By induction, $x_k = T_k(x_1)$ where $T_n(x)$ is the $n$th Chebyshev polynomial. Since $x_k$ is an integer polynomial in $x_1 = c$, and $c$ is an integer, so is $x_k$.

kerryberry wrote: Consider a sequence of real numbers defined by: \begin{align*} x_{1} & = c \\ x_{n+1} & = cx_{n} + \sqrt{c^{2} - 1}\sqrt{x_{n}^{2} - 1} \quad \text{for all } n \geq 1. \end{align*}Show that if $c$ is a positive integer, then $x_{n}$ is an integer for all $n \geq 1$. (South Africa) If $c=1$, then $x_n=1$ for all $n$. If $c>1$, then $x_{n+1}>x_n$ for all $n$. squaring, we get $E(n) \quad 2cx_nx_{n+1}=x_n^2+x_{n+1}^2+c^2-1$ for all $n$. $E(n+1)-E(n) \implies x_{n+2}=2cx_{n+1}-x_n$, then by induction, $x_n$ is integer for all $n$ ($x_2=2c^2-1$)

Nice solution! With this solution, using the characteristical equation we can find the exact form of the general therm of this sequence!

Let $c = \frac{e^x + e^{-x}}{2}$ for some non-negative real number $x$. Then by induction, $x_n = \frac{e^{nx} + e^{-nx}}{2}$. The easiest way to show that this implies that $x_n$ is always an integer is to note that it gives us the recurrence relation $x_{n + 2} = 2cx_{n + 1} - x_n$. This is how I originally discovered that $x_n$ satisfies the simpler recurrence relation, but as the post before mine shows, it is possible to derive directly. (Yes, this is essentially the Chebyshev polynomial solution.)

We can prove it by induction that if (Xn)²-1=(c²-1)k² such that k is a positive integer then (Xn+1)²-1=(ck+√((c²-1)k²+1))² and we get the result

Medali wrote: We can prove it by induction that if (Xn)²-1=(c²-1)k² such that k is a positive integer then (Xn+1)²-1=(ck+√((c²-1)k²+1))² and we get the result Very good answer Medali,It's like this I did it.

Honestly once you write down the xns for times you will find the rules behind them.

Medali wrote: We can prove it by induction that if $(x_n)^2-1=(c^2-1)k^2$ such that k is a positive integer then $(x_n+1)^2-1=(ck+\sqrt{((c^2-1)(k^2+1))^2}$ and we get the result

kerryberry wrote: Consider a sequence of real numbers defined by: \begin{align*} x_{1} & = c \\ x_{n+1} & = cx_{n} + \sqrt{c^{2} - 1}\sqrt{x_{n}^{2} - 1} \quad \text{for all } n \geq 1. \end{align*}Show that if $c$ is a positive integer, then $x_{n}$ is an integer for all $n \geq 1$. (South Africa)

Attachments:

kerryberry wrote: Consider a sequence of real numbers defined by: \begin{align*} x_{1} & = c \\ x_{n+1} & = cx_{n} + \sqrt{c^{2} - 1}\sqrt{x_{n}^{2} - 1} \quad \text{for all } n \geq 1. \end{align*}Show that if $c$ is a positive integer, then $x_{n}$ is an integer for all $n \geq 1$. (South Africa)

Attachments:

kerryberry wrote: Consider a sequence of real numbers defined by: \begin{align*} x_{1} & = c \\ x_{n+1} & = cx_{n} + \sqrt{c^{2} - 1}\sqrt{x_{n}^{2} - 1} \quad \text{for all } n \geq 1. \end{align*}Show that if $c$ is a positive integer, then $x_{n}$ is an integer for all $n \geq 1$. (South Africa)

Attachments:

kerryberry wrote: Consider a sequence of real numbers defined by: \begin{align*} x_{1} & = c \\ x_{n+1} & = cx_{n} + \sqrt{c^{2} - 1}\sqrt{x_{n}^{2} - 1} \quad \text{for all } n \geq 1. \end{align*}Show that if $c$ is a positive integer, then $x_{n}$ is an integer for all $n \geq 1$. (South Africa) We can prove by induction that $$x_{n+1}=2cx_{n}-x_{n-1}$$ Chebyshev polynomials complete the solution