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The Collatz Conjecture and Linear Indefinite Equation

Li Jiang

Beijing, China

Abstract: For the collatz conjecture, we define an iterative formula of odd integers according to the basic theorem of arithmetic, and give the concept of iterative exponent. On this basis, a continuous iterative general formula for odd numbers is derived. With the formula, the equation of cyclic iteration is deduced and get the result of the equation without a positive integer solution except 1. On the other hand, the general formula can be converted to linear indefinite equation. The solution process of this equation reveals that odd numbers are impossible to tend to infinity through iterative operations. Extending the result to even numbers, it can be determined that all positive integers can return 1 by a limited number iterations.

Keywords: Collatz conjecture, 3x + 1 problem, Syracuse problem, iteration

Received June 09, 2020; Revised July 10, 2020; Accepted July 19, 2020

Citation: Li Jiang, The Collatz Conjecture and Linear Indefinite Equation. Turkish Journal of Analysis and Number Theory, 8 (2). 49-51. January 2020. doi:10.12691/tjant-8-2-5

Mathematics Subject Classification (2000): 11T30, 26A18

1. Introduction

The Collatz Conjecture is an unsolved problem in number theory at present. Its summary is that to take a integer greater than 1, if it is even, divide it by 2, if it is odd, multiply it by 3 and plus 1. With such a transformation, we get a new positive integer. Repeating such transformations (iterations) will produce a string of natural numbers and eventually get 1.

As of 2017, the computer verified all natural numbers less than 2⁶⁴. Without exception, all test values greater than 1 eventually return to 1.

People speculate that whether all natural numbers have such a property [1,2,3,4].

The conjecture has also been called the 3x+1 Conjecture, Kakutanis problem, the Syracuse problem, and Ulams problem.

This paper gives the equivalent definition of the problem based on the fundamental theorem of arithmetic, and analyzes the problem from an odd number point of view.

All alphanumeric characters shown below are positive integers unless otherwise noted.

2. Iteration of Positive Odd Number

According to the fundamental theorem of arithmetic, the number of prime factors with value equal to 2 in each positive integer is unique. If x is a positive odd number then 3x + 1 is an even number greater than 2. Divide by all factors which equals 2 in 3x + 1, the quotient is an odd number not less than 1.

Definition 1.

Let be a positive odd number and (α is a positive integer). If is an odd number not less than 1, then the operation is referred to as an iteration of . Denote as the iteration result of , α as the iterative exponent.

Definition 2. If the iteration result of is still , then is called the origin.

Theorem 1. Of all the odd numbers, only 1 is the origin.

Proof. Let odd number be the origin. By the definition 1, we have

[picture] | (2.1)

Since is a positive integer, the solution of formula (2.1) is as follows

[picture] | (2.2)

According to the definition 1, can only be a positive odd number. Among the above three results, since only conforms to the definition, only 1 is the origin.

Let denote the result of the -th iteration of .

If i.e. after a limited number iterations for , the end result is to return to the origin, then is called convergence.

Let be the iterative exponent of the i-th iteration of .

Let

Theorem 2. The -th iteration formula of is as follows.

[picture] | (2.3)

Proof. For . Since , so, according to (2.3), we have

[picture] | (2.4)

If (2.3) holds for , then for we have

3. Possibility of

If , i.e. get itself again through continuous iterations, then is called cyclic.

Theorem 3. The condition that the following equation has a positive integer solution is .

[picture] | (3.1)

(where: is integer, )

Proof. For . It is the same as equation (2.2), so substituting into this equation, we have the positive integer solution .

If (3.1) holds for , then for , the equation can be written as

It is known that and , let ( is a positive integer). We can refer to the following equation

[picture] | (3.2)

1. If , the above function can be written as

and

It shows that, for all and is monotonically decreasing. The calculation result shows that and are not integers, .

So, in this case, cannot be a positive integer.

2. If , substituting it into (3.2), then we get an identity that ;

3. If then, for all , .

Summarizing the above analysis, the theorem is true.

If the result of successive iterations of can constitute a loop, i.e. , then according to theorem 2, we have

From the formula above we get

[picture] | (3.3)

Let and . According to theorem 3, equation (3.3) has no other positive integer solutions except for . Therefore, continuous iterations of is impossible to constitute a loop.

4. Possibility of

If , i.e. can be iterated infinitely and the result tends to infinity, then is called divergent.

Let [A] be the integer part of the real number A.

Theorem 4. When , if is finite, then cannot be infinite.

Proof. Formula (2.3) can be written as follows

[picture] | (4.1)

Let

See the following linear equation

[picture] | (4.2)

Since and are both positive integers and , this equation has an infinite number of integer solutions:

[picture] | (4.3)

The process of determining and is as follows [5]:

a. If , we can construct the following equations

[picture] | (4.4)

Since , so, there are integers and such that the following formula holds

[picture] | (4.5)

According to (4.4) and (4.5), a special solution for the equation is

[picture] | (4.6)

and

[picture] | (4.7)

Thus, we obtain

[picture] | (4.8)

For all ,

[picture] | (4.9)

Since , so, . When is a finite value, then is also a finite value. In this case, according to (4.9), can only be a finite value. It means that if is finite, then is also finite.

b. If , we can construct the following equations in the same way as above

[picture] | (4.10)

According to (4.10), a special solution for the equation is

[picture] | (4.11)

Since , so, if is finite, then cannot be infinite. It means that if is finite, then cannot be infinite.

Let and . Thus, the equation (4.1) is equal to (4.2). The above analysis shows that when , if is finite, then cannot be infinite.

5. Conclusion

Analysis of the above two sections shows that all positive odd numbers are neither divergent nor cyclic. So, for an odd integer greater than 1, there is a large integer such that, for all , . After consecutive iterations, we can get odd numbers less than . Since loops do not occur during successive iterations, the values of those integers greater than 1 are different from each other, and the number of these integers is less than . According to theorem 1, the iteration result for 1 is still 1. Thus, the other iteration results are all 1, and the final iteration result can only be 1. So, all positive odd numbers are convergent.

Since each positive even number can be converted to an odd number after divided by factors equal to 2, the above conclusion also applies to even numbers. So all positive integers are convergent.

Acknowledgements

I sincerely thank Prof. Dr. Valentin Blomer of Georg-August-Universität Göttingen for his help in this research.

References

[1]   | J C Lagarias, The 3x+1 problem and its generalizations. Amer. Math. Monthly, 1985(92): 3-23.
[2]   | I Krasikov, J C Lagarias. (2003). Bounds for the 3x+1 problem using difference inequalities. Acta Arithmetica, 109(3): 237-258.
[3]   | Dengguo FENG, Xiubin FAN, Liping DING, Zhangyi WANG. On the nonexistence of non-trivial small cycles of the function in 3x+1 conjecture. Journal of Systems Science and Complexity, 2012, 25: 1215-1222.
[4]   | A Tomas. A non-uniform distribution property of most orbits, in case the 3x + 1 conjecture is true. Journal of Physics A Mathematical and General, 2016, 30(13): 4537-4562.
[5]   | Changfeng Gao. Application of Number Theory Method in Solving Binary Indefinite Equations. China. Journal of Jinan Vocation College, 2009(72): 89-91.