Show that $n^4+4$ is not a prime number

One can also factor directly

$$n^4+4=(n^4+4n^2+4)-4n^2=(n^2+2)^2-4n^2=(n^2-2n+2)(n^2+2n+2)$$

So no need of induction. This also explains the condition, since to get a proper factorization from the above one needs

$$2 \leq n^2-2n+2$$

which reduces to $2\leq n$

Do you know Fermat's little theorem? If $p$ is prime and $\gcd(p, b) = 1$, then $b^{p - 1} \equiv 1 \pmod p$.

So, if $b$ is not a multiple of 5, then $b^4 \equiv 1 \pmod 5$ and $b^4 + 4 \equiv 0 \pmod 5$.

If $b$ is a multiple of 10, then $b^4 + 4 \equiv 4 \pmod{10}$.

This does not address odd multiples of 5; I might or might not figure it out later.