Is $2^{97}+1$ a prime number?

$2^{97}+1\equiv (-1)^{97}+1\equiv 0 \mod 3$

$2^{97}+1$ 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. $2^{97}$ is an odd power of 2, it is one smaller than a number divisible by 3. Add 1 and you get a number divisible by 3.

In response to Marta Reece : Thank you for your proof in 'words'. Below is one part of your proof by induction written in a more familiar way to me. Maybe it is helpful for some other readers interested in proofs by induction.

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Proof by induction that $3 \, \big| \, 2^{2n+1} + 1$ for $n \geq 0$

1. Base case

$n=0 \Rightarrow 2^{2 \cdot 0 + 1} + 1 = 3$ and 3 is clearly divisible by 3.

2. Induction case

Assume the statement is true for $n = m$ , so $3 \, \big| \, 2^{2m+1} + 1$ for $m \geq 0$ and therefore $2^{2m+1} + 1$ can be written as $3k$ for some natural number $k$ .

For $n = m+1$ it follows that:

$2^{2(m+1) + 1} + 1 =$

$2^{(2m+1) + 2} + 1 =$

$4 \cdot 2^{2m+1} + 1 =$

$4 \cdot (2^{2m+1} + 1) - 3 =$

$4 \cdot 3k - 3 = 3 \cdot (4k - 1)$

which is clearly divisible by 3.

For $n = 48$ it follows that $2^{97} + 1$ is divisible by 3 and therefore it is not a prime number.

$2^{79} + 1 = (2^{79} + 2^{78} + 2^{77} + ... +2+1) - (2^{78} + 2^{77} + ... +2)$ $= ( 2^{80} - 1 ) - (2^{79} - 2 )$

Which is not a prime number.