0: 
co 
Or 
NUMERICAL EQUATIONS OF ALL DEGREES HAVING INTEGER ROOTS. 
8 
0 
} 
> 
& 
8 
24 
2 
SmONAMP WHY 8s 
Com k | NAW] Re 
STOeK| wor! aor 
Ora | BRI] We S 
SCaorFl]l ON] aoe & 
Coe] | ee] ee 
Cant! oY] we & 
This means that if the reduct of any number is (for example) 4, that of its fifth 
power will be 7. Vuce versd, if a fifth power is given, whose reduct is 7, that 
of its root can only be 4; in other words, the root will be some number of the 
form 9n + 4. But if the fifth power have 0 for its reduct (or be divisible by 9) 
then the root may have any one of the three forms— 
9n, 9n+3, 9n+6. 
The above table is formed in this way, and may be continued to any extent. 
Suppose we have calculated the first four columns, and want the column headed 
x. Multiply together x and «*, and we get x’, thus: 
H oo bo 
mone 
IOmH 
OIDM 
HEITOR 
Or OD 
Or else, multiply together the columns 2’ and 2’, thus: 
No ke 
KH OoOre 
Io or 
aH Ee oO aT 
Oro @ 
Ore © bw 
being the same result as before. 
Let us now calculate the reduct of a complex polynomial, such as, for 
example, 
a —15 a + 85 #— 225 a? + 274 «4 —120. 
First, replace the coefficients by their reducts: this gives 
ao — 6x + 4¢°—0.a° + 4¢—38. 
And the result of this depends upon the value which we assign to w. First, 
| then, let = 1, the reduct will be 
1—6+4+4-—3=6. 
Wextletz=2. The table gives, if = 2, the reduct of 2° =5, ofa*=7, of #°=8. 
Therefore substituting these values we have 
5—6.7+4.84+4.2—3 
=5—6 +5 +8 —3=0. 
Next let 2 = 3, then all the superior powers of a give 0, so that the polynomial 
