For how many positive integers \(n\) is \(4^n-1\) a prime number?
We recall that a prime number has exactly two factors, \(1\) and itself.
(The number \(1\) is not considered to be a prime number these days.)
We can factorise \(4^n - 1\) as \[4^n-1 = 2^{2n} - 1 = (2^n - 1)(2^n + 1).\]
Now \(2^n-1 < 2^n+1\), so for \(4^n - 1\) to be prime, we must have \(2^n-1=1\), which means \(n\) must be \(1\).
Checking for \(n = 1\), we see that \(4^1-1=3\) is indeed prime.
Thus, there is exactly one value of \(n\) for which \(4^n-1\) is a prime number.
