THE 68th NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD

Let $y$ be a prime such that $y^3=2^x+x^2+25$; clearly, $y$ is odd. Moreover, $x$ can not be a negative integer nor can it belong to the set $\{0,1,2,3\}$. Hence $x\ge 4$ is even.

1. If $x=6k$, with $k\in {\mathbb{N}}^{\ast }$, we get $64^k+36k^2+25=y^3$.

For $k=1$ we get $x=6$ and $y=5$.

For $k\ge 2$ we have no solutions because of the following inequalities: $$(4^k{)}^3<y^3=2^{6k}+(6k{)}^2+25<(4^k+1{)}^3,$$ the second inequality being equivalent with $3k^2+2<4^{k-1}(4^k+1)$, which is true (induction).

2. If $x=6k+2$ or $x=6k+4$ then: $2^x\equiv 1(\mathrm{mod}3)$, $x^2\equiv 1(\mathrm{mod}3)$ hence $2^x+x^2+25\equiv 0(\mathrm{mod}3)$, i.e. $y^3\equiv 0(\mathrm{mod}3)$, which means $y\equiv 0(\mathrm{mod}3)$ and, as $y$ is prime, we get $y=3$, which does not fulfill the condition.

In conclusion, the only solution is $x=6$.