Is $83^{27} +1 $ a prime number? I'm having problems with exercises on proving whether or not a given number is prime. Is $83^{27} + 1$ prime?

We have a chain of divisibilities, based on the fact that $(a-b)\mid(a^n-b^n)$,
$$
83^1-(-1)^1\mid83^3-(-1)^3\mid83^9-(-1)^9\mid83^{27}-(-1)^{27}=83^{27}+1
$$
Using this chain, we get, using $a^3-b^3=(a-b)(a^2+ab+b^2)$,
$$
\begin{align}
83^{27}+1
&=\frac{83^{27}+1}{83^9+1}\times\frac{83^9+1}{83^3+1}\times\frac{83^3+1}{83^1+1}\times\left(83^1+1\right)\\
&=\left(83^{18}-83^9+1\right)\times\left(83^6-83^3+1\right)\times\left(83^2-83^1+1\right)\times\left(83^1+1\right)\\[9pt]
&=34946659039493167203883141969862007\times326939801583\times6807\times84
\end{align}
$$
Thus, $83^{27}+1$ is not prime.
Note: none of these factors are guaranteed to be prime, just factors.
