NUMERICAL EQUATIONS OF ALL DEGREES HAVING INTEGER ROOTS. 311 
so that the coefficients are 
: a=6, b=4, c=0, d=4, e=38, 
and the formula will be 
p= 5n* — 24.n? + 12.0? + 4, 
_ p= dn — 6n® + 3n? + 4, 
whose value depends on the value given to n. 
‘Resuming the reduct equation 
a —6.2* + 4° + 4¢—3=0, 
we find that it is satisfied by each of the digits 1, 2, 3, 4, 5. 
Thus, for instance, the digit 1 gives 1 — 6 + 4 + 4—3, which is plainly = 0. 
The supposition # = 2 gives the same result (remembering that for z* we must 
write 8, for z* 16, that is 7, for 2° 32, that is 5). And so on for the other digits. 
It is therefore probable that the roots of the equation have the form 
Sy+1, 9¥+2, 99+3, 99+ 4, 9Y+5, 
whose reducts would be 1, 2, 3, 4, 5. 
Multiply together the factors 
(x—1) (e—2) (e—3) (w—4) (e@—5) 
and the result is 
xv — 6a + 40° + 4¢—3, 
which agrees with the reduct equation. We may therefore conclude that the 
roots have the forms above specified. 
Let us now calculate the five roots. I will adopt the following notation. 
p, denotes the value which p acquires when the root has the form 9y + 1 
(or has 1 for its reduct). Similarly p, p, p, p, denote its values for the other 
roots, whose 7educts are 2, 3, 4, 5. The formula is 
p = 5n* — 6n? + 3n* + 4, 
and the following little table will be useful :— 
Values of n n? n® nt 
1 1 1 1 
2 4 8 7 
3 0 0 0 
zt Of 1 4 
5 7 8 4 
whence we deduce (putting P as before), for the reduct of p, 
p,=5—64+344=6=P, 
pp=8—34+34+4=3 =P, 
= +4=4=P, 
p,=2—-64+34+4=3=P, 
p,=2—84+34+4=6=P, 
VOL. XXVIf. PART IT. 4M 
