For how many $n$, $x^6+n$ factors? $\textbf{Question}.$ 
i) For how many integers $n$ with |$n$|$<500$, can the polynomial $p_n(x)=x^6+n$ be written as a product of two non-constant polynomials with integer coefficients?
ii) How will number of solutions depend on $k>0$ if we replace the condition |$n$|$<500$ by |$n$|$<k$. 
iii) Does there exist a $m>6$ such that $p_n(x)=x^m+n$ can never be factored for any $n>0$ 
$\textbf{Thoughts.}$ For last part if $n$ is prime, then by Eisenstein's criterion, it will never be factored as it will be irreducible whatever the $m$ be, but for any general integer $n$, it is too hard to say anything?
For first part that is case of |$n$|$<500$ , again for all primes less than $500$, it is false, so if we write $n=ab$, then $p_n(x)=x^6+ab=(x^3+a)(x^3+b) $ if $(a+b=0)$. But there can be many cases, is there a method or check for all?
 A: There are 7 factorizations for $n = 1, ..., 500$.  They pull off an $x^2 + m$ for $m = 1, ..., 7$.
See this Maple worksheet I wrote.  Hope this helps.
Regards,
Matt

As you have asked, I have a second worksheet to consider negative n.

Best of luck to you.
Matt
A: For the last part, if $m$ is odd, then for $n=a^m$ (and $a$ large enough) we have $$x^m+n=(x+a)(x^{m-1}+x^{m-2}a+\ldots +xa^{m-2}+a^{m-1}).$$
The same trick works if $m$ has an odd factor $>1$. Remains the case that $m$ is a power of $2$, hence a multiple of $4$, say $m=4k$. Then with $n=2a^m$ we have
$$ x^m+n = (x^{2k}+2a^{2k})^2-4a^{2k}x^{2k}=(x^{2k}+2a^kx^k+2a^{2k})(x^{2k}-2a^kx^k+2a^{2k}).$$
A: $$x^2-a^2=(x-a)(x+a)$$
$$x^3-a^3=(x-a)(x^2+ax+a^2)$$
$$x^3+a^3=(x+a)(x^2-ax+a^2)$$
In general, Using this identities we can factor $x^6+n$ for suitable $n.$
Note that $6=2\times 3.$
That means you have to find the number of $n$ such that $-500<n<500$ with the forms  $-a^2, -a^3, a^3.$
A: We always have, for all integers $a$
$$
x^6+a^3=(x^4 - ax^2 + a^2)(x^2+a).
$$
Hence the last question has a negative answer: there is no $m>1$ such that $x^6+n$ can never be factored for all $n\ge m$.
A: First, consider two cases: $(A, B, k > 0, k \in \mathbb{N})$
Case 1: k is odd.


*

*$A^k+B^k=(A+B)(A^{k-1}+A^{k-2}B+\dots+AB^{k-2}+B^{k-1})$

*$A^k-B^k=(A-B)(A^{k-1}-A^{k-2}B+\dots-AB^{k-2}+B^{k-1})$


Case 2: k is even.


*

*$A^k+B^k$ cannot be factorized this way

*$A^k-B^k=(A-B)(A^{k-1}-A^{k-2}B+\dots-AB^{k-2}+B^{k-1})$

*$A^k-B^k=(A+B)(A^{k-1}+A^{k-2}B+\dots+AB^{k-2}+B^{k-1})$


Assuming that no other ways of factorizing polynomial applies, the polynomial $(x^6 + n)$ can only be factored into two polynomials with integer coefficients in the following cases: 
(a, b, c are integers larger than $0$)


*

*$n =-a^2$, $a^2 \leq 500 \Rightarrow a \in [1,22]$, for the case $x^6-a^2=(x^3-a)(x^3+a)$

*$n = b^3$, $b^3 \leq 500 \Rightarrow b \in [1,7]$, for the case $x^6+b^3=(x^2+b)(x^4-x^2 b + b^2)$

*$n = -b^3$, $b^3 \leq 500 \Rightarrow b \in [1,7]$, for the case $x^6-b^3=(x^2-b)(x^4+x^2 b + b^2)$

*$n = -c^6$, $c^3 \leq 500 \Rightarrow c \in [1,2]$, for the case $x^6-c^6=(x-c)(x^5+x^4 c+x^3 c^2 + x^2 c^3 +x c^4 + c^5)$

*$n = -c^6$, $c^3 \leq 500 \Rightarrow c \in [1,2]$, for the case $x^6-c^6=(x+c)(x^5-x^4 c+x^3 c^2 - x^2 c^3 +x c^4 - c^5)$


In total, there are $22+7+7+2+2 = 40$ values of n satisfying the conditions. Note that certain values are counted more than once, since we are counting way to distinctly factorize the polynomial. For examples, $n = -64$ is counted twice, as $x^6-64=(x^3-8)(x^3+8)=(x^2-4)(x^4+4x^2+16)$
If $n=0$ is allowed, $x^6=x^5x=x^4x^2=x^3x^3$, adding 3 additional values.

Another way of approaching the problem:
Let $(x^6+n)=(A_5 x^5+A_4 x^4+A_3 x^3+A_2 x^2+A_1 x^1+A_0)(B_5 x^5+B_4 x^4+B_3 x^3+B_2 x^2+B_1 x^1+B_0)$
Solve the system of equations:


*

*$A_5B_5=0$

*$A_5B_4+A_4B_5=0$

*$\dots$

*$A_5B_1+A_4B_2+A_3B_3+A_2B_4+A_1B_5=1$

*$A_5B_0+A_4B_1+A_3B_2+A_2B_3+A_1B_4+A_0B_5=0$

*$\dots$

*$A_0B_0=n$

*$A_i, B_i, n \in \mathbb{Z}, -500<n<500$

