Factorisation of numbers sum with exponent $12$ How can I prove that $n^{12}+64$ has at least four distinct factors other than $1$ and itself?
I applied $a^3+b^3$ identity.
 A: Using the familiar factorization of $a^3+b^3$, we get the factors $n^4+4$ and $n^8-4n^4+16$. Now note the perhaps a little surprising factorization $$n^4+4=(n^2-2n+2)(n^2+2n+2).$$
 If $n\gt 1$, then $n^2-2n+2$, $n^2+2n+2$, $(n^2-2n+2)(n^8-4n^4+16)$ and $(n^2+2n+2)(n^8-4n^4+16)$ are obviously distinct non-trivial factors. There are others.
A: The given statement is false. At $n=1$,
$$1+64=65=5\times 13$$
A: We have to factor $n^{12}+64$. I will show you how to do this
$$n^{12}+64$$
$$=n^{12}+2^6$$
$$=\left(n^4\right)^3+\left(2^2\right)^3$$
Rule: $a^3+b^3=(a+b)(a^2-ab+b^2)$
$$\left(n^4+2^2\right)\left(\left(n^4\right)^2-n^4\cdot 2^2+\left(2^2\right)^2\right)$$
$$=\left(n^4+4\right)\left(n^8-4n^4+16\right)$$
You cannot simplify this any further. There are only two factors. If you must find four factors, you can use the sum of squares, i.e. $a^2+b^2=(a+bi)(a-bi)$, where $i$ denotes the imaginary unit, to factor $n^4+4$
$$\left(n^2+2i\right)\left(n^2-2i\right)\left(n^8-4n^4+16\right)$$
For $n^8-4n^4+16$, let $n^4=x$.
$$n^8-4n^4+16=x^2-4x+16$$
Using quadratic formula:
$$x=\frac{4\pm\sqrt{(-4)^2-4(1)(16)}}{2}$$
$$x=\frac{4\pm\sqrt{16-64}}{2}$$
$$x=\frac{4\pm\sqrt{-48}}{2}$$
$$x=\frac{4\pm 4\sqrt{-3}}{2}$$
$$x=2\pm 2\sqrt{-3}$$
$$n^4=2\pm 2i\sqrt{3}$$
The remaining two factors are: $n^4-2-2i\sqrt{3}$ and $n^4-2+2i\sqrt{3}$
$$\color{green}{n^{12}+64=\left(n^2+2i\right)\left(n^2-2i\right)\left(n^4-2-2i\sqrt{3}\right)\left(n^4-2+2i\sqrt{3}\right)}$$
