<s>[INST] 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 
 [/INST] 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:
$$\begin{array}{rl}
x^4 + 2^2 \quad=&  (x^2 + 2x + 2)\;(x^2 - 2x + 2) \\\\
\frac{x^6 + 3^3}{x^2 + 3} \quad=&  (x^2 + 3x + 3)\;(x^2 - 3x + 3) \\\\
\frac{x^{10} - 5^5}{x^2 - 5} \quad=&  (x^4 + 5x^3 + 15x^2 + 25x + 25)\;(x^4 - 5x^3 + 15x^2 - 25x + 25) \\\\
\frac{x^{12} + 6^6}{x^4 + 36} \quad=&  (x^4 + 6x^3 + 18x^2 + 36x + 36)\;(x^4 - 6x^3 + 18x^2 - 36x + 36) \\\\
\end{array}$$
</s>