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 

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 $n^2-2n+2>1$.
