Given a prime $p$, two of its primitive roots $(h_1, h_2)$, $A_1=(h_1^xh_2^y) \mod p,$ $A_2=(h_1^yh_2^x) \mod p$, find $(x,y)$.
How can it be solved?
Example:
$p=23$, $(h_1=2, h_2=3)$, $A_1=(2^x3^y) \mod 23 = 16,$ $A_2=(2^y3^x) \mod 23 = 3$,
find $(x,y)$.
I made some research and found that in a close future, having a quantum computer with a sufficient number of qubits, in order to implement Shor's algorithm, we'll be able to compute discrete logarithms. So, by multiplying $A_1, A_2$ we can obtain the sum of the exponents using the product of the two primitive roots as base (I omit "mod p"):
$A_{1} = h_{1}^{x}h_{2}^{y}$
$A_{2} = h_{1}^{y}h_{2}^{x}$
$A_{1}A_{2} = h_{1}^{x}h_{2}^{y}h_{1}^{y}h_{2}^{x} = h_{1}^{(x+y)}h_{2}^{(y+x)} = (h_{1}h_{2})^{(x+y)}$
But even in this hypothesis, I can't understand if and how it is possible to find the single exponents.