My Math Forum Proof to Collatz conjecture.

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 December 7th, 2017, 09:50 PM #201 Banned Camp   Joined: Mar 2017 From: . Posts: 338 Thanks: 8 Math Focus: Number theory Another error; this proves that the cycle exists at 1 only for odd integers. It doesn't prove whether a cycle exists for even integers.
 December 17th, 2017, 07:40 AM #202 Banned Camp   Joined: Mar 2017 From: . Posts: 338 Thanks: 8 Math Focus: Number theory Theorem There exists a cycle at $\{1,2,4\}$ only in the Collatz sequence. proof Let $x_0$ be a positive integer from which the Collatz sequence begins.There exists a cycle in the sequence iff there exists an integer $x_n$ such that $x_n=x_0$. Let the next integer after $x_0$ be $$\frac{x_0}{2^{a_0}}$$ such that $a_0$ is zero or a positive integer. Therefore, $\frac{x_0}{2^{a_0}}$ is either even or odd depending on whether $x_0$ is even or odd and the value of $a_0$.If $x_0$ is odd then $2^{a_0}=1$, so $a_0=0$. $x_0$ is even if$2^{a_0}>1$. If $\frac{x_0}{2^{a_0}}$ is odd, then we let the next integer, $x_1$ be $$x_1=\frac{3(\frac{x_0}{2^{a_0}})+1}{2^{a_1}}$$ such that $a_1$ is zero or a positive integer. $x_1$ is even or odd depending on the value of $a_1$. $x_1$ can be phrased as $$x_1=\frac{3x_0+2^{a_0}}{2^{a_0+a_1}}$$ If $x_1$ is odd, then $x_2$ will be \begin{align*} x_2&=\frac{3x_1+1}{2^{a_2}}\\ &=\frac{3(\frac{3x_0+2^{a_0}}{2^{a_0+a_1}})+1}{2^{ a_2}}\\ &=\frac{3^2x_0+3.2^{a_0}+2^{a_0+a_1}}{2^{a_0+a_1+a _2}} \end{align*} $x_3$ will be \begin{align*} x_3&=\frac{3x_2+1}{2^{a_3}}\\ &=\frac{3(\frac{3^2x_0+3.2^{a_0}+2^{a_0+a_1}}{2^{a _0+a_1+a_2}})+1}{2^{a_3}}\\ &=\frac{3^3x_0+3^22^{a_0}+3.2^{a_0+a_1}+2^{a_0+a_1 +a_2}}{2^{a_0+a_1+a_2+a_3}} \end{align*} A pattern emerges from these three examples, which we can use to derive the formula for $x_n$ in terms of $x_0$ $$x_n=\frac{3^nx_0+3^{n-1}2^{a_0}+3^{n-2}2^{a_0+a_1}+\cdots +3^22^{\sum\limits_{i=0}^{n-3}a_i}+3.2^{\sum\limits_{i=0}^{n-2}a_i}+2^{\sum\limits_{i=0}^{n-1}a_i}}{2^{\sum\limits_{i=0}^na_i}}$$ $x_n$ is either an even integer or odd integer depending on the value of $a_n$. Let $x_n=x_0$ and $\sum\limits_{n=0}^na_i=k$ $$x_0=\frac{3^nx_0+3^{n-1}2^{a_0}+3^{n-2}2^{a_0+a_1}+\cdots +3^22^{\sum\limits_{i=0}^{n-3}a_i}+3.2^{\sum\limits_{i=0}^{n-2}a_i}+2^{\sum\limits_{i=0}^{n-1}a_i}}{2^k}$$ $$2^kx_0-3^nx_0=3^{n-1}2^{a_0}+3^{n-2}2^{a_0+a_1}+\cdots +3^22^{\sum\limits_{i=0}^{n-3}a_i}+3.2^{\sum\limits_{i=0}^{n-2}a_i}+2^{\sum\limits_{i=0}^{n-1}a_i}$$ $$x_0(2^k-3^n)=3^{n-1}2^{a_0}+3^{n-2}2^{a_0+a_1}+\cdots +3^22^{\sum\limits_{i=0}^{n-3}a_i}+3.2^{\sum\limits_{i=0}^{n-2}a_i}+2^{\sum\limits_{i=0}^{n-1}a_i}$$ $2^k$ can be phrased as $\{2^{\frac{k}{n}}\}^n$. x_0(\{2^{\frac{k}{n}}\}^n-3^n)=3^{n-1}2^{a_0}+3^{n-2}2^{a_0+a_1}+\cdots +3^22^{\sum\limits_{i=0}^{n-3}a_i}+3.2^{\sum\limits_{i=0}^{n-2}a_i}+2^{\sum\limits_{i=0}^{n-1}a_i} In factoring rules, $$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\cdots +ab^{n-2}+b^{n-1})$$ This is same as $$a^n-b^n=(a-b)(b^{n-1}+b^{n-2}a+b^{n-3}a^2+\cdots +ba^{n-2}+a^{n-1})$$ Therefore the polynomial in equation 1 is regular. We factorise $x_0(\{2^{\frac{k}{n}}\}^n-3^n)$ $$x_0(\!\{\!2^{\frac{k}{n}}\!\}^n\!-3^n)\!=\!x_0\{\!(2^{\frac{k}{n}}\!-3)(3^{n-1}\!+3^{n-2}2^{\frac{k}{n}}\!+3^{n-3}\{\!2^{\frac{k}{n}}\!\}^2+\cdots +3\{\!2^{\frac{k}{n}}\!\}^{n-2}\!+\{\!2^{\frac{k}{n}}\!\}^{n-1})\!\}$$ =\!3^{n-1}\!x_0(2^{\frac{k}{n}}\!-3)+3^{n-2}2^{\frac{k}{n}}\!x_0(2^{\frac{k}{n}}\!-3)+\cdots +3\{\!2^{\frac{k}{n}}\!\}^{n-2}\!x_0(2^{\frac{k}{n}}\!-3)+\{\!2^{\frac{k}{n}}\!\}^{n-1}\!x_0(2^{\frac{k}{n}}\!-3) Thus, the terms in the polynomial in equation 1 and the terms in the polynomial in equation 2 are equivalent. We define an arbitrary term in the polynomial in equation 2 as $$3^{n-f}\{2^{\frac{k}{n}}\}^{f-1}x_0(2^{\frac{k}{n}}-3)$$ such that $f$ is a positive integer. We equate this to the first term in the polynomial in equation 1 \begin{align*} 3^{n-1}2^{a_0}&=3^{n-f}\{2^{\frac{k}{n}}\}^{f-1}x_0(2^{\frac{k}{n}}-3)\\ x_0&=\frac{3^{n-1}2^{a_0}}{3^{n-f}\{2^{\frac{k}{n}}\}^{f-1}(2^{\frac{k}{n}}-3)}\tag{3} \end{align*} We define another arbitrary term in the polynomial in equation 2 as $$3^{n-s}\{2^{\frac{k}{n}}\}^{s-1}x_0(2^{\frac{k}{n}}-3)$$ such that $s$ is a positive integer. We equate this to the second term in the polynomial in equation 1 \begin{align*} 3^{n-2}2^{a_0+a_1}&=3^{n-s}\{2^{\frac{k}{n}}\}^{s-1}x_0(2^{\frac{k}{n}}-3)\\ x_0&=\frac{3^{n-2}2^{a_0+a_1}}{3^{n-s}\{2^{\frac{k}{n}}\}^{s-1}(2^{\frac{k}{n}}-3)} \end{align*} We substitute 3 \begin{align*} \frac{3^{n-1}2^{a_0}}{3^{n-f}\{2^{\frac{k}{n}}\}^{f-1}(2^{\frac{k}{n}}-3)}&=\frac{3^{n-2}2^{a_0+a_1}}{3^{n-s}\{2^{\frac{k}{n}}\}^{s-1}(2^{\frac{k}{n}}-3)}\\ \frac{3^{n-1}}{3^{n-f}\{2^{\frac{k}{n}}\}^{f-1}}&=\frac{3^{n-2}2^{a_1}}{3^{n-s}\{2^{\frac{k}{n}}\}^{s-1}}\\ 2^{a_1}&=\frac{3^{n-1}3^{n-s}\{2^{\frac{k}{n}}\}^{s-1}}{3^{n-2}3^{n-f}\{2^{\frac{k}{n}}\}^{f-1}}\\ 2^{a_1}&=3^{f-s+1}\{2^{\frac{k}{n}}\}^{s-f}\tag{4} \end{align*} Since $a_1$ is an integer, equation 4 will hold iff $3^{f-s+1}=1$. This implies \begin{align*} 3^{f-s+1}&=3^0\\ f-s+1&=0\\ s-f&=1 \end{align*} We substitute this in equation 4 \begin{align*} 2^{a_1}&=\{2^{\frac{k}{n}}\}^1\\ a_1&=\frac{k}{n} \end{align*} We substitute $a_1=\frac{k}{n}$ in equation 3. \begin{align*} x_0&=\frac{3^{n-1}2^{a_0}}{3^{n-f}\{2^{a_1}\}^{f-1}(2^{a_1}-3)}\\ \frac{x_0}{a_0}&=\{\frac{3}{2^{a_1}}\}^{f-1}\frac{1}{2^{a_1}-3} \end{align*} $2^{a_1}=2^{\frac{k}{n}}>3$. Therefore,$\frac{x_0}{a_0}<1$ for $f>1$. This implies \begin{align*} \frac{x_0}{a_0}&=\frac{1}{2^{a_1}-3}\tag{5} \end{align*} The only value of $a_1$ that satisfies the equation such that $\frac{x_0}{a_0}$ is an integer is 2. We define a third arbitrary term in the polynomial in equation 2 as $$3^{n-l}\{2^{\frac{k}{n}}\}^{l-1}x_0(2^{\frac{k}{n}}-3)$$ such that $l$ is a positive integer. We equate this to the last term in the polynomial in equation 1 \begin{align*} 2^{\sum\limits_{i=0}^{n-1}a_i}&=3^{n-l}\{2^{\frac{k}{n}}\}^{l-1}x_0(2^{\frac{k}{n}}-3)\\ x_0&=\frac{2^{\sum\limits_{i=0}^{n-1}a_i}}{3^{n-l}\{2^{\frac{k}{n}}\}^{l-1}(2^{\frac{k}{n}}-3)} \end{align*} We substitute 3 \begin{align*} \frac{3^{n-1}2^{a_0}}{3^{n-f}\{2^{\frac{k}{n}}\}^{f-1}(2^{\frac{k}{n}}-3)}&=\frac{2^{\sum\limits_{i=0}^{n-1}a_i}}{3^{n-l}\{2^{\frac{k}{n}}\}^{l-1}(2^{\frac{k}{n}}-3)}\\ \frac{2^{\sum\limits_{i=0}^{n-1}a_i}}{2^{a_0}}&=\frac{3^{n-1}3^{n-l}\{2^{\frac{k}{n}}\}^{l-1}}{3^{n-f}\{2^{\frac{k}{n}}\}^{f-1}}\\ 2^{\sum\limits_{i=1}^{n-1}a_i}&=3^{n+f-l-1}\{2^{\frac{k}{n}}\}^{l-f}\tag{6} \end{align*} The summation $\sum\limits_{i=1}^{n-1}a_i$ results in an integer, therefore the equation 6 will hold iff $3^{n+f-l-1}=1$. Thus \begin{align*} 3^{n+f-l-1}&=3^0\\ n+f-l-1&=0\\ l-f&=n-1 \end{align*} We substitute this in equation 6 \begin{align*} 2^{\sum\limits_{i=1}^{n-1}a_i}&=\{2^{\frac{k}{n}}\}^{n-1}\\ \sum\limits_{i=1}^{n-1}a_i&=k-\frac{k}{n} \end{align*} $\sum\limits_{n=0}^na_i=k$. Therefore \begin{align*} \sum\limits_{i=1}^{n-1}a_i&=\sum\limits_{n=0}^na_i-\frac{k}{n}\\ \sum\limits_{n=0}^na_i-\sum\limits_{i=1}^{n-1}a_i&=\frac{k}{n}\\ a_0+a_n&=\frac{k}{n}\\ &=a_1=2 \end{align*} We substitute this in 5 $$\frac{x_0}{a_0}=\frac{1}{2^{a_0+a_n}-3}$$ Lemma1 If $x_0$ is an odd integer, $x_n=x_0$ iff $x_0=1$. proof When $x_0$ is an odd integer, $2^{a_0}=1$, therefore $a_0=0$ and $a_n=2$. Thus equation 5 will be $$x_0=\frac{1}{2^2-3}$$ $\therefore x_0=1$. Lemma2 If $x_0$ is an even integer, $x_n=x_0$ iff $x_0=2$ or $x_0=4$. proof When $x_0$ is an even integer, $2^{a_0}>1$, and $a_n$ is zero or a positive integer. Since $a_0+a_n=2$, then either $a_0=1$ and $a_n=1$ or $a_0=2$ and $a_n=0$. If $a_0=1$ and $a_n=1$, \begin{align*} \frac{x_0}{2}&=\frac{1}{2^2-3}\\ x_0&=\frac{2}{2^2-3} \end{align*} $\therefore x_0=2$ If $a_0=2$ and $a_n=0$, \begin{align*} \frac{x_0}{2^2}&=\frac{1}{2^2-3}\\ x_0&=\frac{4}{2^2-3} \end{align*} $\therefore x_0=4$ Thanks from manus Last edited by Mariga; December 17th, 2017 at 07:51 AM.
 December 27th, 2017, 11:34 AM #203 Banned Camp   Joined: Mar 2017 From: . Posts: 338 Thanks: 8 Math Focus: Number theory Full Proof of Collatz conjecture. Let $x_0$ be a positive integer from which the Collatz sequence begins. Let the next integer after $x_0$ be $$\frac{x_0}{2^{a_0}}$$ such that $a_0$ is zero or a positive integer. Therefore, $\frac{x_0}{2^{a_0}}$ is either even or odd depending on whether $x_0$ is even or odd and the value of $a_0$. If $x_0$ is odd then $2^{a_0}=1$, so $a_0=0$. $x_0$ is even if $2^{a_0}>1$, therefore $a_0>0$. If $\frac{x_0}{2^{a_0}}$ is odd, then we let the next integer, $x_1$ be $$x_1=\frac{3(\frac{x_0}{2^{a_0}})+1}{2^{a_1}}$$ such that $a_1$ is zero or a positive integer. $x_1$ is even or odd depending on the value of $a_1$. $x_1$ can be phrased as $$x_1=\frac{3x_0+2^{a_0}}{2^{a_0+a_1}}$$ If $x_1$ is odd, then $x_2$ will be \begin{align*} x_2&=\frac{3x_1+1}{2^{a_2}}\\ &=\frac{3(\frac{3x_0+2^{a_0}}{2^{a_0+a_1}})+1}{2^{ a_2}}\\ &=\frac{3^2x_0+3.2^{a_0}+2^{a_0+a_1}}{2^{a_0+a_1+a _2}} \end{align*} $x_3$ will be \begin{align*} x_3&=\frac{3x_2+1}{2^{a_3}}\\ &=\frac{3(\frac{3^2x_0+3.2^{a_0}+2^{a_0+a_1}}{2^{a _0+a_1+a_2}})+1}{2^{a_3}}\\ &=\frac{3^3x_0+3^22^{a_0}+3.2^{a_0+a_1}+2^{a_0+a_1 +a_2}}{2^{a_0+a_1+a_2+a_3}} \end{align*} A pattern emerges from these three examples, which we can use to derive the formula for $x_n$ in terms of $x_0$ $$x_n=\frac{3^nx_0+3^{n-1}2^{a_0}+3^{n-2}2^{a_0+a_1}+\cdots +3^22^{\sum\limits_{i=0}^{n-3}a_i}+3.2^{\sum\limits_{i=0}^{n-2}a_i}+2^{\sum\limits_{i=0}^{n-1}a_i}}{2^{\sum\limits_{i=0}^na_i}}$$ The Collatz sequence continues indefinitely, thus the equation above represents an $n$th term in a sequence, which exists for any natural number $n$. This therefore is the general pattern for any sequence with undefined length. Theorem For any Collatz sequence, there exists an integer $x_n$ within the sequence such that $x_n$ equals 1 or 2 or 4. proof Let $x_n=x_0+r$ such that $r$ is an integer. Therefore $x_n$ is greater or less than $x_0$ depending on the value of $r$. Let $\sum\limits_{n=0}^na_i=k$ and we substitute these in the equation above. $$x_0+r=\frac{3^nx_0+3^{n-1}2^{a_0}+3^{n-2}2^{a_0+a_1}+\cdots +3^22^{\sum\limits_{i=0}^{n-3}a_i}+3.2^{\sum\limits_{i=0}^{n-2}a_i}+2^{\sum\limits_{i=0}^{n-1}a_i}}{2^k}$$ $$2^k(x_0+r)-3^nx_0=3^{n-1}2^{a_0}+3^{n-2}2^{a_0+a_1}+\cdots +3^22^{\sum\limits_{i=0}^{n-3}a_i}+3.2^{\sum\limits_{i=0}^{n-2}a_i}+2^{\sum\limits_{i=0}^{n-1}a_i}$$ $$x_0(2^k(1+\tfrac{r}{x_0})-3^n)=3^{n-1}2^{a_0}+3^{n-2}2^{a_0+a_1}+\cdots +3^22^{\sum\limits_{i=0}^{n-3}a_i}+3.2^{\sum\limits_{i=0}^{n-2}a_i}+2^{\sum\limits_{i=0}^{n-1}a_i}$$ $2^k(1+\frac{r}{x_0})$ can be phrased as $\{\!\sqrt[n]{2^k(1+\frac{r}{x_0})}\}^n$. Let $t=\sqrt[n]{2^k(1+\frac{r}{x_0})}$. x_0(t^n-3^n)=3^{n-1}2^{a_0}+3^{n-2}2^{a_0+a_1}+\cdots +3^22^{\sum\limits_{i=0}^{n-3}a_i}+3.2^{\sum\limits_{i=0}^{n-2}a_i}+2^{\sum\limits_{i=0}^{n-1}a_i} In factoring rules, $$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\cdots +ab^{n-2}+b^{n-1})$$ This is same as $$a^n-b^n=(a-b)(b^{n-1}+b^{n-2}a+b^{n-3}a^2+\cdots +ba^{n-2}+a^{n-1})$$ Therefore the polynomial in equation 1 is regular. We factorise $x_0(t^n-3^n)$ \begin{align*} x_0(t^n-3^n)&=x_0\{(t-3)(3^{n-1}+3^{n-2}t+3^{n-3}t^2+\cdots +3t^{n-2}+t^{n-1})\} \end{align*} =3^{n-1}x_0(t-3)+3^{n-2}t.x_0(t-3)+\cdots +3t^{n-2}x_0(t-3)+t^{n-1}x_0(t-3) Thus, the terms in the polynomial in equation 1 and the terms in the polynomial in equation 2 are equivalent. We define an arbitrary term in the polynomial in equation 2 as $$3^{n-f}t^{f-1}x_0(t-3)$$ such that $f$ is a positive integer. We equate this to the first term in the polynomial in equation 1 \begin{align*} 3^{n-1}2^{a_0}&=3^{n-f}t^{f-1}x_0(t-3)\\ x_0&=\frac{3^{n-1}2^{a_0}}{3^{n-f}t^{f-1}(t-3)}\tag{3} \end{align*} We define another arbitrary term in the polynomial in equation 2 as $$3^{n-s}t^{s-1}x_0(t-3)$$ such that $s$ is a positive integer. We equate this to the second term in the polynomial in equation 1 \begin{align*} 3^{n-2}2^{a_0+a_1}&=3^{n-s}t^{s-1}x_0(t-3)\\ x_0&=\frac{3^{n-2}2^{a_0+a_1}}{3^{n-s}t^{s-1}(t-3)} \end{align*} We substitute 3 \begin{align*} \frac{3^{n-1}2^{a_0}}{3^{n-f}t^{f-1}(t-3)}&=\frac{3^{n-2}2^{a_0+a_1}}{3^{n-s}t^{s-1}(t-3)}\\ \frac{3^{n-1}}{3^{n-f}t^{f-1}}&=\frac{3^{n-2}2^{a_1}}{3^{n-s}t^{s-1}}\\ 2^{a_1}&=\frac{3^{n-1}3^{n-s}t^{s-1}}{3^{n-2}3^{n-f}t^{f-1}}\\ 2^{a_1}&=3^{f-s+1}t^{s-f}\tag{4} \end{align*} Since $a_1$ is an integer, equation 4 will hold iff $3^{f-s+1}=1$. This implies \begin{align*} 3^{f-s+1}&=3^0\\ f-s+1&=0\\ s-f&=1 \end{align*} We substitute this in equation 4 $$2^{a_1}=t^1$$ We substitute $t=2^{a_1}$ in equation 3. \begin{align*} x_0&=\frac{3^{n-1}2^{a_0}}{3^{n-f}\{2^{a_1}\}^{f-1}(2^{a_1}-3)}\\ \frac{x_0}{2^{a_0}}&=\{\frac{3}{2^{a_1}}\}^{f-1}\frac{1}{2^{a_1}-3} \end{align*} $2^{a_1}=t>3$. Therefore, $\frac{x_0}{2^{a_0}}<1$ for $f>1$. Therefore $f=0$. This implies \begin{align*} \frac{x_0}{2^{a_0}}&=\frac{1}{2^{a_1}-3}\tag{5} \end{align*} The only value of $a_1$ that satisfies the equation such that $\frac{x_0}{2^{a_0}}$ is an integer is 2. Therefore, \begin{align*} 2^{a_1}&=t=2^2\tag{6} \end{align*} We substitute this in equation 5 \begin{align*} \frac{x_0}{2^{a_0}}&=\frac{1}{2^2-3}\\ x_0&=2^{a_0}\tag{7} \end{align*} $t=\sqrt[n]{2^k(1+\frac{r}{x_0})}$ hence substituting 6 gives \begin{align*} 2^2&=\sqrt[n]{2^k(1+\tfrac{r}{x_0})}\\ 2^{2n}&=2^k(1+\tfrac{r}{x_0})\\ 2^{2n-k}&=1+\tfrac{r}{x_0}\tag{8} \end{align*} We define a third arbitrary term in the polynomial in equation 2 as $$3^{n-l}t^{l-1}x_0(t-3)$$ such that $l$ is a positive integer. We equate this to the last term in the polynomial in equation 1 \begin{align*} 2^{\sum\limits_{i=0}^{n-1}a_i}&=3^{n-l}t^{l-1}x_0(t-3)\\ x_0&=\frac{2^{\sum\limits_{i=0}^{n-1}a_i}}{3^{n-l}t^{l-1}(t-3)} \end{align*} We substitute 3 \begin{align*} \frac{3^{n-1}2^{a_0}}{3^{n-f}t^{f-1}(t-3)}&=\frac{2^{\sum\limits_{i=0}^{n-1}a_i}}{3^{n-l}t^{l-1}(t-3)}\\ \frac{2^{\sum\limits_{i=0}^{n-1}a_i}}{2^{a_0}}&=\frac{3^{n-1}3^{n-l}t^{l-1}}{3^{n-f}t^{f-1}}\\ 2^{\sum\limits_{i=1}^{n-1}a_i}&=3^{n+f-l-1}t^{l-f}\tag{9} \end{align*} The summation $\sum\limits_{i=1}^{n-1}a_i$ results in an integer, therefore the equation 9 will hold iff $3^{n+f-l-1}=1$. Thus \begin{align*} 3^{n+f-l-1}&=3^0\\ n+f-l-1&=0\\ l-f&=n-1 \end{align*} We substitute this in equation 9 $$2^{\sum\limits_{i=1}^{n-1}a_i}=t^{n-1}$$ From 6, $t=2^2$ \begin{align*} 2^{\sum\limits_{i=1}^{n-1}a_i}&=2^{2n-2}\\ \sum\limits_{i=1}^{n-1}a_i&=2n-2\\ \sum\limits_{n=0}^na_i-a_0-a_n&=2n-2 \end{align*} $\sum\limits_{n=0}^na_i=k$. Therefore \begin{align*} k-a_0-a_n&=2n-2\\ 2^{2-(a_0+a_n)}&=2^{2n-k} \end{align*} From equation 8, $2^{2n-k}=1+\tfrac{r}{x_0}$ \begin{align*} 2^{2-(a_0+a_n)}&=1+\tfrac{r}{x_0}\\ r&=x_0(2^{2-(a_0+a_n)}-1) \end{align*} We substitute $x_0=2^{a_0}$ from 7 \begin{align*} r&=2^{a_0}(2^{2-(a_0+a_n)}-1)\\ r&=2^{2-(a_0+a_n)+a_0}-2^{a_0}\\ r&=2^{2-a_n}-2^{a_0} \end{align*} $r$ is an integer iff $a_n \ngtr 2$. Therefore $a_n=0$ or $a_n=1$ or $a_n=2$. If $a_n=0$ \begin{align*} r&=2^{2-0}-2^{a_0}\\ r&=2^2-2^{a_0}\\ x_0+r&=2^{a_0}+2^2-2^{a_0}=4 \end{align*} If $a_n=1$ \begin{align*} r&=2^{2-1}-2^{a_0}\\ r&=2-2^{a_0}\\ x_0+r&=2^{a_0}+2-2^{a_0}=2 \end{align*} $a_n=2$ \begin{align*} r&=2^{2-2}-2^{a_0}\\ r&=1-2^{a_0}\\ x_0+r&=2^{a_0}+1-2^{a_0}=1 \end{align*} Therefore $x_0+r=x_n$ is equal to 1 or 2 or 4. Last edited by skipjack; December 27th, 2017 at 02:19 PM.
 January 3rd, 2018, 12:59 AM #204 Banned Camp   Joined: Mar 2017 From: . Posts: 338 Thanks: 8 Math Focus: Number theory There is one statement that makes this proof wrong. That the polynomial in equation 1 is regular. That's incorrect.
 March 6th, 2018, 11:49 PM #205 Banned Camp   Joined: Mar 2017 From: . Posts: 338 Thanks: 8 Math Focus: Number theory Mathematics was not ready for this problem. It is now. Coming up soon
 March 10th, 2018, 10:02 PM #206 Banned Camp   Joined: Mar 2017 From: . Posts: 338 Thanks: 8 Math Focus: Number theory There is a relationship between collatz and goldbach conjecture...
 March 17th, 2018, 11:39 AM #207 Banned Camp   Joined: Mar 2017 From: . Posts: 338 Thanks: 8 Math Focus: Number theory Indeed this problem is a little bit difficult. Needs thorough understanding since there are very many ideas coming up from its data analyses to a point where it may almost seem random. However, almost every piece of idea you may have can be expressed mathematically, therefore so far mathematics can solve any problem in mathematics. Still coming soon in less than a week.
 March 20th, 2018, 09:32 PM #208 Banned Camp   Joined: Apr 2016 From: Australia Posts: 244 Thanks: 29 Math Focus: horses,cash me outside how bow dah, trash doves Make an account with arxiv. Learn how to write a paper in the formal manner. Submit your proof there. If it gets rejected, then you could bring it back here to ask for assistance. Yes people can try and plagiarize other people's work it is relatively easy to express something using equivalent expressions and go under the radar that way, but at the end of the day if you respect the proof of the person that plagiarized your result then really who gives a shit. Well I don't personally anyway. But it looks like you have put a lot of effort in here. I'm too tired and going to need to sleep for a week, but if the problem is very difficult and you know it is open, don't bother sharing it here or anywhere that isn't going to keep a record of your correspondence. Alan Baker died recently. Fucking Genius. Heard of him? probably not. Yep see its the specialist's specialization mathematics, and problem with that of course is nobody cares about other people's fields unless it concerns their work in the present time. Anyway I have a friend that has studied the Collatz quite a lot if you would like someone to collaborate with, pm me and ill give you a crazy person's contact details. Last edited by Adam Ledger; March 20th, 2018 at 09:41 PM.

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