How do you factor #2^12 + 1#?

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Answer 1 · 2016-12-06T00:30:47

#2^12+1 = 17*241#

The sum of cubes identity can be written:

#a^3+b^3=(a+b)(a^2-ab+b^2)#

Use this with #a=2^4# and #b=1# as follows:

#2^12+1 = (2^4)^3+1^3#

#color(white)(2^12+1) = (2^4+1)(2^8-2^4+1)#

#color(white)(2^12+1) = (16+1)(256-16+1)#

#color(white)(2^12+1) = 17*241#

You are probably aware that #17# is prime.

What about #241#?

#241# ends with an odd digit, so is not divisible by #2#.
#2+4+1 = 7# is not divisible by #3#, so #241# is not divisible by #3#.
#241# does not end in #0# or #5#, so is not divisible by #5#.
#241 = 210+28+3 = 34*7+3# is not divisible by #7#
#241 = 220 + 22 - 1 = 22*11 - 1# is not divisible by #11#
#241 = 260 - 13 - 6 = 19*13-6# is not divisible by #13#
#241 < 289 = 17^2#, so we have checked all the primes we need to and have established that #241# is prime.