15
$\begingroup$

Prove that there exists a real number $ a>1 $, such that $ \{a^n\} $ belongs to $[\frac{1}{3},\frac{2}{3}]$ for all positive integers $n$ and $\lfloor a^n\rfloor$ is even iff $n$ is a prime.

$a^n=\{a^n\}+\lfloor a^n\rfloor$, where $ \{a^n\} $ is the fractional part of $a^n$ and $\lfloor a^n\rfloor$ is the largest integer not greater than $ a^n $

$\endgroup$
  • 4
    $\begingroup$ "$(\{a^n\} \in [\frac{1}{3},\frac{2}{3}] \wedge \lfloor a^n \rfloor \text{ is even} ) \leftrightarrow n\text{ is a prime}$" or "$\{a^n\} \in [\frac{1}{3},\frac{2}{3}] \wedge (\lfloor a^n \rfloor \text{ is even} \leftrightarrow n\text{ is a prime})$"? $\endgroup$ – JiK Aug 30 '14 at 20:51
  • 1
    $\begingroup$ the second one: $\{a^n\} \in [\frac{1}{3},\frac{2}{3}] \wedge (\lfloor a^n \rfloor \text{ is even} \leftrightarrow n\text{ is a prime})$ $\endgroup$ – john Aug 30 '14 at 21:39
  • 3
    $\begingroup$ Where is this from? $\endgroup$ – Mayank Pandey Aug 30 '14 at 21:40
  • $\begingroup$ Is this problem easy if we remove the second condition? I can't figure out a solution to even the first condition. $\endgroup$ – dshin Oct 14 '14 at 19:08
  • 1
    $\begingroup$ This would suggest a possible primality tester. Take a, raise it to the n'th power and check parity. Can be loose with precision because of first condition. $\endgroup$ – dshin Oct 14 '14 at 20:24
4
$\begingroup$

We can't find an exact value of $a$ but it is possible to construct it using the nested intervals theorem.

Let $I_1=[15+\frac13,15+\frac23]$. Assume $I_k=[a_k+1/3,a_k+2/3]$ is defined and also satisfy $I_k'=[(a_k+1/3)^{1/k},(a_k+2/3)^{1/k}]=[b_k,c_k]\subset I_1$. Then since $15<(a_k+1/3)^{1/k}=b_k<c_k$, $$(a_k+\frac23)^{1+\frac1k}-(a_k+\frac13)^{1+\frac1k}=c_k^{k+1}-b_k^{k+1}>(c_k^k-b_k^k)c_k>\frac13\cdot15=5$$ So the interval $J_k=[(a_k+\frac13)^{1+\frac1k},(a_k+\frac23)^{1+\frac1k}-1]$ contain at least $4$ integers which means there are both even number and odd number in $J_k$. Let $$a_{k+1}=\cases{\text{even number in }J_k&\text{,if }k+1\text{ is prime}\\\text{odd number in }J_k&\text{,otherwise}}$$ and define $I_{k+1}=[a_{k+1}+1/3,a_{k+1}+2/3]$.

From $$(a_{k+1}+\frac13)^{\frac1{k+1}}>a_{k+1}^{\frac1{k+1}}\ge((a_k+\frac13)^{1+\frac1k})^{\frac1{k+1}}=(a_k+\frac13)^{\frac1k}$$ and $$(a_{k+1}+\frac23)^{\frac1{k+1}}<(a_{k+1}+1)^{\frac1{k+1}}\le ((a_k+\frac23)^{1+\frac1k})^{\frac1{k+1}}=(a_k+\frac23)^{\frac1k}$$ we see that $I'_{k+1}\subset I'_k$. Thus by the nested intervals theorem, there is a real number $\displaystyle a\in\bigcap_{k=1}^\infty I'_k$. Now $a^k\in I_k=[a_k+1/3,a_k+2/3]$ where $a_k$ is even iff $k$ is prime. Thus $a$ satisfy the conditions in the problem.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.