Somemathhere

For what values of

$n$ is the polynomial

$x^n+64$ reducible over

$\mathbb{Z}$ ? Using the factorizations of

$x^3+4^3$ and

$x^4+4\cdot 2^4$ we see that if

$n$ is divisible by

$4$ or

$3$ the polynomial is reducible. Let us show no other cases exist. Let

$\alpha_i$ be the

$n$ -th roots of

$-1$ (in some order). Observe that a product of thеse roots is always of modulus

$1$ hence if it is real number it is either

$1$ or

$-1$ . Assume that

$x^n+64=g(x)h(x)$ ,

$g$ and

$h$ - rational, nonconstant. Then

$g(x)$ corresponds to some roots of

$x^n+64$ , say

$2^{\frac{6}{n}}\alpha_i$ ,

$i=1,2,\ldots,k$ ,

$k\le n$ . Since

$g$ is rational, then

$\prod_{i=1}^{k}2^{\frac{6}{n}}\alpha_i\in \mathbb{Q}$ . So

$2^{\frac{6k}{n}}$ is rational, and moreover it is a divisor of

$2^6$ . Hence

$6k=ns$ for some

$s=0,1,\ldots,6$ . Cases

$s=0,6$ are trivially impossible (the product of these roots cannot be

$1$ because they are all with modulus

$>1$ ). Cases

$s=2,4,5$ imply that

$3$ divides

$n$ which is indeed a solution, a...

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