Prime Factors of 4,194,305Date: 09/20/1999 at 17:02:38 From: Carolyn Offutt Subject: Prime Factoring 2^22+1 Dr. Math, I am in 8th grade and taking Algeom (algebra and geometry in one year). Our teacher assigned the problem: Prime Factor 2^22+1 I have spent about 2 hours on the problem so far, and have talked to 2 of my classmates about it. I have been unable to make any progress other than finding out that one of the factors is 5. I worked out the number, which is 4194305, which is obviously divisible by 5. I then found a program on the Internet that gives the prime factorization of a number. I entered my number and got 5*397*2113. We are supposed to show how we got our answer. I would like to know if you have a clever way to do this problem. I would appreciate it greatly. Thanks! Carolyn Offutt Date: 09/20/1999 at 18:44:00 From: Doctor Schwa Subject: Re: Prime Factoring 2^22+1 Interesting question, and I like your approach. I tried to do it with a little algebra. Maybe you've learned by now that (x+1)^2 = x^2 + 2x + 1. Well, 2^22 + 1 looks sort of like that, except there's no 2x there. Rats. I thought for a while about how to put a 2x in there, and finally decided on this wishful thinking strategy: 2^22 = (2^11)^2, so if x = 2^11, then I have x^2 + 1. How do I get the 2x? I just stuck it in there. But wait, that's not fair, so I better subtract it, too, like this: 2^22 + 1 = 2^22 + 2*2^11 + 1 - 2*2^11. Now I have that perfect square factoring pattern, but with a -2*2^11 left outside: = (2^11 + 1)^2 - 2*2^11. And 2*2^11 = 2^12, so = (2^11 + 1)^2 - 2^12 But 2^12 = (2^6)^2, so = (2^11 + 1)^2 - (2^6)^2 and now I use the 'Difference Of Two Squares' factoring pattern: = (2^11 + 1 + 2^6)(2^11 + 1 - 2^6) which equals = (2113)(1985) or (2113)(397)(5), the same answer you got. Thanks for a fun problem. I wonder if my method is what your teacher had in mind? - Doctor Schwa, The Math Forum http://mathforum.org/dr.math/ |
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