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

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?

• $$n^4+4 = n^4+4 + 4n^2 - 4n^2 = (n^2+2)^2 - (2n)^2 = (n^2+2n+2)(n^2-2n+2)$$ The factorization I employed here is known as the Sophie Germain Identity – r9m Jan 27 '15 at 5:13
• @r9m That should just be the answer. – Emily Jan 27 '15 at 5:14
• Exploration and a little reflection shows that, for odd $n$, if $5\nmid n,$ then $5|(n^4+4)$. But the general answer is r9m's factorization. – Joffan Jan 27 '15 at 5:27
• – Martin Sleziak Jan 27 '15 at 9:10

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.

Let us prove by induction. For $n=2$ , the result holds. Let the result hold for any $k$,

ie P$(k):k^4+4$ is not prime.

Consider $P(k+1)$=$(k+1)^4+4=((k+1)^2+4)^2-8(k+1)^2=(k^2+2k+5)^2-8k^2-16k-8$ Simplify and use the fact that $P(k)$ is true to conclude that $P(k+1)$ is true and hence the result.

• I think you want $((k+1)^2+2)^2-4(k+1)^2$, and the induction hypothesis is not used. – André Nicolas Jan 27 '15 at 5:28