Prove that 10101...10101 is NOT a prime. 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.
 A: Suppose $n + 1$ is odd. Let $a$ be a real number such that $|a| \neq 1$. 
Then $X = \sum_{k=0}^{n}a^{2k} = \frac{a^{2(n+1)} - 1}{a^{2}-1} = \frac{a^{n+1} - 1}{a - 1} \cdot \frac{(-a)^{n+1} - 1}{(-a) - 1} = \sum_{k=0}^{n}a^{k} \cdot \sum_{k=0}^{n} (-a)^{k}$. 
A: 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 $a\in\mathbb Z_{\ge 2}$).
A: Note that $1010101....10101$ is $\frac{10^{4034}-1}{99}$.
Also, $10^{4034}-1$ is $(10^{2017}-1)(10^{2017}+1)$, both of which are larger than $99$. 
This implies that the number is not prime. 
A: And since you have asked this question, here is an interesting piece of information, that in the sequence $101,10101,1010101,....$ none of the numbers are prime EXCEPT the first one. This can be proved quite easily and in your case the number is $${{(10^{4034}-1)}/99}$$, and the numerator can be written as 
$${{(10^{2017}}-1)} \times {{(10^{2017}}+1)}$$ 
and the first multiplicand has a factor 10-1=9 and the second multiplicand has a factor 10+1=11 so their product is divisible by 9 and 11, which are coprime,hence divisible by 99, so the number is not prime, because it has 2 factors now both greater than 1.
So it is a prime which is anyway supported by the  result above.
