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.
It's an impressive result, but shouldn't be seen as "correction". Framing it as a (drastic) reduction in mistakes is more useful here.
If the model is productionized (read: dumbed down so it isn't as expensive to run), the reasoning abilities drastically decline again.
And these reasoning abilities are still around a language model, rather than around abstract models.
This is a very effective party trick for general math, whose language quite directly maps onto these abstract concepts, but there are some holes. Information about e.g. which values may be zero isn't encoded in the language, and so this approach is liable to blundering around division-by-zero issues.
If you want a particular example to toy around with, LLMs are not fond of quaternions and their conversion to other representations.
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