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If $$f(n) = (3+\sqrt{5})^n + (3-\sqrt{5})^n$$ show that $f(n)$ is an integer and that $$f(n+1)= 6f(n) - 4f(n-1).$$ Deduce that the next integer greater than $(3+\sqrt{5})^n$ is divisible by $2^n.$ I am not getting a way to approach such a question. So, please help me regarding that...

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    $\begingroup$ Have you tried it for small $n$? This screams for an induction proof. The point is that the $\sqrt 5$'s cancel out. I recommend Aurturo Magidin's answer to this question for a good intro $\endgroup$ – Ross Millikan Apr 20 at 2:00
  • $\begingroup$ Please use MathJax to format your posts. You'll get a lot more help if your questions are easy to read. $\endgroup$ – saulspatz Apr 20 at 3:15
  • $\begingroup$ It is a lot easier to see that $f(n)$ is an integer once you have established the given formula for $f(n+1)$ :)z $\endgroup$ – Erick Wong Apr 20 at 5:18
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Proof for showing $f(n) = (3+\sqrt{5})^n + (3-\sqrt{5})^n$ is an integer :

By binomial expansion we see that :

$$(3+\sqrt{5})^n = C_0 + C_1 \cdot 3^{n - 1} \cdot 5 ^{1/2} + C_2 \cdot 3^{n - 2} \cdot 5^{2/2} + \dots + C_n \cdot 5^{n/2}$$

$$(3-\sqrt{5})^n = C_0 - C_1 \cdot 3^{n - 1} \cdot 5 ^{1/2} + C_2 \cdot 3^{n - 2} \cdot 5^{2/2} + \dots + (-1)^r \cdot 3^{n - r} \cdot 5^{r/2} + \dots + (-1)^n \cdot 5^{n/2}$$

Adding these two equations we notice that for $n$ that is odd the respective terms cancel out in the above equations. So only the terms for which $n$ is even are present in $(3+\sqrt{5})^n + (3-\sqrt{5})^n$ and thus $f(n)$ is an integer.

Here is an induction based proof.

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