Interesting identity

[abskebabs]

I've recently been working through a book called "Mathematical puzzling" and during the time I was tackling a problem investigating whether the identity [math]{2}^{n}-1[/math] could produce primes. During the analysis I came up with and derived an equation that seemed to hold generally, and this made it very interesting. In fact tbh, when I found it and realised it's general validity, it made me so excited I felt like hopping up in the air in some sort of stereotypical Archimedic fashion:-p ! It just seemed to give that kind of joy you can only experience when you find something like this in the field of mathematics or mathematical sciences.

 

Also to note how this relates to primes; all powers of n that are odd can produce primes, though this isn't at all always the case. Whether this is ture or not can be determined by the character of the RHS of the equation. This is as opposed to when the power is even; the only even power of n that produces a prime is 2, and this fact can be proved as well.

 

I guess even though I felt this way at the time, it's strange, I looked at the same equation today and thinking of the way I derived it, I couldn't feel more blase about it. In the end I think most mathematicians would probably think it trivial. Still, I'd like to share it with you:

[math]{2}^{n}-1=\sum_{k=0}^{n-1}{2^{k}}[/math]

 

What do you think? I guess also people could contribute their own results and share how they felt when they found them too;) . I encourage people to do so.

[river_rat]

Hey abskebabs

 

You can do one better then what you have posted so far: if [math] 2^n - 1[/math] is a prime number then [math]n[/math] must be prime.

 

I'll leave the proof up to you, its not difficult

[abskebabs]

I've tried to find a proof but I haven't got it yet. I'll keep trying.... Meanwhile in the process I found an indirect way of proving the sum of a geometric series! In fact I'm surprised I didn't realise this earlier.

[river_rat]

Hi abskebabs

 

Here is a big hint for you - how do you factor [math] a^n - b^n [/math]?