# Proving that $a_n$ is an integer for every $n$

For every $k\ge1$ integer number if we define the sequence : $a_1,a_2,a_3,...,$ in the form of :$$a_1=2$$

$$a_{n+1}=ka_n+\sqrt{(k^2-1)(a^2_n-4)}$$

For every $n=1,2,3,....$ how to prove that $a_n$ is an integer for every $n$

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$a_{n+1}=ka_n+\sqrt{(k^2-1)(a^2_n-4)}\Rightarrow a_{n+1}-ka_n=\sqrt{(k^2-1)(a^2_n-4)}$

Squaring both sides we have,

$a_{n+1}^2+k^2a_n^2-2ka_{n+1}a_n=k^2a_n^2-4k^2-a_n^2+4$

$\Rightarrow a_{n+1}^2+k^2a_n^2-2ka_{n+1}a_n-k^2a_n^2+4k^2+a_n^2-4=0$

$\Rightarrow a_{n+1}^2-2ka_{n+1}a_n+4k^2+a_n^2-4=0 \dots (1)$

Replacing $n+1$ by n we have similarly,

$\Rightarrow a_{n}^2-2ka_{n}a_{n-1}+4k^2+a_{n-1}^2-4=0 \dots (2)$

By $(1)-(2)$ we have,

$a_{n+1}^2-a_{n-1}^2-2ka_{n}(a_{n+1}-a_{n-1})=0$

$(a_{n+1}-a_{n-1})(a_{n+1}+a_{n-1}-2ka_n)=0$

$\Rightarrow$ either $a_{n+1}=a_{n-1}$ or $a_{n+1}=2ka_n-a_{n-1}$

Now we use induction,

Hypothesis: $\{a_{k}\}_{k=1}^{n}$ are all integers.

As $a_{n+1}=a_{n-1}$ or $a_{n+1}=2ka_n-a_{n-1}$ so $a_{n+1}$ is also integer.

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Hint: Chebyshev polynomials ...

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