My Math Forum Proof to Collatz conjecture.

 Number Theory Number Theory Math Forum

 May 22nd, 2017, 11:18 PM #191 Banned Camp   Joined: Mar 2017 From: . Posts: 338 Thanks: 8 Math Focus: Number theory Alright then, I will admit I still don't understand your point if my reply doesn't sufficiently answer you. Let's say we have the following polynomial $$(a^pr)^n-(br) ^n=b^{n-1}+b^{n-2}a^{x_1}+b^{n-3}a^{x_2}+\cdots +ba^{x_{n-2}}+a^{x_{n-1}}$$ Then the polynomial must be regular. Otherwise the equation wouldn't hold. we can factor $$(a^pr)^n-(br)^n=(a^pr-br)(br^{n-1}+br^{n-2}a^p+br^{n-3}a^{2p}+\cdots +bra^{p(n-2)}+a^{p(n-1)}$$ to make the two polynomi(als equivalent, then multiply each term by $(a^pr-br)$ so that $$(a^pr)^n-(br)^n=br^{n-1}(a^pr-br)+br^{n-2}a^p(a^pr-br)+br^{n-3}a^{2p}(a^pr-br)+\cdots +bra^{p(n-2)}(a^pr-br)+a^{p(n-1)}(a^pr-br)$$ since two polynomials are regular and have equal number of terms, then the first terms in both are equal, so are the rest corresponding terms. we wont mind if $(a^pr-br)$ is $b^w$. The fact is, $br^{n-1}=br^{n-1}(a^pr-br)$ Also, if $(a^pr-br)=b^w$ then the $(a^pr)^n-(br)^n$ would not be as it is but would be $(a^pr)^{n+w}-(br)^{n+w}$. That's the only way it would hold.
 May 22nd, 2017, 11:31 PM #192 Member   Joined: Jul 2014 From: israel Posts: 76 Thanks: 3 the BIG different is that when you are saying "polynomial" $$b^{n-1}+b^{n-2}a^{x_1}+b^{n-3}a^{x_2}+\cdots +ba^{x_{n-2}}+a^{x_{n-1}}$$ there is no other $$b^{n-1},b^{n-2},b^{n-3} \cdots$$ this is given to you as a fact! after it is a given then you can use what you wanted to do make sure that $$a^{x_1},a^{x_2}, \cdots +a^{x_{n-2}},a^{x_{n-1}}$$ all are different from b first! then you can use it
 May 23rd, 2017, 12:24 AM #193 Banned Camp   Joined: Mar 2017 From: . Posts: 338 Thanks: 8 Math Focus: Number theory Now I get your point. I would give you a thank you but I still don't know how they are given. I will just write it. Thank you.
 May 23rd, 2017, 05:01 AM #194 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 $x_0$ will 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*} $x_3$ will 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 for $x_n$ in terms of $x_0$, which will be $$x_n=\frac{3^nx_0+3^{n-1}+3^{n-2}2^{a_1}+3^{n-3}2^{a_1+a_2}+\cdots +3.2^{\sum\limits_{n=1}^{n-2}a_i}+2^{\sum\limits_{n=1}^{n-1}a_i}}{2^{\sum\limits_{n=1}^na_i}}$$ Theorem If $x_n=x_0$, then $x_n=1$. Proof 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 $k$ $$2^kx_0-3^nx_0=3^{n-1}+3^{n-2}2^{a_1}+3^{n-3}2^{a_1+a_2}+\cdots +3.2^{\sum\limits_{n=1}^{n-2}a_i}+2^{\sum\limits_{n=1}^{n-1}a_i}$$ $2^k$ can be expressed as $(2^{\frac{k}{n}})^n$ and $x_0$ as $(\sqrt[n]{x_0})^n$ $$\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^n-\{3\!\sqrt[n]{x_0}\}^n=3^{n-1}+3^{n-2}2^{a_1}+3^{n-3}2^{a_1+a_2}+\cdots +3.2^{\sum\limits_{n=1}^{n-2}a_i}+2^{\sum\limits_{n=1}^{n-1}a_i}\tag{1}$$ When factoring, $$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\cdots +ab^{n-2}+b^{n-1})$$ This is same as $$a^n-b^n=(a-b)(b^{n-1}+b^{n-2}a+b^{n-3}a^2+\cdots +ba^{n-2}+a^{n-1})$$ We therefore notice equation (1) holds iff the polynomial is regular. We factorise $\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^n-\{3\!\sqrt[n]{x_0}\}^n$. \begin{align*} \{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^n-\{3\!\sqrt[n]{x_0}\}^n&=(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})\ (\{3\!\sqrt[n]{x_0}\}^{n-1}+\{3\!\sqrt[n]{x_0}\}^{n-2}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}\\ &\quad+\{3\!\sqrt[n]{x_0}\}^{n-3}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^2+\cdots +\{3\!\sqrt[n]{x_0}\}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^{n-2}\\ &\quad+\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^{n-1}) \\ \{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^n-\{3\!\sqrt[n]{x_0}\}^n&=\{3\!\sqrt[n]{x_0}\}^{n-1}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})\\ &\quad+\{3\!\sqrt[n]{x_0}\}^{n-2}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})\\ &\quad+\{3\!\sqrt[n]{x_0}\}^{n-3}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^2(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})+\cdots\\ &\quad+\{3\!\sqrt[n]{x_0}\}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^{n-2}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})\\ &\quad+\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^{n-1}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0}))\tag{2} \end{align*} Thus, each term in the polynomial in equation (1) is equal to each corresponding term in the polynomial in equation (2), for example the first term in polynomial (1) is equal to either the first or last term in polynomial (2). The second term in polynomial (1) is equal to either the second or the second last term in polynomial (2) and so forth. If the first term in polynomial (1) equals the last term in polynomial (2), then \begin{align*} 3^{n-1}&=\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^{n-1}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})\\ \frac{3^{n-1}}{(2^{\frac{k}{n}})^{n-1}}&=(\sqrt[n]{x_0})^{n-1}\sqrt[n]{x_0}(2^{\frac{k}{n}}-3)\\ (\frac{3}{2^{\frac{k}{n}}})^{n-1}&=x_0(2^{\frac{k}{n}}-3)\\ x_0&=(\frac{3}{2^{\frac{k}{n}}})^{n-1}\frac{1}{(2^{\frac{k}{n}}-3)} \end{align*} But $2^{\frac{k}{n}}>3$. Hence, $x_0<1$ in this case. If the first term in polynomial (1) equals the first term in polynomial (2), then \begin{align*} 3^{n-1}&=\{3\!\sqrt[n]{x_0}\}^{n-1}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})\\ 1&=(\sqrt[n]{x_0})^{n-1}\sqrt[n]{x_0}(2^{\frac{k}{n}}-3)\\ x_0&=\frac{1}{(2^{\frac{k}{n}}-3)}\tag{3} \end{align*} The next pair of equal terms, \begin{align*} 3^{n-2}2^{a_1}&=\{3\!\sqrt[n]{x_0}\}^{n-2}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})\\ 2^{a_1}&=\sqrt[n]{x_0}^{n-2}\sqrt[n]{x_0}\sqrt[n]{x_0}2^{\frac{k}{n}}(2^{\frac{k}{n}}-3)\\ 2^{a_1}&=x_02^{\frac{k}{n}}(2^{\frac{k}{n}}-3) \end{align*} Substituting (3) \begin{align*} 2^{a_1}&=\frac{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 $2^{a_1}$ in equation (3), $$x_0=\frac{1}{(2^{a_1}-3)}$$ The only value that satisfies this equation such that $x_0$ and $a_1$ are both positive integers is $a_1=2$ to give $x_0=1$.
May 24th, 2017, 02:35 AM   #195
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Quote:
 Originally Posted by Mariga Thus, each term in the polynomial in equation (1) is equal to each corresponding term in the polynomial in equation (2), for example the first term in polynomial (1) is equal to either the first or last term in polynomial (2). The second term in polynomial (1) is equal to either the second or the second last term in polynomial (2) and so forth. If the first term in polynomial (1) equals the last term in polynomial (2)
It's the same mistake.

You need to check all of them, not 2.

You have infinitely many checks to do.

Last edited by skipjack; July 25th, 2017 at 04:32 AM.

 May 28th, 2017, 10:00 AM #196 Banned Camp   Joined: Mar 2017 From: . Posts: 338 Thanks: 8 Math Focus: Number theory Indeed. That step also didn't seem quite right to me too. It didn't make the proof completely rigorous. But I have a way around it. coming soon
 May 29th, 2017, 01:55 AM #197 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 $x_0$ will 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*} $x_3$ will 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 for $x_n$ in terms of $x_0$, which will be $$x_n=\frac{3^nx_0+3^{n-1}+3^{n-2}2^{a_1}+3^{n-3}2^{a_1+a_2}+\cdots +3.2^{\sum\limits_{n=1}^{n-2}a_i}+2^{\sum\limits_{n=1}^{n-1}a_i}}{2^{\sum\limits_{n=1}^na_i}}$$ Theorem If $x_n=x_0$, then $x_n=1$. Proof 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 $k$ $$2^kx_0-3^nx_0=3^{n-1}+3^{n-2}2^{a_1}+3^{n-3}2^{a_1+a_2}+\cdots +3.2^{\sum\limits_{n=1}^{n-2}a_i}+2^{\sum\limits_{n=1}^{n-1}a_i}$$ $2^k$ can be expressed as $(2^{\frac{k}{n}})^n$ and $x_0$ as $(\sqrt[n]{x_0})^n$ $$\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^n-\{3\!\sqrt[n]{x_0}\}^n=3^{n-1}+3^{n-2}2^{a_1}+3^{n-3}2^{a_1+a_2}+\cdots +3.2^{\sum\limits_{n=1}^{n-2}a_i}+2^{\sum\limits_{n=1}^{n-1}a_i}\tag{1}$$ When factoring, $$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\cdots +ab^{n-2}+b^{n-1})$$ This is same as $$a^n-b^n=(a-b)(b^{n-1}+b^{n-2}a+b^{n-3}a^2+\cdots +ba^{n-2}+a^{n-1})$$ We therefore notice equation (1) holds iff the polynomial is regular. We factorise $\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^n-\{3\!\sqrt[n]{x_0}\}^n$. \begin{align*} \{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^n-\{3\!\sqrt[n]{x_0}\}^n&=(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})\ (\{3\!\sqrt[n]{x_0}\}^{n-1}+\{3\!\sqrt[n]{x_0}\}^{n-2}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}\\ &\quad+\{3\!\sqrt[n]{x_0}\}^{n-3}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^2+\cdots +\{3\!\sqrt[n]{x_0}\}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^{n-2}\\ &\quad+\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^{n-1}) \\ \{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^n-\{3\!\sqrt[n]{x_0}\}^n&=\{3\!\sqrt[n]{x_0}\}^{n-1}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})\\ &\quad+\{3\!\sqrt[n]{x_0}\}^{n-2}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})\\ &\quad+\{3\!\sqrt[n]{x_0}\}^{n-3}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^2(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})+\cdots\\ &\quad+\{3\!\sqrt[n]{x_0}\}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^{n-2}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})\\ &\quad+\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^{n-1}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0}))\tag{2} \end{align*} Thus, the number of terms in the two polynomials is equal. The first term in the polynomial in equation (1) is equal to any term in the polynomial in equation (2). Each term in polynomial 2 can be defined by the following, $$3\!\sqrt[n]{x_0}\}^{n-r}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^{r-1}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})$$ Where $r=1,2,3,...,n$ Hence, \begin{align*} 3^{n-1}&=3\!\sqrt[n]{x_0}\}^{n-r}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^{r-1}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})\\ 3^{n-1}&=3^{n-r}\{2^{\frac{k}{n}}\}^{r-1}x_0(2^{\frac{k}{n}}-3)\\ x_0&=\frac{3^{n-1}}{3^{n-r}\{2^{\frac{k}{n}}\}^{r-1}(2^{\frac{k}{n}}-3)}\\ x_0&=\{\frac{3}{2^{\frac{k}{n}}}\}^{r-1}\frac{1}{(2^{\frac{k}{n}}-3)} \end{align*} But $2^{\frac{k}{n}}>3$. Hence, $x_0<1$ for $r>1$. When $r=1$, \begin{align*} 3^{n-1}&=\{3\!\sqrt[n]{x_0}\}^{n-1}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})\\ 1&=(\sqrt[n]{x_0})^{n-1}\sqrt[n]{x_0}(2^{\frac{k}{n}}-3)\\ x_0&=\frac{1}{(2^{\frac{k}{n}}-3)}\tag{3} \end{align*} The next term in the polynomial in equation (1), is equal to any term in polynomial in equation (2), except $\{3\!\sqrt[n]{x_0}\}^{n-1}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})$. Therefore, $$3^{n-2}2^{a_1}=\{3\!\sqrt[n]{x_0}\}^{n-r}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^{r-1}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})$$ Such that $r=2,3,4,...,n$ \begin{align*} 3^{n-2}2^{a_1}&=3^{n-r}\{2^{\frac{k}{n}}\}^{r-1}x_0(2^{\frac{k}{n}}-3)\\ 2^{a_1}&=\frac{3^{n-r}\{2^{\frac{k}{n}}\}^{r-1}x_0(2^{\frac{k}{n}}-3)}{3^{n-2}} \end{align*} We substitute (3) \begin{align*} 2^{a_1}&=\frac{3^{n-r}\{2^{\frac{k}{n}}\}^{r-1}\frac{1}{(2^{\frac{k}{n}}-3)}(2^{\frac{k}{n}}-3)}{3^{n-2}}\\ 2^{a_1}&=3^{2-r}\{2^{\frac{k}{n}}\}^{r-1}\\ \end{align*} But $a_1$ is an integer. Therefore $r=2$ to eliminate $3^{2-r}$. Hence, $$2^{a_1}=2^{\frac{k}{n}}$$ Substituting $2^{a_1}$ in equation (3), $$x_0=\frac{1}{(2^{a_1}-3)}$$ The only value that satisfies this equation such that $x_0$ and $a_1$ are both positive integers is $a_1=2$ to give $x_0=1$.
 July 25th, 2017, 01:13 AM #198 Banned Camp   Joined: Mar 2017 From: . Posts: 338 Thanks: 8 Math Focus: Number theory I have found an error.. The statement $2^{\frac{k}{n}}>3$. This implies $x_0<1$ for $r>1$. is incorrect. $2^{\frac{k}{n}}$ can be greater than 3 but less than 4. Therefore $x_0$ can be greater than 1 for $r>1$.
 July 25th, 2017, 09:24 PM #199 Banned Camp   Joined: Mar 2017 From: . Posts: 338 Thanks: 8 Math Focus: Number theory Also, I will soon give the complete algebraic proof of collatz conjecture that doesn't rely on a probabilistic model.
 November 4th, 2017, 11:34 PM #200 Banned Camp   Joined: Mar 2017 From: . Posts: 338 Thanks: 8 Math Focus: Number theory So, this is what I have as the proof that there exists only one Cycle in the Collatz sequence. I haven't worked out on the rest of the proof but maybe some time in future. 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 $x_0$ will 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*} $x_3$ will 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 for $x_n$ in terms of $x_0$, which will be $$x_n=\frac{3^nx_0+3^{n-1}+3^{n-2}2^{a_1}+3^{n-3}2^{a_1+a_2}+\cdots +3.2^{\sum\limits_{n=1}^{n-2}a_i}+2^{\sum\limits_{n=1}^{n-1}a_i}}{2^{\sum\limits_{n=1}^na_i}}$$ Theorem If $x_n=x_0$, then $x_n=1$. proof 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 $k$ $$2^kx_0-3^nx_0=3^{n-1}+3^{n-2}2^{a_1}+3^{n-3}2^{a_1+a_2}+\cdots +3.2^{\sum\limits_{n=1}^{n-2}a_i}+2^{\sum\limits_{n=1}^{n-1}a_i}$$ $2^k$ can be expressed as $(2^{\frac{k}{n}})^n$ and $x_0$ as $(\sqrt[n]{x_0})^n$ \{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^n-\{3\!\sqrt[n]{x_0}\}^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} When factoring, $$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\cdots +ab^{n-2}+b^{n-1})$$ 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 equation 1 holds iff the polynomial is regular. We factorise $\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^n-\{3\!\sqrt[n]{x_0}\}^n$. \begin{align*} \{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^n-\{3\!\sqrt[n]{x_0}\}^n&=(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})\ (\{3\!\sqrt[n]{x_0}\}^{n-1}+\{3\!\sqrt[n]{x_0}\}^{n-2}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}\\ &\quad+\{3\!\sqrt[n]{x_0}\}^{n-3}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^2+\cdots +\{3\!\sqrt[n]{x_0}\}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^{n-2}\\ &\quad+\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^{n-1}) \\ \{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^n-\{3\!\sqrt[n]{x_0}\}^n&=\{3\!\sqrt[n]{x_0}\}^{n-1}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})\\ &\quad+\{3\!\sqrt[n]{x_0}\}^{n-2}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})\\ &\quad+\{3\!\sqrt[n]{x_0}\}^{n-3}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^2(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})+\cdots\\ &\quad+\{3\!\sqrt[n]{x_0}\}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^{n-2}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})\\ &\quad+\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^{n-1}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})\tag{2} \end{align*} Thus, the polynomial in equation 2 is same as the polynomial in equation 1 such that the first term in the polynomial in equation 1 is equal to any term in the polynomial in equation 2. We define a term in the polynomial in equation 2 as $$\{3\!\sqrt[n]{x_0}\}^{n-r}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^{r-1}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})$$ Such that $r$ is a positive integer. We equate this to the first term in the polynomial in equation 1. Hence, \begin{align*} 3^{n-1}&=3\!\sqrt[n]{x_0}\}^{n-r}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^{r-1}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})\\ 3^{n-1}&=3^{n-r}\{2^{\frac{k}{n}}\}^{r-1}x_0(2^{\frac{k}{n}}-3)\\ x_0&=\frac{3^{n-1}}{3^{n-r}\{2^{\frac{k}{n}}\}^{r-1}(2^{\frac{k}{n}}-3)}\tag{3} \end{align*} We define a different term in the polynomial in equation 2 as $$\{3\!\sqrt[n]{x_0}\}^{n-w}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^{w-1}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})$$ Such that $w$ is a positive integer. We equate this to the next term in the polynomial in equation 1. \begin{align*} 3^{n-2}2^{a_1}&=\{3\!\sqrt[n]{x_0}\}^{n-w}\{2^{\frac{k}{n}}\!\sqrt[n]{x_0}\}^{w-1}(2^{\frac{k}{n}}\!\sqrt[n]{x_0}-3\!\sqrt[n]{x_0})\\ 3^{n-2}2^{a_1}&=3^{n-w}\{2^{\frac{k}{n}}\}^{w-1}x_0(2^{\frac{k}{n}}-3)\\ x_0&=\frac{3^{n-2}2^{a_1}}{3^{n-w}\{2^{\frac{k}{n}}\}^{w-1}(2^{\frac{k}{n}}-3)} \end{align*} We substitute 3 \begin{align*} \frac{3^{n-1}}{3^{n-r}\{2^{\frac{k}{n}}\}^{r-1}(2^{\frac{k}{n}}-3)}&=\frac{3^{n-2}2^{a_1}}{3^{n-w}\{2^{\frac{k}{n}}\}^{w-1}(2^{\frac{k}{n}}-3)}\\ \frac{3^{n-1}}{3^{n-r}\{2^{\frac{k}{n}}\}^{r-1}}&=\frac{3^{n-2}2^{a_1}}{3^{n-w}\{2^{\frac{k}{n}}\}^{w-1}}\\ 2^{a_1}&=\frac{3^{n-1}3^{n-w}\{2^{\frac{k}{n}}\}^{w-1}}{3^{n-r}3^{n-2}\{2^{\frac{k}{n}}\}^{r-1}}\\ 2^{a_1}&=3^{r-w+1}\{2^{\frac{k}{n}}\}^{w-r}\tag{4} \end{align*} $a_1$ is an integer. This implies the equation will hold iff $3^{r-w+1}=1$. \begin{align*} 3^{r-w+1}&=3^0\\ r-w+1&=0\\ r&=w-1 \end{align*} We substitute this in equation 4. \begin{align*} 2^{a_1}&=3^{(w-1)-w+1}\{2^{\frac{k}{n}}\}^{w-(w-1)}\\ 2^{a_1}&=2^{\frac{k}{n}} \end{align*} Substituting $2^{a_1}$ in equation 3, \begin{align*} x_0&=\frac{3^{n-1}}{3^{n-r}\{2^{a_1}\}^{r-1}(2^{a_1}-3)}\\ x_0&=\{\frac{3}{2^{a_1}}\}^{r-1}\frac{1}{2^{a_1}-3} \end{align*} $2^{a_1}=2^{\frac{k}{n}}$ and $2^{\frac{k}{n}}>3$. This implies $x_0<1$ for $r>1$. Therefore $$x_0=\frac{1}{2^{a_1}-3}$$ The only value that satisfies this equation such that $x_0$ and $a_1$ are both positive integers is $a_1=2$ to give $x_0=1$. Last edited by Mariga; November 4th, 2017 at 11:40 PM.

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