Simple explanation and examples of the Miller-Rabin Primality Test

Coming from an understanding of Fermat’s primality test, I’m looking for a clear explanation of the Miller-Rabin primality test.

Specifically: I understand that for some reason, having non-trivial square roots of 1 mod p means that p is definitely composite; and I gather that you can find these non-trivial square roots by squaring x, but I don’t really understand what these reasons are. Specific examples of non-trivial roots of a composite number would be helpful.

Cheers!

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Suppose $p$ was prime, and $y$ was a non-trivial square root of $1$ mod $p$.

Then we must have that $y^2 = 1 \mod p$ and so $(y-1)(y+1) = 0 \mod p$. This implies that either $y = 1 \mod p$ or $y = -1 \mod p$, which implies that $y$ is a trivial square root.

Thus, if there is a non-trivial square root of $1$ mod $p$, then $p$ has to be composite.

For an example of a non trivial square root of a composite, consider $p = 15$. We have that $4^2 = 16 = 1 \mod 15$. Thus $15$ is composite.

Note that the witness in the primality test is not necessarily a non-trivial square root of $1$ mod $p$.

The fact about non-trivial square roots can be used to prove that if $p$ is prime, then for any $a$ relatively prime to $p$, some power of $a$ from a given set of powers (the powers are based on the even factors of $p-1$) must be $-1$ or a specific odd power of $a$ (again based on factor of $p-1$) must be $1$.

If for some $a$ none of the above set of powers is $-1$ and the specific odd power is not $1$, then it must be the fact that $p$ is composite.

It can also be shown that for composite $p$, the chances of finding such $a$ is atleast $3/4$. This $a$ is the witness in the primality test and is not necessarily a non-trivial square root of $1$ mod $p$.

The squaring that is done is to get the powers described above which are based on the factors of $p-1$.

The wiki page has really got a lot of good information (including the exact powers of $a$ that need to be taken): Miller Rabin Primality Test

Below is little-known general yet simple answer to your question about nontrivial sqrts $\rm (mod\ m)\:$.

Theorem $ $ One may quickly factor $\rm m>1\,$ given a polynomial with more roots mod $\rm\, m\,$ than its degree. For suppose that, modulo $\rm m,\;$ the polynomial $\rm\, f(x)\ne 0\,$ has degree $\rm\,n\,$ but has $\rm\,n+1 \,$ distinct roots $\rm\,r_{\,i}.\,$ Then one of $\rm\;gcd(m,\,r_{\,i} – r_{\,j}),\; i\ne j \,$ must yield a proper factor of $\rm\,m.\,$ For if that failed, then all of the gcds must be improper hence $1,\,$ not $\rm\;m \;$ since $\rm\; i\ne j\;\Rightarrow\; r_{\,i} \not\equiv r_{\,j}\ (mod\ m).\,$ Now an induction using the Factor Theorem yields $\rm\;f(x) = (x-r_1)\cdots(x-r_{n+1})\; g(x),\;\; g(x) \ne 0 \;\;$ contra $\rm\,\deg\, f = n.$

Example $\rm\;(deg\ f = 1)\;\,$ Suppose, modulo $\rm\,m,\,$ that $\rm\; 0 \,\ne\, f \,=\, a\,x\;$ has a “nontrivial” root $\rm\, b\ne 0.\,$ Then $\,\rm gcd(m,b)\,$ is a proper factor of $\rm\, m.\ $ More directly: $\rm\; m\mid ab,\ m\nmid a,b \;\Rightarrow\; gcd(m,b)\;$ is a proper factor of $\rm\,m\,,\,$ i.e. $\rm\, m\,$ fails the Prime Divisor Property $\rm\,\Rightarrow m\,$ is “constructively” composite.

Example $\rm\;(deg\ f = 2)\;\,$ Suppose, modulo $\rm\,m\,$, $\rm\; x^2 = 1\,$ has a “nontrivial” square root $\rm\, b\ne \pm1.\,$ Then $\;\rm gcd(m,b\pm 1)\;$ yields a proper factor of $\rm\, m\,$ when $\rm\, m\,$ is odd. This idea lies at the heart of many integer factorization algorithms.

The above proof easily extends to yield a proof of the following

Theorem $ $ Ring $\rm\, R$ is a domain $\iff$ every $\rm\, 0 \ne f(x)\in R[x]\,$ has at most $\rm\, deg\ f(x)\,$ roots in $\rm R$

Proof $\ $ $(\Rightarrow)\;$ Employ induction on the polynomial degree, as in the proof of the above Theorem. $(\Leftarrow)\ \ $ If $\rm\; a \ne 0\;$ then the polynomial $\rm\, a\,x\,$ has only the root $\rm\; x = 0,\,$ hence $\rm\ ab = 0 \ \Rightarrow\ b = 0,\,$ therefore $\rm\, R\,$ has no zero divisors.

If $p$ is prime then $\mathbb{Z}_p$ (integers modulo $p$) is a field. It is a basic result in algebra that in a field, a polynomial of degree $n$ has at most $n$ roots, and so the polynomial $x^2-1$ has exactly two roots: $1$ and $-1$ (which exist in every field).

If $n$ is composite, then $\mathbb{Z}_n$ is never a field because not all elements have an inverse; it it well known that $a\in \mathbb{Z}_n$ has an inverse if and only if $a$ is relatively prime to $n$. Let’s look at the case $n$ is the product of two primes, $n=pq$. In this case we can do arithmetic in $\mathbb{Z}_n$ by doing arithmetic in $\mathbb{Z}_p$ and $\mathbb{Z}_q$ and combining the results using the Chinese remainder theorem (which basically states that $\mathbb{Z}_n\cong\mathbb{Z}_p\times\mathbb{Z}_q$. Since $\mathbb{Z}_p$ and $\mathbb{Z}_q$ are both fields, $1$ has two roots in each of them. For every combination of a root from $\mathbb{Z}_p$ and a root from $\mathbb{Z}_q$ we’ll get a root of 1 in $\mathbb{Z}_n$, meaning we’ll get 4 roots of 1.

The major challenge of the Miller-Rabin test is to show that there is a “large” chance to stumble upon a non-trivial root while squaring random elements of $\mathbb{Z}_n$, and the proof, although not difficult, is not immediate either.

A test based on Fermat’s Theorem ($a^{n-1} \equiv 1 \pmod n$ if n is prime)

-For a candidate odd integer n (3), consider $(n-1)$ s.t. $n-1=2^kq$ with $k>0$, $q$ odd

That is, we divide (n-1) by 2 until the result is an odd number, for a total of $k$ divisions.

-Next we choose an integer $a$ ($1 < a < n – 1 $).

Then involve computation of the residues modulo n of the following sequence of powers

-$a^q, a^{2q}, …, a^{2^{k-1}q}$, $a^{2^kq}$

-If $n$ is prime, $a^{(2k-1)q} \equiv a^{n-1} \equiv 1 \pmod n$.

-There may or may not be an earlier element of the sequence that has a residue 1

-If n is prime, there is a smallest value of j ($0<j<k$) such that $a^{2jq}\equiv 1 \pmod n$.

-There are two cases to consider

-Case 1 ($j=0$) : We have $a^{q–1} \equiv 0 \pmod n$

-Case 2 ($1<j<k$) : $(a^{2^jq}-1) \equiv (a^{2^{j-1}q}-1)(a^{2^{j+1}q}+1) \equiv 0 \pmod n$.

Because j is the smallest integer s.t. n divides $(a^{2^jq}-1)$, n divides $(a2^{(j+1)q}+1)$

A test based on Fermat’s Theorem ($a^{n-1} \equiv 1 \pmod n$ if $n$ is prime)

-Algorithm is:
TEST (n) is:

  1. Find integers $k$, $q$, $k > 0$, $q$ odd, so that $(n–1)=2^kq$

  2. Select a random integer a, 1

  3. if $a^q \equiv 1 \pmod n$ then return (“maybe prime”);

  4. for j = 0 to k –1 do

  5. if($a^{2^jq} \mod n = n-1$)
    then return(” maybe prime “)

  6. return (“composite”)