We have a natural number $n>1$. We want to determine whether there exist natural numbers $a, k>1$ such that $n = a^k$.
Please suggest a polynomialtime algorithm.
We have a natural number $n>1$. We want to determine whether there exist natural numbers $a, k>1$ such that $n = a^k$. Please suggest a polynomialtime algorithm. 


This can be done in "essentially linear time." Check out Daniel Bernstein's website: http://cr.yp.to/arith.html Especially note his papers labeled [powers] and [powers2]. 


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. A pseudocode implementation of this algorithm can be found on page 21 of Dietzelbinger's Primality Testing in Polynoial Time. Its complexity is about $O(log^3(n))$. 


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. 


The computer algebra system GAP performs this test and determines a smallest root $a$ of a given integer $n$ quite efficiently. The following is copied directly from its source code (file gap4r6/lib/integer.gi), and should be selfexplaining:


