$1^\circ \ 5p + x \mid 5p^n + x^n \text{ for all } n \ge 1;$
$2^\circ \ 5p + x \mid 30p^2.$
To prove $1^\circ \Rightarrow 2^\circ$, set $n = 2$ to obtain $5p+x \mid 5p^2+x^2 \Rightarrow 5p+x \mid (x+5p)(x-5p) + 30p^2$, hence $5p+x \mid 30p^2$, as needed. For the second implication, observe that $5p+x \mid 30p^2$ implies $5p+x \mid (x+5p)(x-5p) + 30p^2$, so $5p+x \mid 5p^2+x^2$. The identity $5p^{n+1} + x^{n+1} = (5p^n + x^n)(p+x) - px(5p^{n-1} + x^{n-1})$ holds for all integers $n \ge 2$. Consequently, by induction, the claim $1^\circ$ is proved.
The divisors of $30p^2$ greater than $5p$ are $6p, 10p, 15p, 30p, p^2, 2p^2, 3p^2, 5p^2, 6p^2, 10p^2, 15p^2$ and $30p^2$. Subtracting $5p$ from the numbers listed above yields the required values for $x$.