is $324+455^n$ ever prime

Question

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?

Answer 1

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 $n$.

Answer 2

I've just checked for small $n$, and I can state that $$5 \mid 324+455^0\\17\mid 324+455^2\\89\mid324+455^4\\17\mid 324+455^6$$ I think these examples help to see a path...in fact $$324+455^{2+4k}\equiv1+455^2\cdot455^{4k}\equiv1-1\cdot(455^4)^k\equiv1-1\cdot1^k\equiv1-1\equiv0 \pmod{17}$$