Obviously, the discrete logarithm problem is thought to be hard.
But what if you are also given $g^\left(a^{-1}\right)$?
Does that give enough information to solve for $a$? Or is that also intractable?
2026-02-23 13:43:11.1771854191
Is finding "a" given g, g^a, and g^(a^-1) mod p intractable?
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