15. For how many positive integers $n$ is $4^{n}-1$ a prime number?
A 0
B 1
C 2
D 3
E infinitely many
SOLUTION
B We give two methods for answering this question. Both methods rely on the following property of prime numbers: a positive integer $p$ is a prime number if, and only if, the only way $p$ can be expressed as a product of two positive integers, $m$ and $n$, is where exactly one of $m$ and $n$ is 1 .
MeTHOD 1
The first method uses the fact that when $n$ is a positive integer $4^{n}-1$ is the difference of two squares and so may be factorized as $4^{n}-1=\left(2^{n}\right)^{2}-1^{2}=\left(2^{n}+1\right)\left(2^{n}-1\right)$. So $4^{n}-1$ is not a prime, unless one of the factors $2^{n}+1$ and $2^{n}-1$ is equal to 1 . Now $2^{n}+1$ cannot be equal to 1 and $2^{n}-1=1$ if, and only if, $n=1$. When $n=1$, we have $4^{n}-1=3$ and so $4^{n}-1$ is prime. So there is just one positive integer $n$, namely 1 , for which $4^{n}-1$ is a prime number.
Method 2
The second method uses the fact when $n$ is a positive integer $x-1$ is a factor of $x^{n}-1$, since, for $n \geq 2$,
$$
x^{n}-1=(x-1)\left(x^{n-1}+x^{n-2}+\cdots+x+1\right) \text {. }
$$

Hence, putting $x=4$, we deduce that, for each positive integer $n, 3$ is a factor of $4^{n}-1$. Therefore $4^{n}-1$ is not prime except when $n=1$ and $4^{n}-1=3$. So there is just one positive integer $n$ for which $4^{n}-1$ is a prime number.