Saudi Arabia Mathematical Competitions 2012

If $(2p{)}^n+1=a^3$, then we have $$(2p{)}^n=a^3-1=(a-1)(a^2+a+1)=(a-1)[(a-1{)}^2+3(a-1)+3].$$ Since $a$ is odd, it follows that $a^2+a+1$ is also odd, hence $2^n\nmid a-1$. This means that we have $a=2^n{p}^k+1$ for some integer $k\ge 0$.

We obtain $$2^n{p}^n=2^n{p}^k(2^{2n}{p}^{2k}+3\cdot 2^n{p}^k+3).(1)$$

Case 1. $k>0$. Clearly, we have $k<n$, hence from (1) it follows $p|2^{2n}{p}^{2k}+3\cdot 2^n{p}^k+3$. We get $p|3$, not possible since $p\ge 5$.

Case 2. $k=0$. From (1) we obtain $$p^n=4^n+3\cdot 2^n+3.(2)$$ For $n=1$ we have $p=13$.

For $n=2$ we have $p^2=31$, not possible.

For $n=3$ we have $p^3=64+24+3<5^3$, not possible.

For $n\ge 3$ we prove by induction that $$5^n>4^n+3\cdot 2^n+3=p^n,$$ and so $p<5$, which is not possible.

The unique solutions are $n=1,p=13$.