Estonian Math Competitions

Note that $1001=11\cdot 91$. If $11\mid a$ then also $11\mid a^{2020}+10a^{1010}+1001$. Otherwise, consider $a^5$ modulo $11$. A case study shows that if $a$ is congruent to $1$, $3$, $4$, $5$, or $9$, then $a^5$ is congruent to $1$, and in all other cases, $a^5$ is congruent to $-1$. Hence $a^{10}\equiv 1(\mathrm{mod}11)$ whenever $11\nmid a$. This implies that $a^{1010}\equiv 1(\mathrm{mod}11)$ and $a^{2020}\equiv 1(\mathrm{mod}11)$ because $a^{1010}=(a^{10}{)}^{101}$ $(\mathrm{mod}11)$ and $a^{2020}=(a^{10}{)}^{202}$. Consequently, $10a^{1010}\equiv 10(\mathrm{mod}11)$, implying that $a^{2020}+10a^{1010}+1001\equiv 0(\mathrm{mod}11)$. Thus $a^{2020}+10a^{1010}+1001$ cannot be prime since obviously $a^{2020}+10a^{1010}+1001\ge 1001>11$.