Comment: Paper on the progress of pure mathematics "proof of 3x + 1 conjecture"

The unresolved problem in number theory: the 3x+1 problem, deeply loved by math enthusiasts. I saw a paper titled "Proof of 3x+1 Conjecture" in the Journal of Pure Mathematical Progress (ISSN Print: 2160-0368), and its proof was incorrect.

Introduction

The 3x+1 problem [1,2] is one of the unsolved problems in number theory.

A lot of people have been attracted to solving the problem.

Paper in the Journal of Pure Progress in Mathematics on "Proof of the 3X+1 Conjecture" [3], the proof of it [3] is incorrect.

There are two errors, the first is a correctable error and another is a fatal mistake.

Detailed comments

Modifiable errors

Extraction part (i): See the top section on page 15 [3].

$\text{Proposition 1}:{\text{ 4}}^r\in C_{\text{4}}(r\in Z^{+}),\text{ and its row number }n={\text{ 4}}^{r-\text{1}}-\frac{{\text{4}}^{r-\text{1}}-\text{1}}{3}\cdot$

$\text{Proof}:\because {\text{4}}^r-4=4({\text{4}}^{r-\text{1}}-1)=4[{({\text{2}}^{r-\text{1}})}^2-1]=4({\text{2}}^{r-\text{1}}+1)({\text{2}}^{r-\text{1}}-1)$

$\text{Proof}:\because {\text{4}}^r-4=4({\text{4}}^{r-\text{1}}-1)=4[{({\text{2}}^{r-\text{1}})}^2-1]=4({\text{2}}^{r-\text{1}}+1)({\text{2}}^{r-\text{1}}-1)$

$\therefore \text{3|}\left({\text{2}}^r{}^{-\text{1}}-\text{1}\right)\left({\text{2}}^r{}^{-\text{1}}+\text{1}\right)\cdot$

$\therefore \text{6|4}\left({\text{2}}^r{}^{-\text{1}}-\text{1}\right)\left({\text{2}}^r{}^{-\text{1}}+\text{1}\right)\cdot$

${\text{Let 4}}^r-4=6(n-1)(n\in Z^{+})\cdot$

$\therefore {\text{4}}^r=6(n-1)+4\cdot$

$\therefore {\text{4}}^r\in C_{\text{4}}\cdot \text{ (i)}$

$\because r\in Z^{+}$

$\therefore \text{{3}│\left({\text{2}}^{\text{r-1}}\text{+1}\right)\text{×}\left({\text{2}}^{\text{r-1}}\right)\text{×}\left({\text{2}}^{\text{r-1}}\text{-1}\right)\text{}}$ Inaccurate

$\because \text{r}=1\in \{\text{r}\in {\text{ Z}}^{+}\}$

$\Rightarrow \left(2^{1-1}+1\right)\times \left(2^{1-1}\right)\times \left(2^{1-1}-1\right)\ne \text{An integral multiple of 3}$

$\text{Only}:1＜\text{r}\in {\text{ Z}}^{+}$

$\Rightarrow \{3│\left(2^{\text{r}-1}+1\right)\times \left(2^{\text{r}-1}\right)\times \left(2^{\text{r}-1}-1\right)$

Correction method: setting 1

Non-modifiable fatal errors

Extraction part (ii): See the lower end of page 11 and the upper end of page 12.

$\left(\begin{array}{l}123456\\ 789101112\\ \mathrm{...}\mathrm{...}\mathrm{...}\mathrm{...}\mathrm{...}\mathrm{...}\\ 6n-56n-46n-36n-26n-16n\\ \mathrm{...}\mathrm{...}\mathrm{...}\mathrm{...}\mathrm{...}\mathrm{...}\end{array}\right)\Rightarrow \left(\begin{array}{l}4\\ 10\\ \mathrm{...}\\ 6n-2\\ \mathrm{...}\end{array}\right)\text{ (ii)}$

The fourth column in Figure (ii): 6n-2

{4,10,16,22,28,34,40,46,52,58,64,…,(6n1-2),…}∈(6n-2) (1)

The first line, n=1, (6n-2)=4

The second line, n=2, (6n-2)=10

The third line, n=3, (6n-2)=16

……..

The author takes n as the serial number of each line.

The original author added: (6n-2) = 6(n-1)+4, This is correct

Author's formula: 4^r = 6(n-1)+4, There will be mistakes.

Extraction part (iii): See the top section on page 15.

∴4^r = 6(n-1)+4

∴4^r ∈ C₄

The following mathematical induction proves that row number of 4^r is $4^{r-1}-\frac{4^{r-1}-1}{3}(r\in Z^{+})\cdot$

Proof: 1) $r=1,n=4^{1-1}-\frac{4^{1-1}-1}{3}=1,$ As the conclusion is correct.

2) It is assumed that the conclusion is correct as r=s(s∈Z+, s≥1), that is

$4^s=6(4^{s-1}-\frac{4^{s-1}-1}{3}-1)+4\cdot \text{ (iii)}$

Let's look at n=2. The second line gets: 4^r =6(n-1)+4=10

⇒4^r =10⇒r∉Z+

Conflict with r∈Z+. See: (i).

Let's look at n=3. The second line gets: 4^r = 6(n-1)+4 = 16

⇒ 4^r = 16 ⇒ 2 = r∈Z+

Let's look at n=4. The second line gets: 4^r =6(n-1)+4=22

⇒4^r =22⇒r∉Z+

Conflict with r∉Z+.See: (i).

The truth is:

From formula (1):

{4,10,16,22,28,34,40,46,52.58,64,…,(6n₁-2),…}∈(6n-2)

{4,10,16,22,28,34,40,46,52.58,64,…,(6n₁-2),…}∈(6n-2) ∉4^r.

(6n-2)=6(n-1)+4 {4,16,64,…4ⁿ,…} ∈4^r

Many numbers are missing: {10,22,28,34,40,46,52,58,…}

{10,22,28,34,40,46,52,58,…} ∉4^r.

∴4^r ≠ 6(n-1)+4=6n-2

∴4^r ∉ C₄

When the author [1] chooses n as the serial number and (1

6(n-1)+4=4^r∈ C₄

Get: The author did not prove (3X+1).

Conclusion

If in [3] the author corrects the second error, then [3] the author's method cannot prove (3X+1).

1. Guy RK. Unsolved Problems in Number Theory: The 3x + 1 Problem. Springer Verlag, New York. 2007; 330-336.
2. Lagarias JC. The 3x + 1 Problem and Its Generalizations. Am Math Monthly. 1985; 92:3-23. https://doi.org/10.1080/00029890.1985.11971528.
3. Wang MZ, Yang YB, He ZX, Wang MY. The Proof of the 3X + 1 Conjecture. Adv Pure Math. 2022; 12:10-28. DOI: https://doi.org/10.4236/apm.2022.121002.