The DANUBE Mathematical Competition

The equation can be written as $x^4-x^2+1=21^y$, i.e., $(x^2-x+1)(x^2+x+1)=21^y$. The number in the left hand side is an integer, therefore $y$ must be non-negative. It is easy to see that $x^2-x+1$ and $x^2+x+1$ are co-prime. As $0<x^2-x+1<x^2+x+1$, we can only have $x^2-x+1=1$, $x^2+x+1=21^y$ or $x^2-x+1=3^y$, $x^2+x+1=7^y$. The first case leads to $x\in \{0,1\}$ and then to the solution $x=y=0$. In the second case, from the first equation we have $(2x-1{)}^2=4\cdot 3^y-3$, hence $3\mid (2x-1{)}^2$. It follows that $9\mid 4\cdot 3^y-3$, i.e., $3\mid 4\cdot 3^{y-1}-1$, and therefore $y=1$. We immediately obtain $x=\pm 2$. Thus, the equation has three solutions: $x=y=0$ and $x=\pm 2$, $y=1$.

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Alternative solution.

The equation can be written $(2x^2+1{)}^2=4\times 21^y-3$. The number in the left hand side is an integer, therefore $y$ must be non-negative. If $y=0$ we immediately get $x=0$. For $y=1$, we get $x=\pm 2$. For $y\ge 2$, the number $4\cdot 21^y-3$ is divisible by 3 but not by 9, and so it can not be a perfect square.