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A positive divisor $d$ of a positive integer $n$ is bi-unitary when the greatest common unitary divisor of $d$ and $n/d$ is $1$. The comparison uses unitary divisors, so it differs from requiring the ordinary greatest common divisor to be one.

For $n=p^e$ with $p$ a prime number and $e\ge 1$, consider $d=p^a$, where $0\le a\le e$. The only possible nontrivial common unitary divisor of $p^a$ and $p^{e-a}$ occurs when $a=e-a>0$. Consequently every divisor is bi-unitary except $p^{e/2}$ when $e$ is even.

For example, the bi-unitary divisors of $64=2^6$ are $1,2,4,16,32,64$; only $8$ is excluded. Its unitary divisors are just $1$ and $64$, showing that these two notions differ substantially. For a general positive integer, apply the prime-power condition independently to each prime factor in its prime factorization.

A unitary divisor $d$ of a positive integer $n$ satisfies $\gcd (d,n/d)=1$. The greatest common unitary divisor of $a,b$ is the largest integer that is a unitary divisor of both. In their prime factorizations, it includes precisely those whole prime powers whose positive exponents agree in $a$ and $b$.

For $a=12=2^2\cdot 3$ and $b=18=2\cdot 3^2$, it is $1$, although the ordinary greatest common divisor is $6$. For $12$ and $20=2^2\cdot 5$, it is $4$. Requiring unitary divisibility on both inputs is essential: merely being a unitary divisor of their ordinary greatest common divisor is a weaker condition.