For what positive integers n does 15|M

[maximus101]

Homework Statement

For what positive integers n does $$15|2^{2n}-1$$

Homework Equations

We know $2^{2n}\equiv1mod15$

I was thinking this might be helpful but not sure
$x^{2} ≡ −1 (mod p)$ is solvable if and only if $p ≡ 1 (mod 4)$

The Attempt at a Solution

I think that the answer is for all $n=2k$ where $k$ is an integer
from plugging in various values of n, however I am not sure how to prove it? Any suggestions

 

[mighty2000]

?



You are correct in thinking that the answer is all n=2k where k is an integer. This can be proven by using the fact that 2^{2n}\equiv1mod15. Since 15 is a multiple of 3 and 5, we can rewrite this as 2^{2n}\equiv1mod3 and 2^{2n}\equiv1mod5. By using the Chinese Remainder Theorem, we can combine these two congruences to get 2^{2n}\equiv1mod15. This means that for any positive integer n, 15 divides 2^{2n}-1. Therefore, the answer is all n=2k where k is an integer. I hope this helps!