172 PROFESSOR M. J. M. HILL ON THE LOCUS OF SINGULAR POINTS 
Therefore, 
00 _ 
dx 
/ 9R , -n 30 
Now the substitutions x — y = y,z=£ make 0 = 0, 0 = 0, dxp/dx = 0. 
Therefore they make 
R 
00 
dx 
= 0 . 
Now all the values of x, ■y , z which make 0=0 cannot make d(f>/dx = 0 . Hence 
x — y = y, z — £ must make R vanish. 
Therefore R is divisible by 0 without remainder. 
Therefore 0 is divisible by 0 3 without remainder. 
To prove (iii.) proceed by induction. 
Suppose that the theorem is true for a given value of m. viz., that if x = £, y = y, 
z—t, make 0 = 0 , dxfj/dx = 0 , . . . d m ~ l 0/ dx m ~ l = 0 , then 0 contains 0 "' as a factor. 
Let it now be given that x = £, y — y, z = £ also make d m \p/dx m = 0 . 
Then by the assumption 
0 = p.0™, 
where p is some rational integral function of x, y, z. 
Therefore 
0™0 _ 0™0"' 0p 0™“ 1 0™ 
0a;™ ^ dx m m dx dx m ~ 1 
0™ 0™ 
= p + fo- 
where y is some rational integral function of x, y, z. 
Now 
d m 0™ 
hr™ 
= m ! 
+ 0 -°‘> 
where o- is some rational integral function of x, y, z. 
Therefore 
I } = m '-p^£) +*o> <r +x)- 
Hence the substitutions x — y = y, z — l, make 
but they do not make d(f)/dx = 0. 
