Suppose I wanted to express a number $N$ as a difference of squares. For large $N$ this is in general difficult, as finding $N=a^2-b^2$ leads to the factorization $N=(a+b)(a-b)$. Even if the problem is weakened to searching for $a\neq b$ with $a^2\equiv b^2\pmod N$ the problem is still hard (though not as hard), since enough congruences could be used to factor $N$ with Dixon's method or any of its modern versions (in particular, the number field sieve).

So I am curious about the difficulty of these weak versions of the problem. Are any of these easier than finding relations with the NFS?

*Weak form.*Given $N$, $k$, and a factoring oracle, find $k$ distinct nontrivial congruences $a^2\equiv b^2\pmod N$.*Semi-weak form.*Given $N$, $k$, and the complete factorization of $N$, find $k$ distinct nontrivial congruences $a^2\equiv b^2\pmod N$.*Strong form.*Given $N$, $k$, and a partial factorization of $N$, find $k$ distinct nontrivial congruences $a^2\equiv b^2\pmod N$.