Divisibilty of $n^3+3$ and $n^5+5$ by a prime. Given that $n^3+3$ and $n^5+5$ are divisible by a prime $p$. How many values can $p$ take, with $n$ an integer?
The best I could do is $p=2$ for odd n, and I showed that $p$ cannot be equal to $3$. I know by subtracting the first equation to second, we will get 
$n^5-n^3+2=n^3(n-1)(n+1)+2$. The expression is clearly not divisible by $3$ which means $p$ cannot take value of $3$.
My only question is: Are there any values that $p$ can take? I've tried to solve it various ways, none worked.
If you can help me, I greatly appreciate it, thank you

Hint $\ {\rm mod}\ p\!:\ \begin{align}\color{#0a0}{n^3\equiv -3}\\\color{#c00}{n^5\equiv -5}\end{align}\Rightarrow\, (\color{#0a0}{-3})^5 \equiv(\color{#0a0}{n^3})^5\equiv(\color{#c00}{n^5})^3\equiv(\color{#c00}{-5})^3\Rightarrow\, 0\equiv 3^5\!-5^3\equiv 2\cdot 59 $
