is $324+455^n$ ever prime Another question that I can only solve in part. 
Is there an $n$ such that $324+455^n$ is prime?
When $n$ is odd, this is false since 
$$ 324+455^n = (2\cdot 3^2)^2+(5\cdot 91)^n \equiv (-1)^2+(5\cdot 15)^n \equiv 1+(6\cdot 3)^n \equiv 1+(-1)^n\pmod{19}$$
so that $19\mid 324+455^n$ whenever $n$ is odd. Now, what about $n$ even?

Let's analyse three possible cases:
If $n\equiv 1\pmod 2$, we have $455^n+324 \equiv 0\pmod {19}$ because
$$\begin{eqnarray}
(455^1+324) & = & 19\times 41 \\
(455^2-1) & = & 19\times 10896\\
\end{eqnarray}$$
If $n\equiv 2\pmod 4$, we have $455^n+324 \equiv 0\pmod {17}$ because
$$\begin{eqnarray}
(455^2+324) & = & 17\times 12197\\
(455^4-1) & = & 17\times 2521138272\\
\end{eqnarray}$$
Finally, if $n=4k$, we can apply the equality $$x^4+4y^4 = (x^2 + 2xy + 2y^2)(x^2 - 2xy + 2y^2)$$ to $x=455^k$ and $y=3$ to obtain $$455^{4k}+4\cdot3^4 = (455^{2k} + 6\cdot 455^k + 18)(455^{2k}-6\cdot 455^k + 18)$$
Since both factors are strictly greater than $1$, their product is certainly a composite number.
Thus, the quantity $455^n+324$ is not a prime for any natural number $n$.
