how to quickly determine whether a given natural number is a power of any other natural number? - MathOverflow most recent 30 from http://mathoverflow.net2013-05-24T10:35:51Zhttp://mathoverflow.net/feeds/question/13843http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/13843/how-to-quickly-determine-whether-a-given-natural-number-is-a-power-of-any-other-nhow to quickly determine whether a given natural number is a power of any other natural number? Oleg2010-02-02T18:06:08Z2010-02-02T19:03:04Z
<p>We have a natural number $n>1$. We want to determine whether exists any natural numbers $a, k>1$ such that $n = a^k$. </p>
<p>Please suggest a polynomial-time algorithm.</p>
http://mathoverflow.net/questions/13843/how-to-quickly-determine-whether-a-given-natural-number-is-a-power-of-any-other-n/13845#13845Answer by Felipe Voloch for how to quickly determine whether a given natural number is a power of any other natural number? Felipe Voloch2010-02-02T18:26:17Z2010-02-02T18:26:17Z<p>For each $k \le \log n/\log 2$, compute an approximation to the positive real $k$-th root of $n$ using Newton's method to enough precision to check if it is an integer. Alternatively, use $p$-adic roots for a suitable $p$, with Newton turning into Hensel.</p>
http://mathoverflow.net/questions/13843/how-to-quickly-determine-whether-a-given-natural-number-is-a-power-of-any-other-n/13849#13849Answer by Ben Linowitz for how to quickly determine whether a given natural number is a power of any other natural number? Ben Linowitz2010-02-02T18:40:06Z2010-02-02T18:40:06Z<p>In order to test whether or not a natural number $n$ is a perfect power, we can conduct a binary
search of the integers {1,2,...,n} for a number a number $m$ such that $n = m^b$ for
some $b>1$. Let $b>1$. If a solution $m$ to $m^b =n$ exists,then
it must lie in some interval $[c_i,d_i]$. When $i = 0$ we may take $[c_0,d_0] = [1,n]$. To
define $[c_{i+1},d_{i+1}]$, consider $\alpha:= \left\lfloor \frac{(ci+di)}{2}\right\rfloor$. If $\alpha^b = n$ then we’re done. If $\alpha^b > n$, let $[c_{i+1}, d_{i+1}] = [c_i, \alpha]$; otherwise $\alpha^b < n$ and we let $[c_{i+1}, d_{i+1}] = [\alpha, d_i]$. We continue in this manner until $|c_i − d_i| \leq 1$. We then increase the value stored in variable $b$ and start the loop again. Performing this loop for all $b \leq log(n)$ completes the algorithm. </p>
<p>A pseudocode implementation of this algorithm can be found on page 21 of Dietzelbinger's <a href="http://books.google.com/books?id=DWdjlaxl_7YC&dq=primality+testing+in+polynomial+time&printsec=frontcover&source=bn&hl=en&ei=HnFoS4XxEMuXtgf21KXUBg&sa=X&oi=book_result&ct=result&resnum=4&ved=0CB4Q6AEwAw#v=onepage&q=&f=false" rel="nofollow">Primality Testing in Polynoial Time</a>. Its complexity is about $O(log^3(n))$.</p>
http://mathoverflow.net/questions/13843/how-to-quickly-determine-whether-a-given-natural-number-is-a-power-of-any-other-n/13852#13852Answer by Pace Nielsen for how to quickly determine whether a given natural number is a power of any other natural number? Pace Nielsen2010-02-02T19:03:04Z2010-02-02T19:03:04Z<p>This can be done in "essentially linear time." Check out Daniel Bernstein's website: <a href="http://cr.yp.to/arith.html" rel="nofollow">http://cr.yp.to/arith.html</a></p>
<p>Especially note his papers labeled [powers] and [powers2].</p>