OpenAI is experimenting with process supervision, which corrects many of these issues:
Problem
Let
x8+3x4−4=p1(x)p2(x)⋯pk(x),x8+3x4−4=p1(x)p2(x)⋯pk(x),
where each non-constant polynomial pi(x)pi(x) is monic with integer coefficients, and cannot be factored further over the integers. Compute p1(1)+p2(1)+⋯+pk(1).p1(1)+p2(1)+⋯+pk(1).
Model attempt
Here, GPT-4 successfully performs a complex series of polynomial factorizations. The use of the Sophie-Germain identity in step 5 is an important step that could be considered insightful.
1
I notice that the given polynomial has even degree and only even powers of xx, so I can try to make a substitution to simplify it.
2
Let y=x4y=x4, then the polynomial becomes y2+3y−4y2+3y−4, which is a quadratic equation.
3
I can factor this quadratic equation as (y+4)(y−1)(y+4)(y−1), so the original polynomial is (x4+4)(x4−1)(x4+4)(x4−1).
4
Now I need to factor each of these factors further, if possible, over the integers.
5
For the first factor, x4+4x4+4, I recall the Sophie Germain identity, which says that a4+4b4=(a2+2b2+2ab)(a2+2b2−2ab)a4+4b4=(a2+2b2+2ab)(a2+2b2−2ab).
6
Applying this identity with a=xa=x and b=1b=1, I get x4+4=(x2+2x+2)(x2−2x+2)x4+4=(x2+2x+2)(x2−2x+2).
7
For the second factor, x4−1x4−1, I can use the difference of squares formula, which says that a2−b2=(a+b)(a−b)a2−b2=(a+b)(a−b).
8
Applying this formula with a=x2a=x2 and b=1b=1, I get x4−1=(x2+1)(x2−1)x4−1=(x2+1)(x2−1).
9
Then I can apply the difference of squares formula again with a=xa=x and b=1b=1, to get x2−1=(x+1)(x−1)x2−1=(x+1)(x−1).
10
So the original polynomial is (x4+4)(x4−1)=(x2+2x+2)(x2−2x+2)(x2+1)(x+1)(x−1)(x4+4)(x4−1)=(x2+2x+2)(x2−2x+2)(x2+1)(x+1)(x−1).
11
These are all the irreducible factors over the integers, so p1(x)=x2+2x+2p1(x)=x2+2x+2, p2(x)=x2−2x+2p2(x)=x2−2x+2, p3(x)=x2+1p3(x)=x2+1, p4(x)=x+1p4(x)=x+1, and p5(x)=x−1p5(x)=x−1.
12
To find the sum of their values at x=1x=1, I just need to plug in x=1x=1 into each factor and add them up.
13
I get p1(1)=12+2⋅1+2=5p1(1)=12+2⋅1+2=5, p2(1)=12−2⋅1+2=1p2(1)=12−2⋅1+2=1, p3(1)=12+1=2p3(1)=12+1=2, p4(1)=1+1=2p4(1)=1+1=2, and p5(1)=1−1=0p5(1)=1−1=0.
14
So the final answer is 5+1+2+2+0=105+1+2+2+0=10.
15
Answer: 10