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 $1 \bmod p$ 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!
On
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.
On
Below is little-known general yet simple answer to your question about nontrivial sqrts $\rm (mod\ m)\:$.
Theorem $ $ We can quickly split $\rm m>1\,$ into two nontrivial factors given a nonzero polynomial with more roots mod $\rm\, m\,$ than its degree.
Proof $ $ By hypothesis $\rm\, \color{#0a0}{0\not\equiv} f(x)\,$ has degree $\rm\,n\,$ and at least $\rm\,n\!+\!1 \,$ distinct roots $\rm\,r,\,r_1,\,\ldots,\,r_n.\,$ Inductively applying the the Factor Theorem as here yields either that some difference $\rm\,r_i-r_j\,$ is a nontrivial zero-divisor (so we're done), or else $\rm\,f(x) \equiv c(x\!-\!r_1)\cdots(x\!-\!r_{n}),\ c\color{#0a0}{\not\equiv 0}.\,$ Since also $\rm\,f(r)\equiv 0\,$ we infer $\rm\,m\mid c(r\!-\!r_1)\cdots (r\!-\!r_n).\,$ If $\rm\,m\,$ were coprime to all those factors it would be coprime to their product by Euclid's Lemma. Since it is not, we deduce that $\rm\,m\,$ is not coprime to some factor $\rm\,k.\,$ Since $\,\rm m\,$ divides none of the factors (by the roots are distinct $\rm\!\bmod m\,$ and $\,c\color{#0a0}{\not\equiv 0}),\,$ we infer $\,\rm\gcd(m,k)\neq m,\,$ thus the gcd is a nontrivial factor of $\,\rm m,\,$ being $\rm\neq 1,m.\ $
Example $\rm\;(deg\ f = 1)\;\,$ Suppose, mod $\rm\,m,\,$ that $\rm\; 0 \,\not\equiv\, f \,=\, a\,x\;$ has a "nontrivial" root $\rm\, b\not\equiv 0.\,$ Then $\rm\; m\mid ab,\,\ m\nmid a,b \;\Rightarrow\,$ $\,\rm gcd(m,b)\,$ is a factor of $\rm\, m\,$ that is nontrivial $(\neq 1,\,m),\,$ i.e. $ $ when $\rm\, m\,$ fails the Prime Divisor Property (Euclid's Lemma) it is constructively composite.
Example $\rm\;(deg\ f = 2)\;\,$ Suppose, modulo $\rm\,m\,$, $\rm\; x^2 \equiv 1\,$ has a "nontrivial" square root $\rm\, b\not\equiv \pm1.\,$ Then $\;\rm gcd(m,b\pm 1)\;$ yields a nontrivial factor of $\rm\, m\,$ when $\rm\, m\,$ is odd. This idea lies at the heart of many integer factorization algorithms. In detail, by unique prime factorization we have $\rm\, m\mid (b\!-\!1)(b\!+\!1)\,\Rightarrow\, m = jk,\,\ j\mid b\!-\!1,\, k\mid b\!+\!1,\,$ so $\rm\,m\nmid b\pm 1\,\Rightarrow\,j,k\mid m\,$ nontrivially.
The above proof easily extends to yield a proof of the following
Theorem $ $ A ring $\rm\, R$ is an (integral) domain, i.e. $\rm\,rs = 0\,\Rightarrow\, r=0\,$ or $\rm\,s =0,\,$ for all $\rm\,r,s\in R$ $\iff$ every nonzero polynomial $\,\rm f(x)\in R[x]\,$ 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 $\rm\; r \ne 0\;$ then the polynomial $\rm\, rx\,$ has only the root $\rm\; x = 0,\,$ hence $\rm\ rs = 0 \ \Rightarrow\ s = 0$.
On
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:
Find integers $k$, $q$, $k > 0$, $q$ odd, so that $(n–1)=2^kq$
Select a random integer a, 1
if $a^q \equiv 1 \pmod n$ then return (“maybe prime");
for j = 0 to k –1 do
if($a^{2^jq} \mod n = n-1$) then return(" maybe prime ")
return ("composite")
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