Balti Tee 2017 · Ülesanne 18

Answer: For all primes $p>5$.

When $p=5$ it is clear that it is not possible to determine $a_{1}$ and $a_{2}$ from the residue of $a_{1} a_{2}$ modulo 5 .

Assume that $p>5$. Now $\frac{p-1}{2} \geq 3$. For all $i \in\left\{1,2, \ldots, \frac{p-1}{2}\right\}$ it is possible to choose $j, k \in\left\{1,2, \ldots, \frac{p-1}{2}\right\}$ such that $i, j$ and $k$ are different. Thus we know

$$a_{i}^{2} \equiv\left(a_{i} a_{j}\right)\left(a_{i} a_{k}\right)\left(a_{j} a_{k}\right)^{-1}(\bmod p) .$$

The equation $x^{2} \equiv a(\bmod p)$ has exactly one solution in $\left\{1,2, \ldots, \frac{p-1}{2}\right\}$, and hence it is possible to determine $a_{i}$ for all $i$.