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.

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