Perfect power ($n=m^k,k>1$)

An integer is a perfect power if it can be written $m^k$ with integers $m$ and $k>1$. In the common nontrivial positive version, both the value and the base exceed one. For example, $64=8^2=4^3=2^6$ qualifies, whereas twelve does not. The degenerate values $0$, $1$ and $-1$ are included or excluded according to the convention being used.

For a positive integer $n>1$ with prime factorization $n={\prod }_i{p}_i^{e_i}$, the exact criterion isIf $n=m^k$, every prime exponent is divisible by $k$. Conversely, a common divisor $k>1$ lets one take $m={\prod }_i{p}_i^{e_i/k}$. This gives a recognition test once the prime factorization is known; it does not imply that factorization is the fastest recognition algorithm.

A negative value requires an odd exponent. Thus $-64=(-4{)}^3$ is a perfect power, while $-16$ is not: its absolute value has only powers of two as common divisors of its exponents. Perfect cubes are the special case $k=3$, whereas a cube-full number imposes a different lower bound on each prime exponent.

An integer equal to the third power of another integer, such as $8=2^3$ or $-27=(-3{)}^3$.