Question

What is the product of the two smallest prime factors of 2^1024-1?

Answer 1

1*2?

Answer 2

1 is not prime.

Answer 3

its prime enough :)

Answer 4

http://en.wikipedia.org/wiki/Prime_number#Primality_of_one

Answer 5

ohno! a wikipedia proof ...

Answer 6

The number 1 is a special case which is considered neither prime nor composite (Wells 1986, p. 31). Although the number 1 used to be considered a prime (Goldbach 1742; Lehmer 1909, 1914; Hardy and Wright 1979, p. 11; Gardner 1984, pp. 86-87; Sloane and Plouffe 1995, p. 33; Hardy 1999, p. 46), it requires special treatment in so many definitions and applications involving primes greater than or equal to 2 that it is usually placed into a class of its own. - from someplace other than my head i hope

Answer 7

Also: 2 is clearly not a prime factor of \[ 2^{1024} - 1 \]

Answer 8

back in my day 1 was a prime ;)

Answer 9

\[2^{1024}-1=(2^{512}-1)(2^{512}+1)\]i think is a nice direction to go
it continues that way

Answer 10

i just simplified to: 2^1023

Answer 11

but keep doing difference of squares...

Answer 12

it has to a reasonable factor evenutally

Answer 13

Fool has some brilliant solution I klnow

Answer 14

According to this: http://www.alpertron.com.ar/ECM.HTM

\[2^{1024} - 1 = 3 \times 5 \times 17 \times 257 \times
641 \times 65537 \times 274177 \times 2 424833 \times 6 700417 \]\[\times 67 280421 310721 \times 1238 926361
552897 \times 59649 589127 497217 \]\[ \times 5704 689200 685129 054721 \]\[ \times 7 455602 825647
884208 337395 736200 454918 783366 342657 \]\[\times 93 461639 715357 977769 163558
199606 896584 051237 541638 188580 280321 \]\[\times 741 640062 627530 801524 787141
901937 474059 940781 097519 023905 821316 144415 759504 705008 092818 711693
940737 \]...

So it would be 3 times 5 = 15.
I dont really know how to do it by hand though.

Answer 15

ya, its 15.

Answer 16

You should probably just somehow prove that 3 and 5 are prime factors of this number. I do not see a clever method to factorize it by hand since it is so gigantic.

Answer 17

it factors by difference of squares down to\[(2^2-1)(2^2+1)\cdot\]is how you do it!

Answer 18

times a bunch of other stuff...

Answer 19

My apologies my answer was wrong :(

Answer 20

keep dividing the exponent by 2 in every difference of squares factoring and you will get my answer\[(2^2-1)(2^2+1)(2^4+1)(2^8+1)\cdots\]

Answer 21

lol, 2^1024-1= (2^512 +1)(2^256 +1)(2^128 +1)(2^64 +1)(2^32+1)(2^16+1)
(2^8+1)(2^4 +1)(2^2 +1)(2-1)(2+1)

Answer 22

yep :D
I used to suck at these things, but with you guys around I guess I'm getting better!

Answer 23

we only need (2^2 +1)(2-1)(2+1) = 15

Answer 24

Turing I am having a bad day : http://openstudy.com/users/foolformath#/updates/4f367ecde4b0fc0c1a0ce793

Answer 25

I wish that was how my bad days were
I can't even add on mine!

Answer 26

haha I don't believe you Turing

Answer 27

Turing here is another way for solving the same problem http://openstudy.com/study#/updates/4f376f36e4b0fc0c1a0d4f7c