Preselection tests for the full-time training

If $a$ is relatively prime with $35$ then it is relatively prime with both $5$, $7$. Since $a$ is relatively prime with $5$ then, by Fermat, $a^4\equiv 1\mathrm{mod}5$ which implies $$\left(a^4-1\right)\left(a^4+15a^2+1\right)\equiv 0\mathrm{mod}5$$ Since $a$ is relatively prime with $7$ then $a^6\equiv 1\mathrm{mod}7$. Hence $$\begin{array}{rl}\left(a^4-1\right)\left(a^4+15a^2+1\right)& \equiv \left(a^2+1\right)\left(a^2-1\right)\left(a^4+a^2+1\right)\\ & \equiv \left(a^2+1\right)\left(a^6-1\right)\equiv 0\mathrm{mod}7\end{array}$$ But $5$, $7$ are relatively prime and therefore $$\left(a^4-1\right)\left(a^4+15a^2+1\right)\equiv 0\mathrm{mod}35.$$