NMO Selection Tests for the Junior Balkan Mathematical Olympiad

$1^{\circ }5p+x\mid 5p^n+x^n\text{ for all }n\ge 1;$ $2^{\circ }5p+x\mid 30p^2.$ 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 $x$.