How about irrational number?

Let $d$ be the gcd of $(8\cdot n - 25)$ and $(n + 5)$ , then $d$ is a divisor of $\boxed{8\cdot (n + 5) - (8\cdot n - 25) = 65}(1)$

Let $A = \frac{a^3}{b^3}$ , for $a, b$ coprime integers. Then $8\cdot n - 25 = d\cdot a^3$ and $n + 5 = d\cdot b^3$ . Hence

$(1)\Rightarrow d\cdot (8\cdot b^3 - a^3) = 65$ , for $d \in \Big\{1, 5, 13, 65\Big\}$ .

$1)$ For $d = 1$ we have $(2\cdot b)^3 - a^3 = 65$ . The only cubes that differ $65$ units are the pairs $(64, -1)$ and $(1, -64)$ . Only the former leads to the solution $\boxed{n = 3}$

$2)$ For $d = 5$ we have $(2\cdot b)^3 - a^3 = 13$ . There are no cubes that differ $13$ units.

$3)$ For $d = 13$ we have $(2\cdot b)^3 - a^3 = 5$ . There are no cubes that differ $5$ units.

$4)$ For $d = 65$ we have $(2\cdot b)^3 - a^3 = 1$ . The only cubes that differ $1$ unit are the pairs $(1, 0)$ and $(0, -1)$ , but none of them leads to a solution.