Mine. This arises from my research with Prof. Xiaoheng Wang and Gian Cordana Sanjaya about density of squarefree value of multivariate polynomials on special inputs. One special corollary of our result needed to calculate the number of residues $(a,b) \in \mathbb{Z}/p^2 \mathbb{Z} \times \mathbb{Z}/p^2\mathbb{Z}$ such that $a^3 + b^4 \equiv 0 \pmod{p^2}$, which I think has a pretty cute and neat solution and the idea of the solution

here is the original intent of the solution for the problem above.

We claim that the answer is 2028. We will use a well-known lemma as follows: If \( p \equiv 2 \pmod{3} \) is a prime number, then the map \( x \mapsto x^3 \) is surjective. To solve the problem, note that \( 2027 \equiv 2 \pmod{3} \) and 2027 is a prime number. We consider two cases: If \( 2027 \mid b \), then \( 2027 \mid b^5 + b^2 \), which implies \( a = b = 2027 \), and this works. Conversely, we have \[ \left( \frac{a^3}{b} \right)^2 + 1 \equiv -b^3 \pmod{2027}. \] Pick any \( \chi = \frac{a^3}{b} \in \{1, 2, \dots, 2027\} \). Then, the right-hand side (RHS) is uniquely defined, and from the claim, \( b \) is uniquely defined. Moreover, \( a^3 = b\chi \), and by the claim, \( a \) is uniquely defined. Therefore, there are 2027 solutions for this case. We must be cautious with the case when RHS \( \equiv 0 \pmod{2027} \), which leads to a contradiction. However, this is impossible because it implies \[ 2027 \mid \left( \frac{a^3}{b} \right)^2 + 1, \]but \( 2027 \equiv 3 \pmod{4} \), a contradiction.