How do you factor #2^12 + 1#?
1 Answer
Explanation:
The sum of cubes identity can be written:
#a^3+b^3=(a+b)(a^2ab+b^2)#
Use this with
#2^12+1 = (2^4)^3+1^3#
#color(white)(2^12+1) = (2^4+1)(2^82^4+1)#
#color(white)(2^12+1) = (16+1)(25616+1)#
#color(white)(2^12+1) = 17*241#
You are probably aware that
What about

#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*136# 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.