In honor of the 52nd Mersenne Prime being found, let's do this one.
For any natural numbers and both greater than 1, prove that:
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In honor of the 52nd Mersenne Prime being found, let's do this one.
For any natural numbers and both greater than 1, prove that:
Focus on the right-hand side, we can distribute the first factor (2a−1) into the second one:
The first term gives
and the second term gives
When adding up these two terms, it's easy to see that all terms but the first and the last cancel out proving the identity.
So this basically proves that one Mersenne number (which isn't prime) can be factored out into another Mersenne number. And it also proves that if the exponent is of the form ab it cannot be a Mersenne prime.
Happy to see other people post puzzles as well :)
Anyone with time and coding skills, check out this issue to add spoiler formatting in SN. In addition to the usual SN contribution reward, I am giving out a 30k sat bounty.
Curious: would it be better to left align math or do you prefer it centered?
It can be controlled by the user
y=f(x)
Oh i didn't realize that. But makes sense after i checked how you did it.
$$y = f(x)$$vs
$$ y = f(x) $$Centered is fine. That's what I'm used to in my publications anyhow. Don't have a strong opinion on this.
#736461
Good job :)
rewrite the equation as:
206! - 1 = (2^a - 1)(2^b - 1) + (2^a - 2) + ... + 2 start by factoring out (2^a - 1) from the right-hand side:
206! - 1 = (2^a - 1)[(2^b - 1) + (2^(a-1) - 1) + ... + 1] (2^b - 1) + (2^(a-1) - 1) + ... + 1 = (2^b + 2^(a-1) + ... + 1) - a
= (2^(a+b) - 1)/(2 - 1) - a
= 2^(a+b) - a - 1 206! - 1 = (2^a - 1)[2^(a+b) - a - 1]I got 3 = 2/(1 - 1/(3^a))