@Seva My formulas are valid for any natural number $p\ge3$.

The number ${\cal P}_{31}(4)= M^{31}$ where $M$ is a number of 163 digits. $M=2^{35} \times 5^9\times 311^4\times 373\times \cdots \times 1620371$ with prime factors $p\equiv 1\bmod31$.

More values of the function $p\mapsto {\cal P}_p(3)$, starting from $p=3$ and giving values for $p=3$, $4$, $5$, $6$, \dots, $31$ $$0, -1, -1, 0, -2^7, -3^8, 0, 11^{10}, 23^{11}, 0, 159^{13}, 464^{14}, 0, -6069^{16}, -24617^{17},0,-611009^{19}, -3438875^{20}, 0, 162222611^{22}, 1265401351^{23}, 0, 113562774001^{25}, 1226797460541^{26}, 0, -209594542523392^{28}, 3134065080817441^{29}, 0, -1019802203023098400^{31}$$

@Lucia What is the problem in the case of the $a_n$ distinct? Do you need all the modules $m$ to get the contradiction?

I like your answer. I have one, similar but a litte more complicate, but you posted your better one, one or two hours before I have mine ready.