20071211, 21:31  #1 
13·751 Posts 
factoring while dealing with exponents
I'm really not sure if what I'm going to say makes sense, I didn't log the part of the chat where the guy taught me the stuff, so I may be going down a blind alley.
k^2*2^{something, maybe 2n or 2^n}1 = (k*2^n+1)(k*2^n1) it doesn't have to do with Fermat numbers, btw. I'm going to print it out in words so people don't think I mistyped. k to the power of 2 times 2 to the power of {something} minus 1 equals (k*2^n+1)(k*2^n1) Is there a whole number answer to this problem, or is my memory whacking out and making me think I'm remembering stuff I'm not? 
20071211, 21:48  #2 
"Nancy"
Aug 2002
Alexandria
2,467 Posts 

20071212, 12:31  #3  
Nov 2003
1D24_{16} Posts 
Quote:
Didn't you see this in beginning algebra??? 

20071213, 04:47  #4 
"Jason Goatcher"
Mar 2005
3507_{10} Posts 
In response to the last two posts:
Thanks akruppa, I'll check it out. Mr. Silverman, beginning algebra was a long time ago, I'm so rusty it's not funny. 
20071213, 04:56  #5 
"Jason Goatcher"
Mar 2005
3·7·167 Posts 
Thanks for the link, akruppa, that helped immensely. I ended up picking some low primes to plug in as the values in order to figure it out.
The thing that threw me was having to deal with the dual instances of exponents, I'm not sure if they ever covered that in high school or not. One possibility is that the fact that my dad was in the military when I was growing up may have meant that I could have learned it, but different schools teach things at different times and the timing of our moves could have caused me to miss out. 
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