Proving an expression is composite I am trying to prove that $ n^4 + 4^n $ is composite if $n$ is an integer greater than 1. This is trivial for even $n$ since the expression will be even if $n$ is even. 
This problem is given in a section where induction is introduced, but I am not quite sure how induction could be used to solve this problem. I have tried examining expansions of the expression at $n+2$ and $n$, but have found no success.
I would appreciate any hints on how to go about proving that the expression is not prime for odd integers greater than 1. 
 A: Hint: calculate this value explicitly for $n=1,3$ (or predict what will happen). Can you see any common factors? Can you prove that there is a number $m$ such that if $n$ is odd, then $m|(n^4 + 4^n)$?
Let me know if you need further hints.
A: $(n^2)^2+(2^n)^2=(n^2+2^n)^2-2^{n+1}n^2$. Since $n$ is odd...
A: I am very impressive with Adam's solution. There is very neat. So, I beg for a chance to write the full description about the proof step-by-step.


*

*We can transform $n^4+4^n$ to $(n^2+2^n)^2-2^{n+1}n^2$ as Adam's suggestion by

*

*$n^{(2^2)}+(2^2)^n = (n^2)^2+(2^n)^2$ associative law

*Now, we mention $(a+b)^2 = (a+b)(a+b) = a^2+2ab+b^2$ algebraic multiplication

*$(n^2)^2+(2^n)^2+2(n^2)(2^n)-2(n^2)(2^n)$ adding $+2ab-2ab$ to expression

*$(n^2)^2+2(n^2)(2^n)+(2^n)^2-2(n^2)(2^n)$ re-arrange the expression

*$(n^2+2^n)^2-2(n^2)(2^n)$ from step 2

*$(n^2+2^n)^2-2^{n+1}n^2$ law of Exponential


*We try to get the $(n^2+2^n)^2-2^{n+1}n^2$ to conform to $a^2-b^2$ because $a^2-b^2=(a+b)(a-b)$ algebraic multiplication, again

*

*Treat $n^2+2^n$ as $a$

*Since $n$ is odd, n+1 is even. So, we can assume $2m=n+1$, where $m$ is integer

*So, re-write the $2^{n+1}n^2$ to be $2^{2m}n^2$

*$2^{2m}n^2=(n2^m)^2$ associative law

*Treat $n2^m$ as $b$


*It implies that both $a$ and $b$ are both positive integer

*From $a^2-b^2=(a+b)(a-b)$ and the result of $n^4 + 2^4$, it implies that $a$ is greater than $b$

*Hence both $(a+b)$ and $(a-b)$ are positive integer, that causes the result of $n^4 + 2^4$ is combination of $(a+b)$ and $(a-b)$

A: Some interesting factorizations of a polynomial of type $x^4+\text{const}$:
$$ x^4+4=(x^2+2x+2)(x^2-2x-2) \tag{1}$$
$$ x^4+1=(x^2+\sqrt[]{2}x+1)(x^2-\sqrt[]{2}x+1) \tag{2}$$
So one can ask, how to select the coefficients $a,b,c,d$ in
$$(x^2+ax+b)(x^2+cx+d) \tag{3}$$
such that all coefficients of the resulting polynomial are zero except the constant term and the coefficient of the 4th power. The latter is $1$.
If we expand $(3)$  we get
$$x^4+(c+a)x^3+(d+a c+b)x^2+(a d+b c)x+b d$$
And the coefficients disappear, if
$$ 
\begin{eqnarray}
c+a &=& 0 \\
d+ ac +b &=& 0 \\
ad+bc &=& 0
\end{eqnarray}
$$
When solving for $b,c,d$ we get
$$
\begin{eqnarray}
c &=& -a \\
b &=& \frac{a^2}{2} \\
d &=& \frac{a^2}{2} 
\end{eqnarray}
$$
and therefore
$$x^4+\frac{a^4}{4} = (x^2+a x+\frac{a^2}{2})(x^2-ax+\frac{a^2}{2})$$
For $a=2$ this gives $(1)$, $a=\sqrt[]{2}$ this gives $82)$ .
Substituting $a=2^{t+1}$ we get
$$x^4+4^{2t+1} = (x^2+2\cdot 2^t x+2^{2t+1})(x^2-2\cdot 2^tx+2^{2t+1})$$
Substituting $x=n=2t+1$ gives the required result for odd $n$.
