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Is it any easier to find $X$ for $a=1$ than some other $a$'s that is smaller than $N$. $a$ is quadratic residue.

Gerry Myerson
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Bojan Vasiljević
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    Suppose that $N$ is the product of two very large primes. Beside the obvious solution $x\equiv \pm 1\pmod{N}$, the congruence has two non-obvious solutions. If we know them, then we can easily factor $N$. The problem of factoring the product of two very large primes is widely believed to be very difficult. – André Nicolas Oct 15 '12 at 21:13
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    If $N$ is prime, either http://en.wikipedia.org/wiki/Tonelli%E2%80%93Shanks_algorithm or http://en.wikipedia.org/wiki/Cipolla%27s_algorithm . If $N$ is composite and you can factor it, there is more work to be done, essentially Chinese Remainder Theorem. If you cannot factor $N,$ as Andre points out, you are out of luck. There is a fairly good trick for square roots of $(-1) \pmod p$ when $p \equiv 1 \pmod 4$ is prime, but still based on finding a nonresidue and playing with it. – Will Jagy Oct 15 '12 at 21:23
  • Special-case of this question, on k'th roots. It's no easier than the general case. – Bill Dubuque Oct 15 '12 at 23:39

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