Number of ways in which a composite number can be resolved into two factors which are prime to each other I read in a book that the total number of ways in which a composite number N can be resolved into two factors which are prime to each other is $2^{n-1}$ , where n is the number of prime factors of N. Suppose we have a number N = $a^p$ x $b^q$ x $c^r$, where a,b,c are the prime factors of the given number N,  and we apply the formula for finding the number of ways in which this number can be resolved into two factors prime to each other. We will get the answer $2^{3-1}$ ,i.e., 4. 
Three of the ways are {$a^p$, $b^q$x $c^r$} ,{$b^q$, $a^p$x$c^r$}, and {$c^r$, $a^p$x $b^r$}. But I cannot figure out what the fourth way is. Maybe it is {$a^p$ x $b^q$ x $c^r $, 1}, but these two numbers are not co-prime. Is the formula wrong? Can anybody help me with this?
 A: The formula also counts $1\times N$ as a way to write it as a product of two coprime factors, since $\gcd(1,N)=1$. If you want to exclude that possibility, the formula should be $2^{n-1}-1$. 
Here is how the formula works.
You are trying to split $N$ into a product of two coprime numbers, $N=x\times y$.  Suppose $p^k$ is one of the prime power factors of $N$. The two numbers $x$ and $y$ should be coprime, so they cannot both have a factor $p$. Therefore $p^k$ must be either included in $x$ or included in $y$, and cannot be split between them. So for each prime power you have $2$ choices.
If $N$ factorises into $n$ prime powers, you have $2$ choices for each prime power, giving $2^n$ possible choices all together.
However, $x\times y = y\times x$, and we consider these the same, so everything was double counted. Therefore there are actually only $\frac{2^n}{2}=2^{n-1}$ possible different ways to write $N$ as a product of two coprime numbers (including $\{1,N\}$).
A: If you define two integers $a,b$ as being coprime if $\gcd(a,b) = 1$, then $1$ is coprime with every integer.
