Is $2^{99}+1$ a prime number?

$\Large (2^{33})^3+1^3=(2^{33}+1)(2^{66}-2^{33}+1)\\~~~~=(2^{11}+1)(2^{22}-2^{11}+1)(2^{66}-2^{33}+1)$ and hence has factors other than1 and itself. $\therefore \boxed{not ~prime}$

$2^{99}+1$ 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. $2^{2}-1$ 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. $2^{99}$ is an odd power of 2, it is one smaller than a number divisible by 3. Subtract 1 and you get a number divisible by 3.

The solution is very simple. $3|2^(99)+1$ and it is not equal to 3. And hence it is not prime