## Subject 1

Solve, in the set of integers, the equation $x^{2}\left(x^{2}+1\right)=21^{y}-1$.

Lucian Petrescu
Solution: (Lucian Petrescu) We can observe that $y \in \mathbb{N}$, and if $\left(x_{0}, y_{0}\right)$ is a solution, then $\left(-x_{0}, y_{0}\right)$ is also a solution of the equation. Moreover, $(0,0)$ is a solution. We rewrite the equation as $\left(x^{2}-x+1\right)\left(x^{2}+x+1\right)=21^{y}$. We notice that $x^{2}-x+1$ and $x^{2}+x+1$ are coprime. It follows that $x^{2}-x+1=1$ and $x^{2}+x+1=21^{y}$ with the solution $(0,0)$ or $x^{2}-x+1=3^{y}$ and $x^{2}+x+1=7^{y}$. From the first relation, we have $(2 x-1)^{2}=3\left(4 \cdot 3^{y-1}-1\right)$, which means $3 \mid (2 x-1)^{2}$, and more specifically, $9 \mid (2 x-1)^{2}$, so $9 \mid \left(4 \cdot 3^{y-1}-1\right)$, which gives the solution $y=1$, then $x=2$. Therefore, the set of solutions is $(-2,1),(0,0),(2,1)$.