Can you prove that 2^n-1 will be divisible by 3 if n is even.
I have generated this:
Prime factors of 2^2-1 = 3 are [3]
Prime factors of 2^3-1 = 7 are [7]
Prime factors of 2^4-1 = 15 are [3, 5]
Prime factors of 2^5-1 = 31 are [31]
Prime factors of 2^6-1 = 63 are [3, 3, 7]
Prime factors of 2^7-1 = 127 are [127]
Prime factors of 2^8-1 = 255 are [3, 5, 17]
Prime factors of 2^9-1 = 511 are [7, 73]
Prime factors of 2^10-1 = 1023 are [3, 11, 31]
Prime factors of 2^11-1 = 2047 are [23, 89]
Prime factors of 2^12-1 = 4095 are [3, 3, 5, 7, 13]
Prime factors of 2^13-1 = 8191 are [8191]
Prime factors of 2^14-1 = 16383 are [3, 43, 127]
Prime factors of 2^15-1 = 32767 are [7, 31, 151]
Prime factors of 2^16-1 = 65535 are [3, 5, 17, 257]
Prime factors of 2^17-1 = 131071 are [131071]
Prime factors of 2^18-1 = 262143 are [3, 3, 3, 7, 19, 73]
Prime factors of 2^19-1 = 524287 are [524287]...
I have noticed that when n is even in 2^n-1 the results are also divisible by 3. But i can't prove it..
Another Idea; take $n=2k$ so $$2^n-1=2^{2k}-1=\\(2^k-1)(2^k+1)$$ between $3$ sequential numbers :one of them is divisible by $3$
Now $2^k \neq 3q $ so one of $2^k-1$ or $2^k+1$ must divided by $3$
Remark :$2^k-1 , 2^k , 2^k+1 $ are sequential