The Method






In ICF Tertius/Quartus we ended with:

T133 =	(4-3) *	(64-(62+57)+(55+50)-(48+43)+(41+36)-(34+29)+(27+22)-(20+15)+(13+8)-(6+1))
Q133 =	(4+3) *	(64-(62-57)-(55-50)-(48-43)-(41-36)-(34-29)-(27-22)-(20-15)-(13-8)-(6-1))

Now how did we arrive at that?

• First of all Q133 = U67+U66 so the degree of the polynom, one less than the index, is 66.
• Then Q133 = Q(7*19) so te first factor is Q7 = (U4+U3) the degree of which is 3.
  The degree of the second factor therefore is 63, meaning that the terms of it start with U64.
• The second factor is related to the corresponding primefactor of the index which is 19.
  The number of steps of the reduction is: (corresponding primefactor index - 1)/2.
  In this case (19-1)/2 = 9.
  The number of terms of the second factor is the corresponding primefactor itself, which is 19.
• Finally the steplength is:
  2*(highest index of U in the second factor - the corresponding primefactor)/(corresponding primefactor - 1).
  In this case 2*(64-19)/(19-1) = 5.

The tertius only differs in the signs. Typically the sum of the indices in any factor of the quartus is equal to the corresponding primefactor of the index, while the sum of the indices in any factor of the tertius equals 1.

So how difficult can it be to factorize T135 and Q135?
The nice thing about the method is that, provided you've got the previous one's, your only concern is the last primefactor of the index - the other one's are already known.
In this case Q135 = Q3*3*3*5, so the first 3 factors are those of Q3*3*3:

Q27 =

(2+1) *

(4-(2-1)) *

(10-(8-1))

The degree of Q27 is 1+3+9 = 13. Since the degree of Q135 is 67, the degree of the last factor is 67-13 = 54, meaning that it starts with U55.
The number of steps of the reduction equals: (corresponding primefactor index - 1)/2.
In this case (5-1)/2 = 2.
The number of terms equals the corresponding primefactor itself.
The steplength equals:
2*(highest index of U in the corresponding factor - the corresponding primefactor)/(corresponding primefactor - 1).
In this case 2*(55-5)/(5-1) = 25.

So we get:

Q135 =

(2+1) *

(4-(2-1)) *

(10-(8-1)) *

(55-(53-28)-(26-1))

and consequently:

T135 =

(2-1) *

(4-(2+1)) *

(10-(8+1)) *

(55-(53+28)+(26+1))

Easy as fruitpie.