All of them should be prime

Consider the expression $n^3+1$ , it can factorized to $(n+1)(n^2-n+1)$ . Let $n^3+1$ is prime number, either $(n+1)$ or $(n^2-n+1)$ is $1$ .

Now, if $n+1=1$ , then $n=0$ , which means $n^3+1=1$ which is not a prime nunber. If $n^2-n+1=1$ , then $n=0$ or $n=1$ , which means that $n^4+5=21$ (n=1). Hence, the answer is no.

$\begin{aligned}\text{Since }&\text{n is an integer,}\\ n&\equiv 0,1\text{ or }2\pmod{3}\\\\ \underline{Case:1 \hspace{4mm}\color{#3D99F6} n\equiv 0\pmod{3}}\\ n&\equiv 0\pmod{3}\\ \implies n^2+3&\equiv 0\pmod{3}\\ \implies n^2+3 &\text{ is not prime }\small \color{#3D99F6}\hspace{4mm} n=0 \text{ is not valid as it implies }n^3+1=1 \text{ which is not prime.}\\\\ \underline{Case:2 \hspace{4mm}\color{#3D99F6} n\equiv 1\pmod{3}}\\ n&\equiv 1\pmod{3}\\ \implies n^4+5&\equiv 0\pmod{3}\\ \implies n^4+5 &\text{ is not prime }\\\\ \underline{Case:3 \hspace{4mm}\color{#3D99F6} n\equiv 2\pmod{3}}\\ \text{We have,}\\ n^{2x}&\equiv 1\pmod{3}\\ n^{2x+1}&\equiv 2\pmod{3}\\ \implies n^4&\equiv 1\pmod{3}\\ \implies n^4+5&\equiv 0\pmod{3}\\ \implies n^4+5 &\text{ is not prime }\\\\ \implies \text{At least one of } &n^2+3,n^3+1,n^4+5 \text{ will be composite for all } n \end{aligned}$

take $n^4+5$ modulo 3.If,n is not divisible by 3 then this is 0 (mod 3).Again,if n is divisible by 3,then the first term is divisible by 3.For that to be prime n should be 0,which contradicts that n^3+1 is a prime.