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?
Best Answer
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 $2\leq n$