The age of a person (in years) in $$2025$$ is a perfect square. His age (in years) was also a perfect square in 2012.
His age (in years) will be a perfect cube 𝑚 years after $$2025$$. Determine the smallest value of $$m$$.

Let the person’s age in the year $$2012$$ be $$n^2$$ years, where $$n$$ is a positive integer.

Thirteen years later, in $$2025$$, the age becomes $$n^2+13$$. We are told this is also a perfect square, say $$k^2$$. Hence

$$k^2 - n^2 = 13 \quad -(1)$$

Factorising the left side using the identity $$k^2-n^2=(k-n)(k+n)$$ gives

$$(k-n)(k+n)=13 \quad -(2)$$

Because $$13$$ is prime, the only pair of positive integers whose product is $$13$$ is $$1$$ and $$13$$. Thus

$$k-n = 1 \quad\text{and}\quad k+n = 13$$

Adding the two equations: $$2k = 14 \;\Longrightarrow\; k = 7$$.
Substituting back: $$n = k-1 = 6$$.

Therefore

Age in $$2012 = n^2 = 6^2 = 36 \text{ years}$$
Age in $$2025 = k^2 = 7^2 = 49 \text{ years}$$.

Let $$m$$ be the number of years after $$2025$$ when the age first becomes a perfect cube. We require

$$49 + m = p^3$$ for some integer $$p$$, with $$m \gt 0$$ and as small as possible.

The cubes just above $$49$$ are

$$4^3 = 64,\quad 5^3 = 125, \ldots$$

The first one that exceeds $$49$$ is $$64$$, and

$$64 - 49 = 15$$.

Thus the earliest perfect-cube age is $$64$$ years, reached $$m = 15$$ years after $$2025$$.

Smallest value of $$m = 15$$.

Answer: 15