I have little math background and I just do some random observations on prime numbers and notice that:
for $p > k$ where $p$ is a prime and $k$ is a positive integer: $$\sum_{i=0}^{k-1} 2^{ip} = 0\pmod{2^k-1}$$
For example, pick $p = 5$ and $k = 3$: $$\sum_{i=0}^2 2^{5i} = 2^{10}+2^5+1=1057=7\times151=(2^3-1)\cdot151 $$
pick $p = 7$ and $k = 4$: $$\sum_{i=0}^3 2^{7i} = 2^{21}+2^{14}+2^{7}+1=2113665=15 \cdot 140911=(2^4-1) \cdot 140911 $$
pick $p = 11$ and $k = 7$: $$\sum_{i=0}^6 2^{11i} = 2^{66}+2^{55}+2^{44}+2^{33}+2^{22}+2^{11}+1= 73823022692637345793 \\ =127\times581283643249112959=(2^7-1)\cdot581283643249112959 $$
I cannot find any similar observations related to this elsewhere, and it would be really appreciated if anyone could provide some guidance on whether this always holds. I am very curious about that. Thank you!
For any positive integers $k$ and $p$, $x^{kp}-1$ is divisible (as a polynomial) by both $x^k-1$ and $x^p-1$. Moreover, if $k$ and $p$ are coprime the gcd of $x^p-1$ and $x^k-1$ is $x-1$. Thus if $k$ and $p$ are coprime, $x^{kp}-1$ is divisible by $(x^k-1)(x^p-1)/(x-1)$. Taking $x=2$, $2^{kp}-1$ is divisible by $(2^k-1)(2^p-1)$ in these cases.