<s>[INST] Show that $n^{12} + 64$ has at least 4 non-trivial factors. 
Show that for every integer $n>1$, $n^{12} + 64$ has at least four factors different from $1$ and the number.(It has at least four non-trivial factors.)

My attempt:
We can expand $n^{12} + 64$  in the following way :
$=>n^{12} + 64 = (n^4)^3 + (2^2)^3$
$=>n^{12} + 64 = (n^4 + 2^2)((n^4)^2 + (2^2)^2 - (2^2)(n^4))$
$=>n^{12} + 64 = ((n^2 + 2)^2 - 2(2)(n^2))((n^4)^2 + (2^2)^2 - (2^2)(n^4)) $
$=>n^{12} + 64 = (n^2  + 2 + 2n)(n^2 + 2 - 2n)(n^8 + 2^4 - 4n^4)$
i.e there are 3 different factors.
How can I prove that there is one more factor ?.
Update: I think we are already done, because we got the following factors:
1.$(n^2  + 2 + 2n)$
2.$(n^2 + 2 - 2n)$
3.$(n^8 + 2^4 - 4n^4)$
4.$(n^2  + 2 + 2n)(n^2 + 2 - 2n)$
5.$(n^2  + 2 + 2n)(n^8 + 2^4 - 4n^4)$
6.$(n^2 + 2 - 2n)(n^8 + 2^4 - 4n^4)$
Clearly, we have got 6 non-trivial factors.(None of them can be one.)
 [/INST] Well we can write-
$n^{12}+64=(n^4)^3+4^3$
$=(n^4+4)(n^8-4n^4+16)$
$=[(n^2)^2+2(n^2)(2)+2^2-(2n)^2](n^8-4^2+4n^6-4n^6+12n^4-16n^4+16n^2-16n^2)$
$=[(n^2+2)^2-(2n)^2][(n^8+4^2+4n^4+4n^6+8n^4+16n^2)-(4n^6+16n^2+16n^4)]$
$=(n^2+2n+2)(n^2-2n+2)[(n^4+2n^2+4)^2-(2n^3+4n)^2]$
$=(n^2+2n+2)(n^2-2n+2)(n^4+2n^3+2n^2+4n+4)(n^4-2n^3+2n^2-4n+4)$
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