Prove $a^4+4$ is composite for $a>1$ not using its algebraic factorization There is a famous problem that says that $a^4+4, a\in\Bbb N$ cannot be prime except than $a=1$. The solution based on factoring:
$$a^4+4=(a^2+2)^2-4a^2=(a^2+2-2a)(a^2+2+2a)$$
Usually, a problem in mathematics has more than one solution method.  But for this question, I couldn't find any way other than factoring.  As far as I can tell, it's nowhere to be found. So, I have a natural question. Is there any known way around this problem other than factoring?
This way can be induction.  It can be modular arithmetic or some techniques of number theory.
Clearly $a$ can not be even, and if $a=5k+n, 1\leq n\leq 4$, then $5\mid a^4+4$. But this doesn't work when $a\not\mid 5$.
But, I haven't seen any helpful hints that these work yet.
 A: If $a$ is divisible by $2$ then $a^4 + 4$ is a multiple of $4$.
If $a$ is not divisible by $5$ then $a^4 + 4$ is a multiple of 5 by modular arithmetic: $a^4 \equiv 1 \bmod 5$, so $a^4 + 4 \equiv 0 \bmod 5$.
If $a$ is an odd multiple of $5$ then I think this is a more subtle issue if you don't "see" that algebraic factorization of $a^4+4$ because there is no universal common prime factor, and in fact the two factors from the algebraic factorization can often both be prime:
$5^4 + 4 = 17 \cdot 37$, $15^4 + 4 = 197 \cdot 257$, $25^4 + 4 = 577 \cdot 677$, $55^4 + 4 = 2917 \cdot 3137$, $125^4 + 4 = 15377 \cdot 15877$
The difference of these prime factors has an easily spotted pattern: it is $4a$ when $a$ is an odd multiple of $5$, and armed with this information you might be led to discover the algebraic factorization of $a^4 + 4$ for all $a$.
Remark. As far as I am aware, the only reason $a^4 + 4$ is famous in math is due to its "unexpected" algebraic factorization. So that factorization is a good thing and not something to avoid knowing.
