3x + 1 problem: Iterative operations neither cause loops nor diverge to innity

The 3x+1 problem is a problem of continuous iteration for integers. According to the basic theorem of arithmetic and the way of iteration, we derive a general formula for continuous iteration for odd integers. Through this formula, we can construct a loop iteration equation and obtain the result of the equation: the equation has only one positive integer solution. In addition, this general formula can be converted into a linear indeterminate equation. The process of solving this equation shows that the relationship between the iteration result and the odd number being iterated is linear. Extending this result to all positive even numbers, we get the answer to the 3 x + 1 question.


Introduction
The 3x + 1 problem is currently an unsolved problem in number theory. According to the description of this problem: take a positive integer, if it is an even number, divide it by 2, and if it is an odd number, multiply it by 3 and add 1 to get a positive integer. Repeat this operation, and then get a string of natural numbers, and finally return 1.
People guess that same result will be obtained for all natural number iterations. [2]− [5] This problem is also known as the Collatz conjecture, Syracuse problem or Ulams problem. Since this issue was raised, no substantial progress has been made in proving this issue. As of 2017, more than 2 64 natural numbers have been verified by the computer.
As a result, all tested integers will eventually return 1 without exception. Until September 2019, Terence Tao proved that this conjecture was almost correct for all integers.
This article, without violating the original intent of the problem, improves the original description of the problem and analyzes the three possible results of successive iterations of odd numbers, thereby solving this problem.

Iterative formula for positive odd integers
It is known that the number of factors equal to 2 contained in a positive integer is constant. Let x be a positive integer, then 3x + 1 is an even. After dividing by all factors equal to 2, the quotient is a positive odd integer. Let x be a positive odd integer, T = 3x + 1 2 β (β is a positive integer). If T is a positive odd integer, then call it an iteration of x. Let T (x) be the iteration result of x, β be the iterative exponent.
Let T (n) (x) be the result of the n-th iteration of x. If T (n) (x) = 1 (x > 1) then x is convergence. Let β n be the iterative exponent of the n-th iteration of x.
Theorem 1. The n-th iteration formula of x is as follows. .

T (n) (x) tends to infinity of possibilities
Let [A] be the integer part of the real number A.
If T (n) (x) tends to infinity when n → ∞, then x is divergent.

See the following indefinite equation
Since A n , B n and C n are both positive integers and (A n , B n ) = 1, there are infinitely many integer solutions to this equation: The following is the process of getting Y 1 and X 1 : [1] (i). For B n > A n , the following equation can be constructed Since (q 1 , q 2 ) = 1, so, there are integers θ and λ such that the following formula holds Through (3.2) and (3.3), we get a special solution of the equation Thus, we obtain For all n > 4, Since q 2 < A n , so, X 1 > C n A n + θ. So, if X 1 is a finite, then C n A n is also a finite. In this case, according to (3.4), B n A n can only be a finite value, as is Y 1 .
(ii). For A n > B n , same as above, the following equation can be constructed (3.5) Through (3.5), we get a special solution of the equation , so, if X 1 is finite, then Y 1 is also finite.
Both cases show that when X is a finite value, Y can only be a finite value.
If both ends of the equal sign are multiplied by 2 s(n) at the same time, then formula (2.1) can be written as  ≤ 1/5, r(1) > 2.
The above formula shows that, only f (1) = 1 is a positive integer solution of this formula.

Refer to the equation below
1. If d = 1, the above function can be written as If , then x is said to be cyclic.
After continuous n iterations for x, if T (n) (x) = x, then, according to formula (2.1), we have .
From the formula above we get Let s(n) = r(n) and x = f (n). Then the above equation is the same as equation (4.1). According to Theorem 3, apart from x = 1, the equation has no positive integer solutions. Therefore, x cannot form a cycle after successive iterations.

Conclusion.
The above analysis shows that all positive integers are neither cyclic nor divergent. Therefore, no matter how many iterations, every odd integer will return 1, that is, all positive odd numbers converge.
Because after dividing by all factors equal to 2, every positive even number is an odd integer, so this conclusion also applies to even numbers. Therefore, all positive integers are Converge.