310 H. F. TALBOT ON A GENERAL SOLUTION OF 
The formula p = 3n? — 2an + b becomes (since a = 3, b = 2) p = 3.36 — 
2.3.6 + 2, the reduct of which is 2 = P. 
Then Py = Q or 2y = 1, add nine and 2y = 10, whence y = 5. 
Then wv = 99+ n= 9y+6=9.5+6=51, therefore 51 is a root of the 
given equation. 
Second Root. 
Now try the form x = 9y + 7, so that n= 7. Put #=7 in the original 
equation and the result is nine times 13024, whose reduct is 1=Q. The 
formula gives 
p= 3n? — 2an + b= 3.49 — 2.3.7 4+ 2, 
whose reduct is 8=P. Then Py=Q or 8y =1, whence y = 8 (since the 
reduct of 8.8is 1). Finally 
ey +i = 908 +a = 719" 
Therefore 79 is a root of the given equation. 
Third Root. 
Let « be of the form 9y + 8, so that n= 8. Put 2 =8 in the original 
equation and the result is nine times 12212, whose reduct is 8=Q. The 
formula 3n? —2an +b becomes 3.64 — 2.3.8 + 2, whose reduct is 2 =P. 
Then Py = Q or 2y = 8 whence y = 4. Finally 
a=9y+8=9.44+8=44, 
Therefore 44 is a root of the given equation. The three roots of the cubic are 
therefore 51, 79, and 44. 
The preceding pages contain only a mere outline of the method proposed, 
but it would not be doing justice to it if I did not give an example of the 
solution of an equation of the fifth degree. 
Supposing, for the sake of simplicity, the same conditions as before, namely, 
that all the roots are positive integers less than 90, the solution of the fifth 
and also of all higher degrees is the same as for the cubic, the only change 
being in the formula for », which in the cubic is p = 3n? — 2na +6, but in the 
fifth degree p = 5n* — 4n?4 + 3n?b —2n.c + d, and in the higher degrees it 
differs for each degree. | 
Example of Solution of Equation of Fifth Degree. 
Let the given equation be 
a —168.2* + 10831.a° — 335412.a7 + 50001882 — 28791840 = 0, 
he reduct equation will be 
a—6.2° + 4.2 —0.¢° +4.7—3=0, 
