Higher residuosity problem


In cryptography, most public key cryptosystems are founded on problems that are believed to be intractable. The higher residuosity problem is one such problem. This problem is easier to solve than integer factorization, so the assumption that this problem is hard to solve is stronger than the assumption that integer factorization is hard.

Mathematical background

If n is an integer, then the integers modulo n form a ring. If n=pq where p and q are primes, then the Chinese remainder theorem tells us that
The group of units of any ring form a group, and the group of units in is traditionally denoted.
From the isomorphism above, we have
as an isomorphism of groups. Since p and q were assumed to be prime, the groups and are cyclic of orders p-1 and q-1 respectively. If d is a divisor of p-1, then the set of dth powers in form a subgroup of index d. If gcd = 1, then every element in is a dth power, so the set of dth powers in is also a subgroup of index d. In general, if gcd = g, then there are / dth powers in, so the set of dth powers in has index dg.
This is most commonly seen when d=2, and we are considering the subgroup of quadratic residues, it is well known that exactly one quarter of the elements in are
quadratic residues.
The important point is that for any divisor d of p-1 the set of dth powers forms a subgroup of

Problem statement

Given an integer n = pq where p and q are unknown, an integer d such that d divides p-1, and an integer x < n, it is infeasible to determine whether x is a dth power modulo n.
Notice that if p and q are known it is easy to determine whether x is a dth residue modulo n because x will be a dth residue modulo p if and only if
When d=2, this is called the quadratic residuosity problem.

Applications

The semantic security of the Benaloh cryptosystem and the Naccache-Stern cryptosystem rests on the intractability of this problem.