# How to solve the following Number Theory problem? [closed]

I've been practicing Number Theory lately and I've stumbled upon a problem that I cannot solve. I Googled a little bit on the Internet and I've found that I should be using Fermat's Little Theorem.

Here is the problem:

For positive integers $n\in \mathbb{N}$ find which of the two numbers $a_n=2^{2n+1}-2^{n+1}+1$ and $b_{n} = 2^{2n +1} + 2^{n + 1} + 1$ is divisible by $2$.

Any help would be appreciated.

## closed as off-topic by TheGeekGreek, Especially Lime, user91500, kingW3, JMPJun 28 '17 at 11:09

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – TheGeekGreek, Especially Lime, user91500, kingW3, JMP
If this question can be reworded to fit the rules in the help center, please edit the question.

• None is divisible by $2$, because $2\nmid 1$. – Dietrich Burde Jun 28 '17 at 9:20
• @Gigaxel, a typo is there in your question as $a_n$ would equal to $1$ if we cancel out +ve and -ve $2^{2n+1}$ , you have mistyped it – Atul Mishra Jun 28 '17 at 9:22
• @projectilemotion Ah yes you are right. My mistake it should have been just like in the second equation. Power of n + 1 not on 2n + 1. – Gigaxel Jun 28 '17 at 9:24
• @AtulMishra You're right. Edited it. – Gigaxel Jun 28 '17 at 9:26
• Even after the edit both $a_n$ and $b_n$ are odd. – kingW3 Jun 28 '17 at 9:41