# How do you factor 2^12 + 1?

Dec 6, 2016

${2}^{12} + 1 = 17 \cdot 241$

#### Explanation:

The sum of cubes identity can be written:

${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$

Use this with $a = {2}^{4}$ and $b = 1$ as follows:

${2}^{12} + 1 = {\left({2}^{4}\right)}^{3} + {1}^{3}$

$\textcolor{w h i t e}{{2}^{12} + 1} = \left({2}^{4} + 1\right) \left({2}^{8} - {2}^{4} + 1\right)$

$\textcolor{w h i t e}{{2}^{12} + 1} = \left(16 + 1\right) \left(256 - 16 + 1\right)$

$\textcolor{w h i t e}{{2}^{12} + 1} = 17 \cdot 241$

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 $5$.

• $241 = 210 + 28 + 3 = 34 \cdot 7 + 3$ is not divisible by $7$

• $241 = 220 + 22 - 1 = 22 \cdot 11 - 1$ is not divisible by $11$

• $241 = 260 - 13 - 6 = 19 \cdot 13 - 6$ 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.