TITLE: Show that $n^4+4$ is not a prime number
QUESTION [6 upvotes]: How do you show that for all $n ∈ N, n ≥ 2,$
$n^4 + 4$ is not a prime number?
My attempt: 
I see that whatever number $n^4+4$ makes when $n$ is an even number would result to an even number. Thus it is not a prime since it can be at least divisible by 2. 
When I tried $n = x3$ where $x$ us any integer $n > 1$, then $n^4 + 4$ would at least be divisible by 5 which is also not a prime number. The $x$ means the number before 3, it does not mean it's being multiplied. The same rule applies for $n = x1$ and $n = x7$ where it results in $n^4 + 4$ being divisible by 5. 
Sorry if I'm not explaining with proper mathematical notation. 
For any number $n = x5$ (eg. $n = 5, n = 15 , n = 25 $…etc), the number is hard to determine if it's a prime. Eg. $5^4 + 4 = 629$, which is sometimes difficult to determine if it's a prime. But I think that any number $n = x5$ is going to make $n^4 + 4$ a composite number from the products of prime numbers. (Eg. $5^4 + 4 = 629 = 17*37$)
Is there another way of proving this?

REPLY [2 votes]: 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.