Is $83^{27} +1 $ a prime number?

$83$ is odd, so is any power of $83$. Hence $83^{27}+1$ is even, but the only even prime number is $2$ and this number is not $2$.

More generally, if $a,k\in\mathbb N$ and $k$ is odd, then $$a^k+1\equiv (-1)^k+1\equiv 0\pmod{a+1}$$ So $a+1\mid a^k+1$. In this case this yields $84=2^2\cdot 3\cdot 7$ as divisor.

Well, it is an even number, so...

$$ 83^{27} + 1 = \Big(83^9\Big)^3 + 1^3 = a^3+b^3 = (a+b)(a^2-ab+b^2) = \Big(83^9+1\Big)\Big((83^9)^2-83^9+1\Big). $$

So, no, it's not prime.

PS (added later): Some point out that it's obviously an even number, so it's not prime. But what I do above would work just as well if it were $84$ rather than $83$.