<s>[INST] 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. 
 [/INST] 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.


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

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

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*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 $(a-b)$

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