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Let $S_{n}=\left (3+\sqrt{5}\right)^{n}+\left(3-\sqrt{5}\right)^{n}$then, by mathematical induction, show that $S_{n}$ is an integer. Also, prove that the next integer greater than $\left(3+\sqrt{5}\right)^{n}$ is divisible by $2^{n}$. Note: please do not use binomial theorem to prove the first part.

Akshay Hegde
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    For the first part, use recursion for two variable symmetric polynomials. In simpler terms, just express $S_n$ in terms of $S_{n-1},S_{n-2}\cdots$. For the next part, think about what you need to add to $(3+\sqrt{5})^n$ to get the next integer. – rah4927 Jul 02 '15 at 07:00
  • Thanks... Got the first part since $S_{n+2}=6 S_{n+1} -4 S_{n}$ – Akshay Hegde Jul 02 '15 at 07:03
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    Note that the reason other people do not like your question and want to close it is because you did not mention what you've tried. The only reason I gave a solution sketch to your question is because it is not at all clear how to tackle the problem without the binomial theorem if one hasn't seen the recursive nature of powers of polynomial roots before. – user21820 Jul 02 '15 at 07:17
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    Related : http://math.stackexchange.com/questions/936479/proving-that-frac-phi4001-phi200-is-an-integer – lab bhattacharjee Jul 02 '15 at 08:10
  • @user21820 Thank you very much for your advise.. I'm new to this community, I've no idea about the culture... So it'll take some time get used to it. – Akshay Hegde Jul 02 '15 at 08:15
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    See http://meta.math.stackexchange.com/questions/9959/how-to-ask-a-good-question. – user21820 Jul 02 '15 at 10:35
  • Thanks for those who have downvoted so that I can rectify..... – Akshay Hegde Jul 02 '15 at 11:24
  • See also: http://math.stackexchange.com/questions/906584/the-number-3-sqrt5n3-sqrt5n-is-an-integer – Martin Sleziak Nov 05 '15 at 20:56

2 Answers2

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Hint: $$A^{n+1} + B^{n+1} = (A + B)(A^n + B^n) - AB(A^{n-1} + B^{n-1})$$

corindo
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  • Your hint makes me wonder whether it is easier to think of it this way for higher-order recurrence relations. Seems a bit mysterious though without resorting to the underlying factorization of the linear operators involved.. – user21820 Jul 02 '15 at 07:05
  • If you use the fact $3-\sqrt5=\dfrac{1}{3+\sqrt5}$ it will be more nice. Any way +1 for nice work. – Bumblebee Jul 02 '15 at 08:51
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$3+\sqrt{5}$ and $3-\sqrt{5}$ are roots of the polynomial $r \mapsto r^2 - 6r + 4$, and by the form of $S_n$ it tells us that it is a solution to the recurrence relation $f(n+2) - 6f(n+1) + 4f(n) = 0$. Now all you need to do is to check that $S_0$ and $S_1$ are integers to prove that $S_n$ is an integer for any natural $n$. The next part is equally simple as you can see a factor of $2$ popping out every time you turn the recurrence crank. All you need to check is that $(3-\sqrt{5})^n < 1$.

user21820
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