Show that the sequence $$\{a_n\}$$ is composed of terms of integers if $$a_0=1$$ and $$a_{n+1}=\dfrac{3a_n+\sqrt{5a^2_n-4} }{2} $$

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Show that the sequence $$\{a_n\}$$ is composed of terms of integers if $$a_0=1$$ and $$a_{n+1}=\dfrac{3a_n+\sqrt{5a^2_n-4}  }{2}  $$

Uniteasy Admin · Accepted Answer

$$(2a_{n+1}-3a_n)^2 =(\sqrt{5a^2_n-4})^2$$
$$4a^2_{n+1}-6a_na_{n+1}+9a^2_n = 5a^2_n-4$$
$$4a^2_{n+1} -6a_na_{n+1} +4a^2_n+4=0$$n
 $$4a^2_{n-1} -6a_na_{n-1} +4a^2_n+4=0$$n
It's found that $$a_{n+1}$$ and $$a_{n-1}$$ are the two roots of the following quadratic equation
$$4x^2-6a_nx +4a^2_n+4 = 0$$
Using Vieta's formulas gives
$$a_{n+1}+a_{n-1} = 6a_n $$n
Given that
$$a_0=1$$,
$$a_2 = \dfrac{3a_1+\sqrt{5a^2_1-4} }{2} = 2$$
Using the equation (3), it can be deduced that all other terms of the sequence are also integers by using the method of mathematical induction.

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