Prove a number is composite How can I prove that $$n^4 + 4$$ is composite for all $n > 5$? 
This problem looked very simple, but I took 6 hours and ended up with nothing :(. I broke it into cases base on quotient remainder theorem, but it did not give any useful information.
Plus, I try to factor it out:
$$n^4 - 16 + 20 = ( n^2 - 4 )( n^2 + 4 ) - 5\cdot4$$
If a composite is added to a number that is a multiple of $5$, is there anything special? A hint would suffice.
Thanks,
Chan 
 A: Factoring yields
$$
\begin{align}
n^4+4
&=(n^2+2i)(n^2-2i)\\
&=(n+1+i)(n-1-i)(n+1-i)(n-1+i)\\
&=(n+1+i)(n+1-i)(n-1-i)(n-1+i)\\
&=((n+1)^2+1)((n-1)^2+1)
\end{align}
$$
So for $n>1$, $n^4+4$ is composite.
A: $$ x^4+4=\\
[(x^2)^2+4x^2+4]-4x^2\\
=(x^2+2)^2-(2x)^2\\
=(x^2+2x+2)(x^2-2x+2)\ldots $$
A: In fact, it is true for $n>1$. You just need a clever way of factoring the expression:
$$n^4 + 4 = n^4 + 4 + 4n^2 - 4n^2 = \left ( \cdots + \cdots - \cdots \right ) \cdot \left ( \cdots + \cdots + \cdots \right )$$
A: You can factor out $n^{4}+4$ algebraically by finding the four roots of $n^{4}+4=0$.
Since $n^{4}+4=0\Leftrightarrow n^{4}=4e^{i\pi }$, we have
$$\begin{eqnarray*}
n &=&4^{1/4}e^{i (\pi +2k\pi)/4}\quad k=0,1,2,3 \\
&& \\
n &=&\sqrt{2}e^{i\pi /4 }=1+i\quad \left( k=0\right) 
\\[2ex]
n &=&\sqrt{2}e^{i 3\pi /4 }=-1+i\quad \left( k=1\right) 
\\[2ex]
n &=&\sqrt{2}e^{i 5\pi/4 }=-1-i\quad \left( k=2\right) 
\\[2ex]
n &=&\sqrt{2}e^{i 7\pi/4}=1-i\quad \left( k=3\right). 
\end{eqnarray*}$$
Now combining the complex conjugates factors, we get
$$\begin{eqnarray*}
n^{4}+4 &=&\left( n-1-i\right) \left( n+1-i\right) \left( n+1+i\right)
\left( n-1+i\right)  \\
&=&\left( \left( n+1-i\right) \left( n+1+i\right) \right) \left( \left(
n-1-i\right) \left( n-1+i\right) \right)  \\
&=&\left( n^{2}+2n+2\right) \left( n^{2}-2n+2\right). 
\end{eqnarray*}$$
Note: for $n>1$, $n^2+2n+2>5$ and $n^2-2n+2>1$.
A: This is a special case of a class of cyclotomic factorizations due to Aurifeuille, Le Lasseur and Lucas, the so-called Aurifeuillian factorizations of cyclotomic polynomials $\rm\;\Phi_n(x) = C_n(x)^2 - n\ x\  D_n(x)^2\;$. These play a role in factoring numbers of the form $\rm\; b^n \pm 1\:$, cf. the Cunningham Project. Below are some simple examples of such factorizations:
$$\begin{array}{rl}
x^4 + 2^2 \quad=&  (x^2 + 2x + 2)\;(x^2 - 2x + 2) \\\\
\frac{x^6 + 3^3}{x^2 + 3} \quad=&  (x^2 + 3x + 3)\;(x^2 - 3x + 3) \\\\
\frac{x^{10} - 5^5}{x^2 - 5} \quad=&  (x^4 + 5x^3 + 15x^2 + 25x + 25)\;(x^4 - 5x^3 + 15x^2 - 25x + 25) \\\\
\frac{x^{12} + 6^6}{x^4 + 36} \quad=&  (x^4 + 6x^3 + 18x^2 + 36x + 36)\;(x^4 - 6x^3 + 18x^2 - 36x + 36) \\\\
\end{array}$$
A: Try factoring as the product of two quadratic expressions: $n^4+4=(n^2+an+b)(n^2+cn+d)$.
