Baltic Way 2019

No such numbers $x$ can ever exist. To see this, let us recall that when $x$ is an integer, the right-hand side of the equation is always $\equiv 0(\mathrm{mod}3)$ or $1(\mathrm{mod}3)$. We recall also that when $x$ and $y$ are integers, we have $$P(x)\equiv P(y)(\mathrm{mod}3),\text{whenever}x\equiv y(\mathrm{mod}3).$$ Therefore $$P(x)\equiv P(-3)\equiv -40\equiv 2(\mathrm{mod}3)\text{when}x\equiv 0(\mathrm{mod}3),$$ and similarly $$P(x)\equiv P(-5)\equiv -156\equiv 0(\mathrm{mod}3)\text{when}x\equiv 1(\mathrm{mod}3),$$ and $$P(x)\equiv P(-1)\equiv -4\equiv 2(\mathrm{mod}3),\text{when}x\equiv 2(\mathrm{mod}3).$$ In particular, $P(x)$ is always $\equiv 0(\mathrm{mod}3)$ or $\equiv 2(\mathrm{mod}3)$, and $P(P(x))$ is always $\equiv 2(\mathrm{mod}3)$. Therefore we always have $P(P(x))\not\equiv x^2(\mathrm{mod}3)$ for every integer $x$.