Peter wrote: The sequence $\{a_{n}\}_{n \ge 1}$ is defined by \[a_{1}=1, \; a_{2}=12, \; a_{3}=20, \; a_{n+3}= 2a_{n+2}+2a_{n+1}-a_{n}.\] Prove that $1+4a_{n}a_{n+1}$ is a square for all $n \in \mathbb{N}$. We define the sequence of positive numbers $b_{n}$ as $b_{1}=1,b_{2}=12$ and $b_{n+1}^{2}+(b_{n}-b_{n-1})^{2}=2b_{n+1}(b_{n}+b_{n-1})+1$ (it is not hard to see that $b_{n+1}$ is defined by $b_{n}$ and $b_{n-1}$ uniquely). Now let's prove that $a_{n}$ and $b_{n}$ are identical. $b_{n+1}^{2}+(b_{n}-b_{n-1})^{2}=2b_{n+1}(b_{n}+b_{n-1})+1$ $\Leftrightarrow$ $(b_{n+1}+b_{n}+b_{n-1})^{2}+1=2(b_{n+1}^{2}+b_{n}^{2}+b_{n-1}^{2})$ $\Rightarrow$ $(b_{n}+b_{n-1}+b_{n-2})^{2}+1=2(b_{n}^{2}+b_{n-1}^{2}+b_{n-2}^{2})$ Subtracting the last two equations, we find that \[(b_{n+1}-b_{n-2})(b_{n+1}+2b_{n}+2b_{n-1}+b_{n-2})=2(b_{n+1}^{2}-b_{n-2}^{2})\] $\Leftrightarrow$ \[b_{n+1}=2b_{n}+2b_{n-1}-b_{n-2}\] Also it is easy to check that $b_{3}=20$,therefore $a_{n}=b_{n}$ for all $n\in{N}$. Consequently, we have that \[a_{n+1}^{2}+(a_{n}-a_{n-1})^{2}=2a_{n+1}(a_{n}+a_{n-1})+1\] $\Leftrightarrow$ \[1+4a_{n}a_{n+1}=(a_{n+1}+a_{n}-a_{n-1})^{2}\] Which means that $1+4a_{n}a_{n+1}$ is always a perfect square!

Peter wrote: The sequence $\{a_{n}\}_{n \ge 1}$ is defined by \[a_{1}=1, \; a_{2}=12, \; a_{3}=20, \; a_{n+3}= 2a_{n+2}+2a_{n+1}-a_{n}.\]Prove that $1+4a_{n}a_{n+1}$ is a square for all $n \in \mathbb{N}$. $$\bf\color{blue}My \ solution$$$\color{green}\textbf{Claim.}$ \begin{align*} &1+4a_na_{n+1}=(a_{n+2}-a_{n+1}-a_n)^2 \end{align*}for all $n \in \mathbb{N}.$ $\color{green}\textbf{Proof.}$ \[1+4a_na_{n+1}=(a_{n+2}-a_{n+1}-a_n)^2\]$\iff$ \[1+4a_na_{n+1}=a^2_{n+2}+(a^2_{n+1}+a^2_n+2a_{n+1}a_n)-2a_{n+2}(a_{n+1}+a_n)\]$\iff$ \begin{align*} &a^2_{n+2}+a^2_{n+1}+a^2_n-2a_{n+2}a_{n+1}-2a_{n+1}a_n-2a_na_{n+2}=1.\ (\bigstar) \end{align*}We will prove $(\bigstar)$ by induction. For $n=1,$ the result is obvious. Now, let us assume that $(\bigstar)$ is true and then prove it for $n+1.$ Or, using mathematical symbols, we just need to prove the following equation: \[a^2_{n+3}+a^2_{n+2}+a^2_{n+1}-2a_{n+3}a_{n+2}-2a_{n+2}a_{n+1}-2a_{n+1}a_{n+3}=1.\]$\iff$ \begin{align*} &a^2_{n+2}+a^2_{n+1}+a^2_n-2a_{n+2}a_{n+1}-2a_{n+1}a_n-2a_na_{n+2}=\\ &=a^2_{n+3}+a^2_{n+2}+a^2_{n+1}-2a_{n+3}a_{n+2}-2a_{n+2}a_{n+1}-2a_{n+1}a_{n+3}\\ \end{align*}$\iff$ \[a^2_n-2a_{n+1}a_n-2a_na_{n+2}=a^2_{n+3}-2a_{n+3}a_{n+2}-2a_{n+1}a_{n+3}\]$\iff$ \[a_n(a_n-2a_{n+1}-2a_{n+2})=a_{n+3}(a_{n+3}-2a_{n+2}-2a_{n+1}).\]But the given condition \[a_{n+3}= 2a_{n+2}+2a_{n+1}-a_{n}\]implies \[a_n-2a_{n+1}-2a_{n+2}=-a_{n+3}\]and \[a_{n+3}-2a_{n+2}-2a_{n+1}=-a_n,\]so \[a^2_n-2a_{n+1}a_n-2a_na_{n+2}=a^2_{n+3}-2a_{n+3}a_{n+2}-2a_{n+1}a_{n+3}\]$\iff$ \[a_n\cdot (-a_{n+3})=a_{n+3}\cdot (-a_n),\]which is true. $\color{red}\textbf{We are done!}$

henderson wrote: Peter wrote: The sequence $\{a_{n}\}_{n \ge 1}$ is defined by \[a_{1}=1, \; a_{2}=12, \; a_{3}=20, \; a_{n+3}= 2a_{n+2}+2a_{n+1}-a_{n}.\]Prove that $1+4a_{n}a_{n+1}$ is a square for all $n \in \mathbb{N}$. $$\bf\color{blue}My \ solution$$$\color{green}\textbf{Claim.}$ \begin{align*} &1+4a_na_{n+1}=(a_{n+2}-a_{n+1}-a_n)^2 \end{align*}for all $n \in \mathbb{N}.$ $\color{green}\textbf{Proof.}$ \[1+4a_na_{n+1}=(a_{n+2}-a_{n+1}-a_n)^2\]$\iff$ \[1+4a_na_{n+1}=a^2_{n+2}+(a^2_{n+1}+a^2_n+2a_{n+1}a_n)-2a_{n+2}(a_{n+1}+a_n)\]$\iff$ \begin{align*} &a^2_{n+2}+a^2_{n+1}+a^2_n-2a_{n+2}a_{n+1}-2a_{n+1}a_n-2a_na_{n+2}=1.\ (\bigstar) \end{align*}We will prove $(\bigstar)$ by induction. For $n=1,$ the result is obvious. Now, let us assume that $(\bigstar)$ is true and then prove it for $n+1.$ Or, using mathematical symbols, we just need to prove the following equation: \[a^2_{n+3}+a^2_{n+2}+a^2_{n+1}-2a_{n+3}a_{n+2}-2a_{n+2}a_{n+1}-2a_{n+1}a_{n+3}=1.\]$\iff$ \begin{align*} &a^2_{n+2}+a^2_{n+1}+a^2_n-2a_{n+2}a_{n+1}-2a_{n+1}a_n-2a_na_{n+2}=\\ &=a^2_{n+3}+a^2_{n+2}+a^2_{n+1}-2a_{n+3}a_{n+2}-2a_{n+2}a_{n+1}-2a_{n+1}a_{n+3}\\ \end{align*}$\iff$ \[a^2_n-2a_{n+1}a_n-2a_na_{n+2}=a^2_{n+3}-2a_{n+3}a_{n+2}-2a_{n+1}a_{n+3}\]$\iff$ \[a_n(a_n-2a_{n+1}-2a_{n+2})=a_{n+3}(a_{n+3}-2a_{n+2}-2a_{n+1}).\]But the given condition \[a_{n+3}= 2a_{n+2}+2a_{n+1}-a_{n}\]implies \[a_n-2a_{n+1}-2a_{n+2}=-a_{n+3}\]and \[a_{n+3}-2a_{n+2}-2a_{n+1}=-a_n,\]so \[a^2_n-2a_{n+1}a_n-2a_na_{n+2}=a^2_{n+3}-2a_{n+3}a_{n+2}-2a_{n+1}a_{n+3}\]$\iff$ \[a_n\cdot (-a_{n+3})=a_{n+3}\cdot (-a_n),\]which is true. $\color{red}\textbf{We are done!}$ Hello How can we solve this problem without using \ (\bigstar) \end{align*}

All is true