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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.

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

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If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ None is divisible by $2$, because $2\nmid 1$. $\endgroup$ – Dietrich Burde Jun 28 '17 at 9:20
  • $\begingroup$ @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 $\endgroup$ – Atul Mishra Jun 28 '17 at 9:22
  • $\begingroup$ @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. $\endgroup$ – Gigaxel Jun 28 '17 at 9:24
  • $\begingroup$ @AtulMishra You're right. Edited it. $\endgroup$ – Gigaxel Jun 28 '17 at 9:26
  • $\begingroup$ Even after the edit both $a_n$ and $b_n$ are odd. $\endgroup$ – kingW3 Jun 28 '17 at 9:41
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There's no need of Fermat's Theorem. You can easily notice, in both of the numbers, first two terms are even and are being subtracted (in first number) and added (in the second), this means that first two terms shall give an even number always, then you add 1 to it, making it odd. Thus both of them are indivisible by 2. Further, you could have used mod and this question would be like a piece of cake.

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  • $\begingroup$ Yes, you are right. I see what a trivial question this is, and I feel ashamed of posting it hahaha xD. Thank you :D $\endgroup$ – Gigaxel Jun 28 '17 at 9:41

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