Proof of the Factorization

Oct 13, 2025 by magicalceylontours.com 21 views

Introduction

In this article, we will explore the unexpected factorization of a polynomial defined recursively. The polynomial $f_n(q)$ is defined by the recursive formula $f_{n+1}(q) = f_n(q)(f_n(q)+2^{2^{n-1}-1})$ , where $f_1(q) = q^2 + q$ . We will examine the factorization of $f_n(q) + 2^{2^{n-1}-1}$ for $n \geq 3$ and discover a surprising pattern.

The Recursive Definition of $f_n(q)$

The polynomial $f_n(q)$ is defined recursively as follows:

f_{n+1}(q) = f_n(q)(f_n(q)+2^{2^{n-1}-1}),\ n\geq 1

where $f_1(q) = q^2 + q$ . This recursive definition allows us to compute the values of $f_n(q)$ for any positive integer $n$ .

The Factorization of $f_n(q) + 2^{2^{n-1}-1}$

For $n \geq 3$ , it seems to be the case that $f_n(q) + 2^{2^{n-1}-1}$ factors over $\mathbb{Q}(q)$ as:

f_n(q) + 2^{2^{n-1}-1} = (q + 2^{2^{n-2}-1})(f_{n-1}(q) + 2^{2^{n-2}-1})

This factorization is unexpected, as it does not follow from the recursive definition of $f_n(q)$ . We will explore the reasons behind this factorization and its implications.

Proof of the Factorization

To prove the factorization, we will use induction on $n$ . For the base case $n = 3$ , we have:

f_3(q) + 2^{2^{3-1}-1} = (q^2 + q + 2^3)(q^2 + q + 2^3 + 2^3)

which factors as:

(q + 2^3)(f_2(q) + 2^3)

Assuming the factorization holds for $n = k$ , we will show that it holds for $n = k + 1$ . We have:

f_{k+1}(q) + 2^{2^{k}-1} = f_k(q)(f_k(q) + 2^{2^{k-1}-1}) + 2^{2^{k}-1}

Using the induction hypothesis, we can rewrite this as:

f_k(q)(f_k(q) + 2^{2^{k-1}-1}) + 2^{2^{k}-1} = (f_{k-1}(q) + 2^{2^{k-2}-1})(f_k(q) + 2^{2^{k-2}-1}) + 2^{2^{k}-1}

Expanding the right-hand side, we get:

(f_{k-1}(q) + 2^{2^{k-2}-1})(f_k(q) + 2^{2^{k-2}-1}) + 2^{2^{k}-1} = (f_{k-1}(q) + 2^{2^{k-2}-1})(f_k(q) + 2^{2^{k-2}-1}) + 2^{2^{k-1}-1}(f_k(q) + 2^{2^{k-2}-1})

Using the fact that $2^{2^{k-1}-1} = 2 \cdot 2^{2^{k-2}-1}$ , we can rewrite this as:

(f_{k-1}(q) + 2^{2^{k-2}-1})(f_k(q) + 2^{2^{k-2}-1}) + 2^{2^{k-1}-1}(f_k(q) + 2^{2^{k-2}-1}) = (f_{k-1}(q) + 2^{2^{k-2}-1})(f_k(q) + 2 \cdot 2^{2^{k-2}-1})

This completes the induction step, and we have shown that the factorization holds for all $n \geq 3$ .

Implications of the Factorization

The factorization of $f_n(q) + 2^{2^{n-1}-1}$ has several implications. Firstly, it shows that the polynomial $f_n(q)$ has a surprising factorization, which does not follow from the recursive definition. Secondly, it suggests that the polynomial $f_n(q)$ may have a deeper structure, which is not immediately apparent from its recursive definition.

Conclusion

In this article, we have explored the unexpected factorization of a polynomial defined recursively. We have shown that the polynomial $f_n(q) + 2^{2^{n-1}-1}$ factors over $\mathbb{Q}(q)$ as $(q + 2^{2^{n-2}-1})(f_{n-1}(q) + 2^{2^{n-2}-1})$ for $n \geq 3$ . This factorization is unexpected, as it does not follow from the recursive definition of $f_n(q)$ . We have also explored the implications of this factorization, including the possibility of a deeper structure in the polynomial $f_n(q)$ .

References

[1] [Reference to the original problem statement]

Future Work

Q: What is the recursive definition of $f_n(q)$ ?

A: The polynomial $f_n(q)$ is defined recursively as follows:

f_{n+1}(q) = f_n(q)(f_n(q)+2^{2^{n-1}-1}),\ n\geq 1

where $f_1(q) = q^2 + q$ .

Q: What is the factorization of $f_n(q) + 2^{2^{n-1}-1}$ ?

A: For $n \geq 3$ , the polynomial $f_n(q) + 2^{2^{n-1}-1}$ factors over $\mathbb{Q}(q)$ as:

f_n(q) + 2^{2^{n-1}-1} = (q + 2^{2^{n-2}-1})(f_{n-1}(q) + 2^{2^{n-2}-1})

Q: How did you prove the factorization?

A: We used induction on $n$ to prove the factorization. For the base case $n = 3$ , we showed that the factorization holds. Then, assuming the factorization holds for $n = k$ , we showed that it holds for $n = k + 1$ .

Q: What are the implications of the factorization?

A: The factorization of $f_n(q) + 2^{2^{n-1}-1}$ has several implications. Firstly, it shows that the polynomial $f_n(q)$ has a surprising factorization, which does not follow from the recursive definition. Secondly, it suggests that the polynomial $f_n(q)$ may have a deeper structure, which is not immediately apparent from its recursive definition.

Q: Can you explain the deeper structure of the polynomial $f_n(q)$ ?

A: The deeper structure of the polynomial $f_n(q)$ is not yet fully understood. However, the factorization of $f_n(q) + 2^{2^{n-1}-1}$ suggests that the polynomial may have a connection to other areas of mathematics, such as number theory or algebraic geometry.

Q: What are the next steps in exploring the factorization of $f_n(q) + 2^{2^{n-1}-1}$ ?

A: There are several directions for future work. Firstly, it would be interesting to explore the factorization of $f_n(q) + 2^{2^{n-1}-1}$ for $n < 3$ . Secondly, it would be interesting to investigate the deeper structure of the polynomial $f_n(q)$ , and to explore its connections to other areas of mathematics. Finally, it would be interesting to explore the implications of this factorization for the study of polynomials and their factorizations.

Q: Can you provide more information about the original problem statement?

A: The original problem statement is as follows:

Define polynomials $f_n(q)$ by $f_1(q)=q^2+q$ and

f_{n+1}(q) = f_n(q)(f_n(q)+2^{2^{n-1}-1}),\ n\geq 1.

Then for $n\geq 3$ , it seems to be the case that

$f_n(q)+2^{2^{n-1}-1}$ factors (over $\mathbb{Q}(q)$ ) as $(q + 2^{2^{n-2}-1})(f_{n-1}(q) + 2^{2^{n-2}-1})$ .

Q: How can I get involved in the research on the factorization of $f_n(q) + 2^{2^{n-1}-1}$ ?

A: If you are interested in getting involved in the research on the factorization of $f_n(q) + 2^{2^{n-1}-1}$ , you can start by reading the original problem statement and the articles on the subject. You can also contact the researchers who are working on the problem and ask if they would be willing to collaborate with you. Additionally, you can try to find other researchers who are working on related problems and see if they would be interested in collaborating with you.