Solovay-Strassen test

It is known that an odd number $n$ is prime if and only if for every such that $(a,n)=1$ we have

$$\left(\frac{a}{n}\right)\equiv a^{\frac{n-1}{2}}(modn)$$ (1)

where $\left(\frac{a}{n}\right)$ is the Jacobi symbol. (The only if part is obvious; the if part follows from Theorem 1.) From this we can derive the following algorithm.

1. Choose a random number $a$ between $1$ and $n-1$.

2. Check if $(a,n)=1$ (for example using the Euclidean algorithm). If it is not, then $n$ is not prime and $(a,n)$ is a divisor of $n$.

3. Check if Equation (1) holds. If it does not, then $n$ is not prime. Otherwise $n$ is a candidate for primality.

By repeating this algorithm we can increase the chance that the result is correct. In order to estimate the probability of error, we make use of Theorem 1, which says that every independent iteration of the algorithm has a chance of at most $50\%$ of being wrong. Hence, after $t$ iterations there is at most a $2^{-t}$ chance of getting a wrong result.

Theorem 1.

Let $n\mathrm{\ge }\mathrm{3}$ be an odd composite integer. Then at least half of the elements of ${\mathrm{Z}}_n^{\mathrm{*}}$ do not satisfy Equation (1).

Proof.

It suffices to exhibit one element $a\in {\mathbb{Z}}_n^{*}$ which does not satisfy Equation (1). Indeed, if there exists one such element, then the set of all elements which do satisfy Equation (1) forms a proper subgroup of ${\mathbb{Z}}_n^{*}$, from which we conclude that the elements satisfying Equation (1) number no more than half of the elements of ${\mathbb{Z}}_n^{*}$.

We consider separately the cases where $n$ is squarefree and not squarefree. If $n$ is squarefree, let $p$ be a prime dividing $n$ and let $b$ be a quadratic non-residue mod $n$. Using the Chinese Remainder Theorem, choose an integer $a\in {\mathbb{Z}}_n^{*}$ such that:

$a$ $\equiv b(modp),$

$a$ $\equiv 1(mod{\displaystyle \frac{n}{p}}).$

The Jacobi symbol $\left(\frac{a}{n}\right)$ is given by

$$\left(\frac{a}{n}\right)=\left(\frac{a}{p}\right)\left(\frac{a}{n/p}\right)=\left(\frac{b}{p}\right)\left(\frac{1}{n/p}\right)=(-1)\cdot 1=-1.$$

We will assume that $a^{\frac{n-1}{2}}\equiv -1(modn)$ and derive a contradiction. The equation $a^{\frac{n-1}{2}}\equiv -1(modn)$ implies that $a^{\frac{n-1}{2}}\equiv -1(mod\frac{n}{p})$. However, since $a\equiv 1(mod\frac{n}{p})$, we must have $a^{\frac{n-1}{2}}\equiv 1(mod\frac{n}{p})$, so that $1\equiv -1(mod\frac{n}{p})$, which is a contradiction.

Now suppose that $n$ is not squarefree. Let $p$ be a prime such that $p^2\mid n$, and set $a=1+\frac{n}{p}$. By the binomial theorem, we have

$$a^p={\left(1+\frac{n}{p}\right)}^p=1+\left(\genfrac{}{}{0pt}{}{p}{1}\right)\left(n/p\right)+\left(\genfrac{}{}{0pt}{}{p}{2}\right){\left(n/p\right)}^2+\mathrm{\cdots }+\left(\genfrac{}{}{0pt}{}{p}{p}\right){\left(n/p\right)}^p\equiv 1(modn),$$

so the multiplicative order of $amodn$ is equal to $p$, and hence in particular $a^{n-1}\ne 1(modn)$, since $p\nmid n-1$. On the other hand,

$$\left(\frac{a}{n}\right)=\left(\frac{1+\frac{n}{p}}{n}\right)=\left(\frac{1+\frac{n}{p}}{p}\right)\left(\frac{1+\frac{n}{p}}{n/p}\right)=\left(\frac{1}{p}\right)\left(\frac{1}{n/p}\right)=1,$$

so $a$ does not satisfy Equation (1). ∎

Title	Solovay-Strassen test
Canonical name	SolovayStrassenTest
Date of creation	2013-03-22 14:27:33
Last modified on	2013-03-22 14:27:33
Owner	mathwizard (128)
Last modified by	mathwizard (128)
Numerical id	7
Author	mathwizard (128)
Entry type	Algorithm
Classification	msc 11Y11
Related topic	MillerRabinPrimeTest