TITLE: Prove that 10101...10101 is NOT a prime.
QUESTION [23 upvotes]: So basically we have a number $10101...10101$ that contains $2016$ zeros and can be written as$ \sum _{ k=0 }^{ 2016 }{ 100^{ k } }$ . I want to prove that this number is not a prime without using anything besides a piece of paper and a pen. I'm stuck on this for quite a few days now.

REPLY [5 votes]: In fact, you can generalize this to any base $a\in\mathbb Z_{\ge 2}$.
If $a,k\in\mathbb Z_{\ge 2}$, then ($_a$ denotes 'base $a$'):
$$\underbrace{10101\cdots 101_a}_{k\text{ zeros}}=\sum_{i=0}^k a^{2i}=\frac{a^{2(k+1)}-1}{a^2-1}=\frac{\left(a^{k+1}+1\right)\left(a^{k+1}-1\right)}{a^2-1},$$
$a^{k+1}+1>a^{k+1}-1>a^2-1$, therefore $\underbrace{10101\cdots 101_a}_{k\text{ zeros}}$ is composite.
Therefore, $10101_a, 1010101_a,\ldots$ are all composite (for any base $a\in\mathbb Z_{\ge 2}$).