Is $2^{100}-1$ a prime number?

$\Large (2^{50})^2-1^2=(2^{50}-1)(2^{50}+1)\\~~~~~~ =(2^{25}-1)(2^{25}+1)(2^{50}+1)$ and hence has factors other than1 and itself. $\therefore \boxed{not ~prime}$

$2^{100}-1$ is divisible by $3$ and is therefore not a prime number. More generally any even power of $2$ is one larger than a number divisible by $3$ and any odd power of $2$ is one smaller than a number divisible by $3$ .

To show that this is so, let's start with a number which is one larger than a number divisible by $3$ . This number can be written as $3n+1$ . Multiplying this number by $2$ we get $6n+2$ . This number is $2$ larger than a number divisible by $3$ , therefore it is also one smaller than a number divisible by $3$ .

Let's take this number, one which is one smaller than a number divisible by $3$ and can therefore be written as $3m-1$ , and multiply it by $2$ . We get $6m-2$ , which is $2$ smaller than a number divisible by $3$ , and is therefore $1$ larger than a number divisible by $3$ .

So if we repeatedly multiply by $2$ we go from being one below a number divisible by $3$ to being one over, to being one below etc.

To finish the proof by induction we need a starting point. $2^{2}-1=3$ is divisible by $3$ . $2^{2}$ is an even power of $2$ and is one larger than a number divisible by $3$ . Multiply it by $2$ to get $2^{3}$ and it will be one smaller than a number divisible by $3$ , etc. All even powers of $2$ will be one over, all odd will be one below. $2^{100}$ is an even power of $2$ , it is one larger than a number divisible by $3$ . Subtract $1$ and you get a number divisible by $3$ .

101 is a prime number. 2^100 - 1 mod 101 = 0 (Fermat's little theorem)

According to the definition of prime numbers a prime number is 2 raised to the powers odd minus one.

2^10=1024 & 2^10 * 2^10=1048576 giving last digit 6. 6-1=5 showing that 2^20 -1 is not prime. In we follow the pattern series of last digits, it is 3,5,3,5... giving the last digit of 2^100 -1 as 5. Hence it is not prime

(2^n)-1,when n is even then it can never be a prime number