Prove expression is not prime $$
(n + 4)^4 + 4
$$
If n is natural number, how to prove that above expression is not prime? 
I am stuck here
$$
(n+4)^2 \cdot (n+4)^2 + 2 . 2
$$
$$
\left(\left(n^2+4^2\right) \cdot 2\right)\left(\left(n^2+4^2\right) \cdot 2\right)
$$
 A: From the context of the question, I assume $n ≥ 0$.
Clearly for even $n$, $(n+4)^4+4$ is even. Hence suppose $n$ is odd, so $n = 2k+1$. Then $(n+4)^4+4 = (2k+5)^4+4 = 16 k^4 + 160 k^3 + 600 k^2 + 1000 k + 629 = (4 k^2 + 16 k + 17) (4 k^2 + 24 k + 37)$.
Clearly this is composite, as for $n≥0$, this is a product of two natural numbers greater than $1$.
A: Try to write
a more general form
as the difference of two squares
which we know how to factor.
$\begin{array}\\
x^4 + a
&=(x^2+u)^2-b^2x^2\\
&=x^4+2ux^2+u^2-b^2x^2\\
&=x^4+(2u-b^2)x^2+u^2\\
\text{and}\\
x^4 + a
&=(x^2+u+bx)(x^2+u-bx)\\
\end{array}
$
so
$u^2=a$
and
$b^2 = 2u$.
Therefore
$b^4 = 4u^2
= 4a
$.
Writing $c^2$ for $a$,
this becomes
$\begin{array}\\
x^4 + c^2
&=(x^2+u)^2-b^2x^2\\
&=x^4+2ux^2+u^2-b^2x^2\\
&=x^4+(2u-b^2)x^2+u^2\\
\text{so}\\
u
&=c\\
\text{and}\\
x^4 + c^2
&=(x^2+c+bx)(x^2+c-bx)\\
\end{array}
$
so
$b^2 = 2c$.
Since $b$ and $c$
are integers,
$c = 2d^2$
and
$b^2 = 4d^2$
so $b = 2d$.
Therefore,
the expression can 
only be factored
if it is of the form
$x^4+4d^4$
when the factorization is
$x^4+4d^4
=(x^2+2d^2)^2-4d^2x^2
=(x^2+2dx+2d^2)(x^2-2dx+2d^2)
$.
This is the case
$d=1$
so $b=2$
and the factorization is
$x^4+4
=(x^2+2x+2)(x^2-2x+2)
$.
A: Hint: $(x+4)^4+4=0$ has roots $x = -4 \pm \sqrt{\pm 2i}=-4 \pm (1 \pm i)\,$, so the expression factors as: $$(x+4)^4+4=(x^2 + 6 x + 10) (x^2 + 10 x + 26)$$
