A natural number $n$ is said to be a perfect square when there exists a natural number $x$ satisfying $n = x ^ { 2 }$. Similarly, $n$ is said to be a perfect cube when there exists a natural number $x$ satisfying $n = x ^ { 3 }$.
In the following two cases, find the natural number $n$ that satisfies the conditions.
(i) $n$ is a perfect square. Furthermore, the number obtained by adding 13 to $n$ is also a perfect square.
(ii) $n$ is a perfect cube. Furthermore, the number obtained by adding 61 to $n$ is also a perfect cube.
First, consider (i). From the definition of a perfect square number, $n$ can be expressed as $n = x ^ { 2 }$, where $x$ is a natural number. In addition, there exists a natural number $y$ such that
$$x ^ { 2 } + 13 = y ^ { 2 } .$$
Since $x < y$, $y - x = \square \mathbf { J }$ and $y + x = \mathbf { K L }$. It follows that
$$x = \mathbf { M } , \quad y = \mathbf { N } ,$$
and finally that $n = \mathbf { O P }$.
Next, consider (ii). Similar to (i), in (ii), there exists a natural number $x$ such that $n = x ^ { 3 }$, and there also exists a natural number $y$ such that
$$x ^ { 3 } + 61 = y ^ { 3 } .$$
When we solve this equation, we obtain
$$x = \mathbf { Q } , \quad y = \mathbf { R } ,$$
and hence the perfect cube $n = \mathbf{ST}$.