
order \(p — 1), by means of a quadratic equation. 
Let r be any root of the given equation, then 
r 6 + r 5 + r 4 + r 3 + r 2 + r + 1 = 0 ; 
the roots of this equation are r 1 , r 2 , r\ r 4 , r 5 , r f \ 
Hence 
afi + x 5 + x A -}- x 3 + x 2 + x + 1 = (x — r 1 ) (x — r 2 ) (x — r 4 ). (x — r 3 ) (x — r 5 ) (x — r 6 ), 
and denoting the two cubic factors by y 1} y. 2 , or writing 
y 2 = (x — r 1 ) (x — r 2 ) (x — r i ), 
y. 2 =(x — r 3 ) (x — r 5 ) (x — r 6 ), 
we have 
and thence 
2/i + y-2 = 2æ 3 + x 2 — x — 2. 
Hence y v , y. 2 are the roots of the quadratic equation 
y 2 — y (2ot? + x 2 — x — 2) + x 3 + x 3 + ¿r 1 + x 3 + x 2 + x + 1 = 0.