My Math Forum Proof to Collatz conjecture.

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 May 3rd, 2017, 11:37 PM #171 Banned Camp   Joined: Mar 2017 From: . Posts: 338 Thanks: 8 Math Focus: Number theory Formal proof coming up.
 May 4th, 2017, 06:27 AM #172 Banned Camp   Joined: Mar 2017 From: . Posts: 338 Thanks: 8 Math Focus: Number theory This is a sample of the farthest I have come when trying to simplify the problem. This is for a 1 step cycle. Let $x_n$ be an odd positive integer. From the sequence's formula, $x_{n+1}=\frac{(3x_n+1)}{2^{a_1}}$, $x_{n+2}=\frac{(3x_{n+1}+1)}{2^{a_2}}$, and so forth, $a_i \in \mathbb{Z^+}$. \\ For there to exist a cycle in the sequence, there must exist an odd integer $x_0$ such that $x_n=x_0$.\\ For an odd integer $x_0$, the next odd integer, $x_1$ will be given by \begin{align*} x_1&=\frac{3x_0+1}{2^{a_1}}\tag{1} \end{align*} \begin{lemma} $x_1\geq x_0$ iff $a_1\leq 2$. \end{lemma} \begin{proof} Let \lambda be any 2^{a_1}.We substitute this in equation (1). \begin{align*} x_1&=\frac{3x_0+1}{\lambda}\\ \lambda x_1&=3x_0+1\\ \lambda x_1-3x_0&=1\\ \end{align*} Whena_1>2$,$\lambda >4$,$\lambda x_1-3x_0=1$will hold iff$x_1x_1$.\\$a_1\neq 0$because the sum of$3x_0+1$is an even integer for any$x_0$. \end{proof} \begin{proposition} For$x_0>1$,$x_1\neq x_0$. \end{proposition} \begin{proof} Assume there exists$x_0>1$such that$x_1=x_0.Then, \begin{align*} \frac{3x_0+1}{2^k}&=x_0\\ 2^kx_0-3x_0&=1\\ x_0&=\frac{1}{2^k-3}\\ \intertext{From lemma 1,when we substitutea_1=2$} x_0&=\frac{1}{2^2-3}\\ &=1 \intertext{When$a_1=1} x_0&=\frac{1}{2^1-3}\\ &=-1\\ \end{align*}x_0\ngtr 1$in both cases. Hence, there is no$x_0>1$such that$x_0=x_1$\end{proof} I realised I might need to prove that for$4x_1-3x_0=1$, why the only values that could satisfy the equation is 1 for both$x_1$and$x_0$. Also, the problem could easily end up either involving already discovered deep mathematics, or new ones to come. Last edited by Mariga; May 4th, 2017 at 06:36 AM.  May 4th, 2017, 08:20 AM #173 Banned Camp Joined: Mar 2017 From: . Posts: 338 Thanks: 8 Math Focus: Number theory copy paste that on your latex editor in case it isn't legible. Last edited by Mariga; May 4th, 2017 at 08:23 AM.  May 4th, 2017, 02:31 PM #174 Newbie Joined: Jan 2017 From: Slovakia Posts: 6 Thanks: 0 Seriously?  May 4th, 2017, 09:25 PM #175 Banned Camp Joined: Mar 2017 From: . Posts: 338 Thanks: 8 Math Focus: Number theory What? I have just found a way for cracking that. Not as complicated as I thought it was.  May 10th, 2017, 11:16 PM #176 Banned Camp Joined: Mar 2017 From: . Posts: 338 Thanks: 8 Math Focus: Number theory Erdős once said mathematics is not ready for such problems. It has always been. Last edited by skipjack; May 15th, 2017 at 09:34 AM. May 10th, 2017, 11:39 PM #177 Senior Member Joined: Feb 2016 From: Australia Posts: 1,714 Thanks: 597 Math Focus: Yet to find out. Quote:  Originally Posted by Mariga Erdős once said mathematics is not ready for such problems. It has always been. What has always been? Ready or not ready? Last edited by skipjack; May 15th, 2017 at 09:34 AM.  May 11th, 2017, 08:49 AM #178 Banned Camp Joined: Mar 2017 From: . Posts: 338 Thanks: 8 Math Focus: Number theory Ready. Here is the proof  May 11th, 2017, 09:16 AM #179 Banned Camp Joined: Mar 2017 From: . Posts: 338 Thanks: 8 Math Focus: Number theory Let$x_n$be an odd positive integer. From the sequence’s formula,$x_{n+1}=\frac{3x_n+1}{2^k}$,$x_{n+2}=\frac{3x_{n+1}+1}{2^m}$and so forth,$m,k\in \mathbb{Z^+}$\\ For there to exist a cycle in the sequence, there must exist an odd integer$x_0$such that$x_n=x_0$.\\ Let$x_1$, the next odd integer after$x_0$,be given by $$x_1=\frac{3x_0+1}{2^{a_1}}$$$x_2$will be given by $$x_2=\frac{3x_1+1}{2^{a_2}}$$$x_2$in terms of$x_0will be \begin{align*} x_2&=\frac{3(\frac{3x_0+1}{2^{a_1}})+1}{2^{a_2}}\\ &=\frac{9x_0+3+2^{a_1}}{2^{a_1+a_2}} \end{align*}x_3will be given by \begin{align*} x_3&=\frac{3x_2+1}{2^{a_3}}\\ &=\frac{3(\frac{3x_1+1}{2^{a_2}})+1}{2^{a_3}}\\ &=\frac{3(\frac{9x_0+3+2^{a_1}}{2^{a_1+a_2}})+1}{2 ^{a_3}}\\ &=\frac{27x_0+9+3.2^{a_1}+2^{a_1+a_2}}{2^{a_1+a_2+ a_3}} \end{align*} From the three examples,we can generate a formula forx_n$in terms of$x_0$,which will be $$x_n=\frac{3^nx_0+3^{n-1}+3^{n-2}2^{a_1}+3^{n-3}2^{a_1+a_2}+\cdots +3.2^{\sum\limits_{n=1}^{n-2}a_i}+2^{\sum\limits_{n=1}^{n-1}a_i}}{2^{\sum\limits_{n=1}^na_i}}$$ Let$x_n=x_0$, hence $$x_0=\frac{3^nx_0+3^{n-1}+3^{n-2}2^{a_1}+3^{n-3}2^{a_1+a_2}+\cdots +3.2^{\sum\limits_{n=1}^{n-2}a_i}+2^{\sum\limits_{n=1}^{n-1}a_i}}{2^{\sum\limits_{n=1}^na_i}}$$ Let$\sum\limits_{n=1}^na_i$be$k$$$2^kx_0-3^nx_0=3^{n-1}+3^{n-2}2^{a_1}+3^{n-3}2^{a_1+a_2}+\cdots +3.2^{\sum\limits_{n=1}^{n-2}a_i}+2^{\sum\limits_{n=1}^{n-1}a_i}$$ $$x_0(2^k-3^n)=3^{n-1}+3^{n-2}2^{a_1}+3^{n-3}2^{a_1+a_2}+\cdots +3.2^{\sum\limits_{n=1}^{n-2}a_i}+2^{\sum\limits_{n=1}^{n-1}a_i}$$$2^k$can be expressed as$(2^{\frac{k}{n}})^n$$$x_0[(2^{\frac{k}{n}})^n-3^n]=3^{n-1}+3^{n-2}2^{a_1}+3^{n-3}2^{a_1+a_2}+\cdots +3.2^{\sum\limits_{n=1}^{n-2}a_i}+2^{\sum\limits_{n=1}^{n-1}a_i}\tag{1}$$ When factoring, $$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\cdots +a.b^{n-2}+b^{n-1})$$ This is same as $$a^n-b^n=(a-b)(b^{n-1}+b^{n-2}a+b^{n-3}a^2+\cdots +b.a^{n-2}+a^{n-1})$$ We therefore notice the polynomial in equation (1) is regular. Hence,$x_0[(2^{\frac{k}{n}})^n-3^n]can be expressed as, $$x_0[(2^{\frac{k}{n}})^n-3^n]=x_0(2^{\frac{k}{n}}-3)(3^{n-1}+3^{n-2}2^{\frac{k}{n}}+3^{n-3}(2^{\frac{k}{n}})^2+\cdots +3(2^{\frac{k}{n}})^{n-2}+(2^{\frac{k}{n}})^{n-1})$$ $$=3^{n-1}(2^{\frac{k}{n}}-3)x_0+3^{n-2}2^{\frac{k}{n}}(2^{\frac{k}{n}}-3)x_0+3^{n-3}(2^{\frac{k}{n}})^2(2^{\frac{k}{n}}-3)x_0+\cdots +3(2^{\frac{k}{n}})^{n-2}(2^{\frac{k}{n}}-3)x_0+(2^{\frac{k}{n}})^{n-1})(2^{\frac{k}{n}}-3)x_0$$ \\ Each term in this polynomial corresponds to the terms in the polynomial in equation (1). \begin{align*} 3^{n-1}\quad &:\quad 3^{n-1}(2^{\frac{k}{n}}-3)x_0\\ 3^{n-2}2^{a_1}\quad &:\quad 3^{n-2}2^{\frac{k}{n}}(2^{\frac{k}{n}}-3)x_0\\ 3^{n-3}2^{a_1+a_2}\quad &:\quad 3^{n-3}(2^{\frac{k}{n}})^2(2^{\frac{k}{n}}-3)x_0\\ \vdots\qquad &:\qquad \vdots \\ 3.2^{\sum\limits_{n=1}^{n-2}a_i}\quad &:\quad 3(2^{\frac{k}{n}})^{n-2}(2^{\frac{k}{n}}-3)x_0\\ 2^{\sum\limits_{n=1}^{n-1}a_i}\quad &:\quad (2^{\frac{k}{n}})^{n-1})(2^{\frac{k}{n}}-3)x_0 \end{align*} Hence, each of these corresponding terms will be equal. Hence, \begin{align*} 3^{n-1}&=3^{n-1}(2^{\frac{k}{n}}-3)x_0\\ 1&=(2^{\frac{k}{n}}-3)x_0\\ x_0&=\frac{1}{(2^{\frac{k}{n}}-3)}\tag{2} \end{align*} Substituting this in the next pair of terms, \begin{align*} 3^{n-2}2^{a_1}&=3^{n-2}2^{\frac{k}{n}}(2^{\frac{k}{n}}-3)\frac{1}{(2^{\frac{k}{n}}-3)}\\ 2^{a_1}(2^{\frac{k}{n}}-3)&=2^{\frac{k}{n}}(2^{\frac{k}{n}}-3)\\ 2^{a_1}&=2^{\frac{k}{n}} \end{align*} Substituting this in equation (2), $$x_0=\frac{1}{(2^{a_1}-3)}$$ The only value that satisfies this equation such thatx_0$and$a_1$are both positive integers is$a_1=2$to give$x_0=1\$.
 May 15th, 2017, 08:41 AM #180 Banned Camp   Joined: Mar 2017 From: . Posts: 338 Thanks: 8 Math Focus: Number theory I wouldn't have come this far without your help guys. Thank you.

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