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, as shown earlier. It remains to consider $s=3$, $n=2k$. If $k$ is even, $4$ divides $n$ which again is a solution. Assume now $k$ is odd. Hence $g$ is of odd power, so it has real root. But that means a real root for $x^n+64$ which has no real roots since $x^n+64=(x^k)^2+64\ge 64$ when $x$ is real.
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, as shown earlier. It remains to consider $s=3$, $n=2k$. If $k$ is even, $4$ divides $n$ which again is a solution. Assume now $k$ is odd. Hence $g$ is of odd power, so it has real root. But that means a real root for $x^n+64$ which has no real roots since $x^n+64=(x^k)^2+64\ge 64$ when $x$ is real.
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