Finding an $n$ such that $n^2 \equiv -1 \mod p$

Question

What is an efficient algorithm to find the first number $n$ such that $n^2 \equiv -1 \mod p$ for a prime $p$, if such an $n$ exists?

Is there anything better than the brute-force approach up to $p-1 \over 2$?

I know this is simple to find for primes of the form $n^2+1$ because $n^2 \equiv -1 \mod (n^2+1)$, resulting in $n$, but is there a fast way for a generic case $n$?

Ex:

$p = 29, n = \pm 12$
$p = 37, n = \pm 6$
$p = 41, n = \pm 9$
$p = 53, n = \pm 23$

Find a $k$ with $\left(\frac{k}{p}\right) = -1$. Then $k^{(p-1)/4}$ does the trick. The smallest quadratic nonresidue is typically small. — Daniel Fischer, Jul 31 '14 at 22:11
You want the smallest, so yes, you take the modulus, and it could be that $p - (k^{(p-1)/4}\bmod p)$ is smaller, so we need a $\pm$. — Daniel Fischer, Jul 31 '14 at 22:22
And, following up on @DanielFischer's comment, the most efficient way to find a quadratic non-residue is guess-and-check. Probabilistic, yes, but, ... :) — paul garrett, Jul 31 '14 at 22:26
http://en.wikipedia.org/wiki/Tonelli%E2%80%93Shanks_algorithm — Will Jagy, Jul 31 '14 at 22:30
Something close to this was already asked and answered here, more or less. (If $a^2 + b^2 = p$ and $b^$ denotes a mod $p$ inverse of $b$, then you can take $n = a b^$.) — user160609, Jul 31 '14 at 23:08

score 1 · Accepted Answer · answered Jul 31 '14 at 22:30

1

Such an $n$ exists iff $p=1 \mod 4$. And also $p=1 \mod 4$ iff $p=x^2+y^2$ for integers $x, y$. Then the solution is $n=\frac{x}{y} \mod p$. This could be faster using $1\le x\le \sqrt{p}$.

answered Jul 31 '14 at 22:30

i. m. soloveichik

4,259

Finding an $n$ such that $n^2 \equiv -1 \mod p$

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