9

Show that if you round $(\sqrt{2} + \sqrt{3})^{2009}$ to the closest integer you get an even number.

I tried without success to write it in binomial form and to multiply with a conjugate.

edit: Maybe they changed the number for the new course of 2009? im not sure. Now volfram alfa gives 0 as rest. http://www.wolframalpha.com/input/?i=round%5B%28sqrt%282%29%2Bsqrt%283%29%29%5E2009%5D+mod+10

Anyone got any ideas? Thanks,

Nathaniel Bubis
  • 33,018
Johan
  • 4,002
  • 3
    Why do you believe it rounds to an even number? Numerically I find that the "final" digits are $11.1364...$ But I could be wrong. – Brad Jun 09 '14 at 13:51
  • If $\alpha = \sqrt 2 + \sqrt 3$, then $\alpha^1, \alpha^3, \alpha^{17}, \alpha^{25} \ldots $ don't round to an even number. – mercio Jun 09 '14 at 13:52
  • It's handed out as one of the exercises in a course at my school (not that im taking). But I started thinking about it. It could be wrong – Johan Jun 09 '14 at 14:03
  • Wolfram says that it is true: http://www.wolframalpha.com/input/?i=Mod[IntegerPart[%28sqrt%282%29%2Bsqrt%283%29%29^2007%2B0.5]%2C2] – Quimey Jun 09 '14 at 14:06
  • I think the exercise got the year wrong. After all, we are in 2014 now. – mercio Jun 09 '14 at 14:08
  • I think the course was created 2007... – Johan Jun 09 '14 at 14:09
  • @Quimey I get WA saying the last digit is 1 http://www.wolframalpha.com/input/?i=round%5B%28sqrt%282%29%2Bsqrt%283%29%29%5E2007%5D+mod+10 – Casteels Jun 09 '14 at 14:33
  • Perhaps use that $\alpha^4=10\alpha^2-1$. – lhf Jun 09 '14 at 14:43
  • 2
    Maple, using Digits = 3000, gives me $(\sqrt{2}+\sqrt{3})^{2009} = \ldots 4789.7172 \ldots$. – Robert Israel Jun 09 '14 at 15:09
  • @RobertIsrael, does Maple also refute any of the smaller values computed by Alpha in the answer just posted? – zyx Jun 09 '14 at 17:26
  • @zyx: Maple confirms your answer. Moreover, I found no pattern in the last 200 entries using the gfun package in Maple. – Robert Israel Jun 10 '14 at 01:31
  • @RobertIsrael, thanks very much. Since the two CAS disagree, according to above comments, for $n=2009$, the point of divergence might indicate how much precision Alpha uses by default, though Wolfram may have published that somewhere. – zyx Jun 10 '14 at 13:38
  • Wolfram Alpha also says it's $...4789.7172...$, if you ask it the right way, e.g. .0001 * (floor(10000 * (sqrt(2)+sqrt(3))^2009) mod 10^8) – Robert Israel Jun 11 '14 at 02:12

2 Answers2

4

This is not a solution, but rather a general way to approach this kind of questions. Maybe you can continue from here:
1) Denote $\alpha=\sqrt{2}+\sqrt{3}$. Find a polynomial over the integers $p(x)$ such that $p(\alpha)=0$, preferably of a small degree. For example, here we have: $$\alpha^2=2+2\sqrt{6}+3=5+2\sqrt{6}\hspace{5pt}\Rightarrow\hspace{5pt}\alpha^4-10\alpha^2+25=(\alpha^2-5)^2=24$$ So we can choose $p(x)=x^4-10x^2+1$. From the construction it is easy to see that the other roots of $p(x)$ are $\bar{\alpha}=\sqrt{2}-\sqrt{3}$, $-\alpha=-\sqrt{2}-\sqrt{3}$ and $-\bar{\alpha}=-\sqrt{2}+\sqrt{3}$.
2) Consider the linear homogeneous recurrence relation such that $p(x)$ is its characteristic polynomial: $a_n-10a_{n-2}+a_{n-4}=0$. The general solution to this equation is given by $a_n=A_1\alpha^n+A_2(-\alpha)^n+A_3(\bar{\alpha})^n+A_4(-\bar{\alpha})^n$.
Now we need to construct a solution (find some $A_2,A_3,A_4$ and fix $A_1=1$) such that the closest integer to $\sqrt{2}+\sqrt{3})^{2009}$ is either $a_{2009}$ or $a_{2009}-1$.
Having the recurrence relation allows us to prove inductively that for all odd $n$, $a_n$ have the same parity - based only on $a_1$ and $a_3$.

Dennis Gulko
  • 15,640
  • The rounded values of $(\sqrt{2} + \sqrt{3})^n$ and the linear recursion sequence are the same only for even $n$. – zyx Jun 09 '14 at 17:09
4

The hope is that for large enough $n$ the rounding of $a_n = (\sqrt{2}+\sqrt{3})^n$ is a linear recursive sequence, but there is a complication. Rounding of even and odd powers will work differently due to one of the conjugates, $-\sqrt{2} - \sqrt{3}$, having the same absolute value as $\sqrt{2}+\sqrt{3}$. Therefore the even and odd subsequences should be considered separately.

$a_{2n} = (5 + 2 \sqrt{6})^n$ is close to $b_n = (5 + 2 \sqrt{6})^n + (5 - 2 \sqrt{6})^n$, an integer, and the second term converges to $0$, so with the possible exception of a few small $n$, the rounding of $a_{2n}$ is $b_n$ whose parity (and periodic pattern mod $k$ for any $k$) is easily determined from the recursion $b_{n+2}=10b_{n+1}-b_n$.

For odd $n$, the values of the parity of the rounded $a_{2n+1}$ for $n=1$ to $900$ are

{1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 1, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 0, 1, 0, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 0, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 0}

Maybe there is a periodic subsequence that would "explain" the appearance of some large odd $n$ in a problem like this, but I don't see one.

zyx
  • 35,436
  • The computation used Wolfram Alpha, which according to the comments under the question might not retain enough precision for the later values to be meaningful. The small values show aperiodicity, such as the 9 consecutive 0's in the second line that is not repeated anywhere in the table (and particularly, not in the first few rows where the Alpha computation is likelier to be correct). – zyx Jun 09 '14 at 17:20
  • The table of parities was kindly confirmed by Robert Israel using Maple (discussion in comments below the question). So the precision of Alpha is sufficient for this problem up to $n=900$ but not necessarily as high as $n=2009$. – zyx Jun 10 '14 at 13:47