A Probable Prime Test With High Confidence

Jon Grantham

arXiv:1903.06823·math.NT·March 19, 2019

A Probable Prime Test With High Confidence

Jon Grantham

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TL;DR

This paper introduces a new probable prime test using quadratic polynomials and Frobenius automorphism, achieving higher confidence by reducing the likelihood of composite numbers passing the test compared to previous methods.

Contribution

A novel probable prime test based on quadratic polynomials and Frobenius automorphism that significantly lowers false positives for composite numbers.

Findings

01

Composite passes less than 1/7710 of the polynomial tests

02

Test runs approximately three times slower than the Strong Probable Prime Test

03

Provides higher confidence in primality testing

Abstract

Monier and Rabin proved that an odd composite can pass the Strong Probable Prime Test for at most $\frac{1}{4}$ of the possible bases. In this paper, a probable prime test is developed using quadratic polynomials and the Frobenius automorphism. The test, along with a fixed number of trial divisions, ensures that a composite $n$ will pass for less than $\frac{1}{7710}$ of the polynomials $x^{2} - b x - c$ with $(\frac{b ^{2} + 4 c}{n}) = - 1$ and $(\frac{- c}{n}) = 1$ . The running time of the test is asymptotically $3$ times that of the Strong Probable Prime Test.

Equations32

N = \frac{1}{4} # {(b, x, y) mod p : x, y \neq \equiv 0 mod p and b^{2} \equiv R (2 x)^{2} + S y^{2} mod p} .

N = \frac{1}{4} # {(b, x, y) mod p : x, y \neq \equiv 0 mod p and b^{2} \equiv R (2 x)^{2} + S y^{2} mod p} .

\multline N = N_{1} = \frac{1}{4} # {(x, y, z) : x, y \neq \equiv 0 mod p and x^{2} + y^{2} \equiv z^{2}} = \frac{1}{4} # {(x, y, z) : x, y \neq \equiv 0 mod p and (z - y) (z + y) \equiv x^{2}} = \frac{1}{4} x = 1 \sum p - 1 # {ab \equiv x^{2} : a \neq \equiv b} = (p - 1) (p - 1 - 2) /4 = (p - 1) (p - 3) /4. \endmultline N = N_{1} = \frac{1}{4} # {(x, y, z) : x, y \neq \equiv 0 mod p and x^{2} + y^{2} \equiv z^{2}} = \frac{1}{4} # {(x, y, z) : x, y \neq \equiv 0 mod p and (z - y) (z + y) \equiv x^{2}} = \frac{1}{4} x = 1 \sum p - 1 # {ab \equiv x^{2} : a \neq \equiv b} = (p - 1) (p - 1 - 2) /4 = (p - 1) (p - 3) /4.

\multline N = N_{1} = \frac{1}{4} # {(x, y, z) : x, y \neq \equiv 0 mod p and x^{2} + y^{2} \equiv z^{2}} = \frac{1}{4} # {(x, y, z) : x, y \neq \equiv 0 mod p and (z - y) (z + y) \equiv x^{2}} = \frac{1}{4} x = 1 \sum p - 1 # {ab \equiv x^{2} : a \neq \equiv b} = (p - 1) (p - 1 - 2) /4 = (p - 1) (p - 3) /4. \endmultline N = N_{1} = \frac{1}{4} # {(x, y, z) : x, y \neq \equiv 0 mod p and x^{2} + y^{2} \equiv z^{2}} = \frac{1}{4} # {(x, y, z) : x, y \neq \equiv 0 mod p and (z - y) (z + y) \equiv x^{2}} = \frac{1}{4} x = 1 \sum p - 1 # {ab \equiv x^{2} : a \neq \equiv b} = (p - 1) (p - 1 - 2) /4 = (p - 1) (p - 3) /4.

\multline N = N_{2} = \frac{1}{4} # {(x, y, z) : x, y \neq \equiv 0 mod p and r x^{2} + r y^{2} \equiv z^{2}} = \frac{1}{4} # {(x, y, z) : x, y \neq \equiv 0 mod p and x^{2} + y^{2} \equiv r (z / r)^{2}} . \endmultline N = N_{2} = \frac{1}{4} # {(x, y, z) : x, y \neq \equiv 0 mod p and r x^{2} + r y^{2} \equiv z^{2}} = \frac{1}{4} # {(x, y, z) : x, y \neq \equiv 0 mod p and x^{2} + y^{2} \equiv r (z / r)^{2}} .

\multline N = N_{2} = \frac{1}{4} # {(x, y, z) : x, y \neq \equiv 0 mod p and r x^{2} + r y^{2} \equiv z^{2}} = \frac{1}{4} # {(x, y, z) : x, y \neq \equiv 0 mod p and x^{2} + y^{2} \equiv r (z / r)^{2}} . \endmultline N = N_{2} = \frac{1}{4} # {(x, y, z) : x, y \neq \equiv 0 mod p and r x^{2} + r y^{2} \equiv z^{2}} = \frac{1}{4} # {(x, y, z) : x, y \neq \equiv 0 mod p and x^{2} + y^{2} \equiv r (z / r)^{2}} .

\multline N_{1} + N_{2} = \frac{1}{4} # {(x, y, z) : x, y \neq \equiv 0 mod p and x^{2} + y^{2} \equiv z^{2} or r z^{2}} \leq \frac{1}{2} # {(x, y) : x, y \neq \equiv 0 mod p} \leq (p - 1)^{2} /2. \endmultline N_{1} + N_{2} = \frac{1}{4} # {(x, y, z) : x, y \neq \equiv 0 mod p and x^{2} + y^{2} \equiv z^{2} or r z^{2}} \leq \frac{1}{2} # {(x, y) : x, y \neq \equiv 0 mod p} \leq (p - 1)^{2} /2.

\multline N_{1} + N_{2} = \frac{1}{4} # {(x, y, z) : x, y \neq \equiv 0 mod p and x^{2} + y^{2} \equiv z^{2} or r z^{2}} \leq \frac{1}{2} # {(x, y) : x, y \neq \equiv 0 mod p} \leq (p - 1)^{2} /2. \endmultline N_{1} + N_{2} = \frac{1}{4} # {(x, y, z) : x, y \neq \equiv 0 mod p and x^{2} + y^{2} \equiv z^{2} or r z^{2}} \leq \frac{1}{2} # {(x, y) : x, y \neq \equiv 0 mod p} \leq (p - 1)^{2} /2.

\multline N = N_{3} = \frac{1}{4} # {(x, y, z) : x, y \neq \equiv 0 mod p and x^{2} + r y^{2} \equiv z^{2}} = \frac{1}{4} # {(x, y, z) : x, y \neq \equiv 0 mod p and x^{2} + r y^{2} \equiv r z^{2}} = \frac{1}{4} # {(x, y) : x, y \neq \equiv 0 mod p} = (p - 1)^{2} /4. \endmultline N = N_{3} = \frac{1}{4} # {(x, y, z) : x, y \neq \equiv 0 mod p and x^{2} + r y^{2} \equiv z^{2}} = \frac{1}{4} # {(x, y, z) : x, y \neq \equiv 0 mod p and x^{2} + r y^{2} \equiv r z^{2}} = \frac{1}{4} # {(x, y) : x, y \neq \equiv 0 mod p} = (p - 1)^{2} /4.

\multline N = N_{3} = \frac{1}{4} # {(x, y, z) : x, y \neq \equiv 0 mod p and x^{2} + r y^{2} \equiv z^{2}} = \frac{1}{4} # {(x, y, z) : x, y \neq \equiv 0 mod p and x^{2} + r y^{2} \equiv r z^{2}} = \frac{1}{4} # {(x, y) : x, y \neq \equiv 0 mod p} = (p - 1)^{2} /4. \endmultline N = N_{3} = \frac{1}{4} # {(x, y, z) : x, y \neq \equiv 0 mod p and x^{2} + r y^{2} \equiv z^{2}} = \frac{1}{4} # {(x, y, z) : x, y \neq \equiv 0 mod p and x^{2} + r y^{2} \equiv r z^{2}} = \frac{1}{4} # {(x, y) : x, y \neq \equiv 0 mod p} = (p - 1)^{2} /4.

\multline N_{ϵ_{1}, ϵ_{2}} (n) = N_{ϵ_{1}, ϵ_{2}} (m) N_{1, 1} (p) + N_{- ϵ_{1}, ϵ_{2}} (m) N_{- 1, 1} (p) + N_{ϵ_{1}, - ϵ_{2}} (m) N_{1, - 1} (p) + N_{- ϵ_{1}, - ϵ_{2}} (m) N_{- 1, - 1} (p) . \endmultline N_{ϵ_{1}, ϵ_{2}} (n) = N_{ϵ_{1}, ϵ_{2}} (m) N_{1, 1} (p) + N_{- ϵ_{1}, ϵ_{2}} (m) N_{- 1, 1} (p) + N_{ϵ_{1}, - ϵ_{2}} (m) N_{1, - 1} (p) + N_{- ϵ_{1}, - ϵ_{2}} (m) N_{- 1, - 1} (p) .

\multline N_{ϵ_{1}, ϵ_{2}} (n) = N_{ϵ_{1}, ϵ_{2}} (m) N_{1, 1} (p) + N_{- ϵ_{1}, ϵ_{2}} (m) N_{- 1, 1} (p) + N_{ϵ_{1}, - ϵ_{2}} (m) N_{1, - 1} (p) + N_{- ϵ_{1}, - ϵ_{2}} (m) N_{- 1, - 1} (p) . \endmultline N_{ϵ_{1}, ϵ_{2}} (n) = N_{ϵ_{1}, ϵ_{2}} (m) N_{1, 1} (p) + N_{- ϵ_{1}, ϵ_{2}} (m) N_{- 1, 1} (p) + N_{ϵ_{1}, - ϵ_{2}} (m) N_{1, - 1} (p) + N_{- ϵ_{1}, - ϵ_{2}} (m) N_{- 1, - 1} (p) .

\multline N_{ϵ_{1}, ϵ_{2}} (n) \leq [\frac{1}{4} ϕ (m)^{2} + \frac{ϵ _{1}}{2} μ (m) ϕ (m)] [\frac{( p - 1 ) ^{2}}{4} - \frac{p - 1}{2}] + \frac{1}{16} ϕ (m)^{2} (p - 1)^{2} + \frac{1}{16} ϕ (m)^{2} (p - 1)^{2} + [\frac{1}{4} ϕ (m)^{2} - \frac{ϵ _{1}}{2} μ (m) ϕ (m)] [\frac{( p - 1 ) ^{2}}{4} + \frac{p - 1}{2}] = \frac{1}{4} ϕ (m)^{2} (p - 1)^{2} - \frac{ϵ _{1} μ ( m ) ϕ ( m ) ( p - 1 )}{2} = \frac{1}{4} ϕ (m p)^{2} + \frac{ϵ _{1}}{2} μ (m p) ϕ (m p) . \endmultline N_{ϵ_{1}, ϵ_{2}} (n) \leq [\frac{1}{4} ϕ (m)^{2} + \frac{ϵ _{1}}{2} μ (m) ϕ (m)] [\frac{( p - 1 ) ^{2}}{4} - \frac{p - 1}{2}] + \frac{1}{16} ϕ (m)^{2} (p - 1)^{2} + \frac{1}{16} ϕ (m)^{2} (p - 1)^{2} + [\frac{1}{4} ϕ (m)^{2} - \frac{ϵ _{1}}{2} μ (m) ϕ (m)] [\frac{( p - 1 ) ^{2}}{4} + \frac{p - 1}{2}] = \frac{1}{4} ϕ (m)^{2} (p - 1)^{2} - \frac{ϵ _{1} μ ( m ) ϕ ( m ) ( p - 1 )}{2} = \frac{1}{4} ϕ (m p)^{2} + \frac{ϵ _{1}}{2} μ (m p) ϕ (m p) .

\multline N_{ϵ_{1}, ϵ_{2}} (n) \leq [\frac{1}{4} ϕ (m)^{2} + \frac{ϵ _{1}}{2} μ (m) ϕ (m)] [\frac{( p - 1 ) ^{2}}{4} - \frac{p - 1}{2}] + \frac{1}{16} ϕ (m)^{2} (p - 1)^{2} + \frac{1}{16} ϕ (m)^{2} (p - 1)^{2} + [\frac{1}{4} ϕ (m)^{2} - \frac{ϵ _{1}}{2} μ (m) ϕ (m)] [\frac{( p - 1 ) ^{2}}{4} + \frac{p - 1}{2}] = \frac{1}{4} ϕ (m)^{2} (p - 1)^{2} - \frac{ϵ _{1} μ ( m ) ϕ ( m ) ( p - 1 )}{2} = \frac{1}{4} ϕ (m p)^{2} + \frac{ϵ _{1}}{2} μ (m p) ϕ (m p) . \endmultline N_{ϵ_{1}, ϵ_{2}} (n) \leq [\frac{1}{4} ϕ (m)^{2} + \frac{ϵ _{1}}{2} μ (m) ϕ (m)] [\frac{( p - 1 ) ^{2}}{4} - \frac{p - 1}{2}] + \frac{1}{16} ϕ (m)^{2} (p - 1)^{2} + \frac{1}{16} ϕ (m)^{2} (p - 1)^{2} + [\frac{1}{4} ϕ (m)^{2} - \frac{ϵ _{1}}{2} μ (m) ϕ (m)] [\frac{( p - 1 ) ^{2}}{4} + \frac{p - 1}{2}] = \frac{1}{4} ϕ (m)^{2} (p - 1)^{2} - \frac{ϵ _{1} μ ( m ) ϕ ( m ) ( p - 1 )}{2} = \frac{1}{4} ϕ (m p)^{2} + \frac{ϵ _{1}}{2} μ (m p) ϕ (m p) .

\multline N_{ϵ_{1}, ϵ_{2}} (n) \leq \frac{1}{4} ϕ (m)^{2} \topsmash [\frac{( p - 1 ) ^{2}}{4} - \frac{p - 1}{2}] + [\frac{1}{4} ϕ (m)^{2} - \frac{ϵ _{1}}{2} μ (m) ϕ (m)] \frac{( p - 1 ) ^{2}}{4} + [\frac{1}{4} ϕ (m)^{2} + \frac{ϵ _{1}}{2} μ (m) ϕ (m)] \frac{( p - 1 ) ^{2}}{4} + \frac{1}{4} ϕ (m)^{2} [\frac{( p - 1 ) ^{2}}{4} + \frac{p - 1}{2}] = \frac{1}{4} ϕ (m)^{2} (p - 1)^{2} = \frac{1}{4} ϕ (m p)^{2} . \endmultline N_{ϵ_{1}, ϵ_{2}} (n) \leq \frac{1}{4} ϕ (m)^{2} \topsmash [\frac{( p - 1 ) ^{2}}{4} - \frac{p - 1}{2}] + [\frac{1}{4} ϕ (m)^{2} - \frac{ϵ _{1}}{2} μ (m) ϕ (m)] \frac{( p - 1 ) ^{2}}{4} + [\frac{1}{4} ϕ (m)^{2} + \frac{ϵ _{1}}{2} μ (m) ϕ (m)] \frac{( p - 1 ) ^{2}}{4} + \frac{1}{4} ϕ (m)^{2} [\frac{( p - 1 ) ^{2}}{4} + \frac{p - 1}{2}] = \frac{1}{4} ϕ (m)^{2} (p - 1)^{2} = \frac{1}{4} ϕ (m p)^{2} .

\multline N_{ϵ_{1}, ϵ_{2}} (n) \leq \frac{1}{4} ϕ (m)^{2} \topsmash [\frac{( p - 1 ) ^{2}}{4} - \frac{p - 1}{2}] + [\frac{1}{4} ϕ (m)^{2} - \frac{ϵ _{1}}{2} μ (m) ϕ (m)] \frac{( p - 1 ) ^{2}}{4} + [\frac{1}{4} ϕ (m)^{2} + \frac{ϵ _{1}}{2} μ (m) ϕ (m)] \frac{( p - 1 ) ^{2}}{4} + \frac{1}{4} ϕ (m)^{2} [\frac{( p - 1 ) ^{2}}{4} + \frac{p - 1}{2}] = \frac{1}{4} ϕ (m)^{2} (p - 1)^{2} = \frac{1}{4} ϕ (m p)^{2} . \endmultline N_{ϵ_{1}, ϵ_{2}} (n) \leq \frac{1}{4} ϕ (m)^{2} \topsmash [\frac{( p - 1 ) ^{2}}{4} - \frac{p - 1}{2}] + [\frac{1}{4} ϕ (m)^{2} - \frac{ϵ _{1}}{2} μ (m) ϕ (m)] \frac{( p - 1 ) ^{2}}{4} + [\frac{1}{4} ϕ (m)^{2} + \frac{ϵ _{1}}{2} μ (m) ϕ (m)] \frac{( p - 1 ) ^{2}}{4} + \frac{1}{4} ϕ (m)^{2} [\frac{( p - 1 ) ^{2}}{4} + \frac{p - 1}{2}] = \frac{1}{4} ϕ (m)^{2} (p - 1)^{2} = \frac{1}{4} ϕ (m p)^{2} .

L \in V \sum \frac{n ϕ ( n )}{2 ^{k} B ^{2}} < \frac{n ϕ ( n )}{B ^{2}} .

L \in V \sum \frac{n ϕ ( n )}{2 ^{k} B ^{2}} < \frac{n ϕ ( n )}{B ^{2}} .

\frac{r _{1} r _{2} r _{3}}{j _{1} j _{2} j _{3}} = \frac{( p _{2} p _{3} - 1 ) ( p _{1} p _{3} - 1 ) ( p _{1} p _{2} - 1 )}{( p _{1}^{2} - 1 ) ( p _{2}^{2} - 1 ) ( p _{3}^{2} - 1 )} .

\frac{r _{1} r _{2} r _{3}}{j _{1} j _{2} j _{3}} = \frac{( p _{2} p _{3} - 1 ) ( p _{1} p _{3} - 1 ) ( p _{1} p _{2} - 1 )}{( p _{1}^{2} - 1 ) ( p _{2}^{2} - 1 ) ( p _{3}^{2} - 1 )} .

\frac{r _{1} r _{2} r _{3}}{j _{1} j _{2} j _{3}} = \frac{n ^{2} - p _{1}^{2} p _{2} p _{3} - p _{1} p _{2}^{2} p _{3} - p _{1} p _{2} p _{3}^{2} + p _{1} p _{2} + p _{2} p _{3} + p _{1} p _{3} - 1}{n ^{2} - p _{1}^{2} p _{2}^{2} - p _{1}^{2} p _{3}^{2} - p _{2}^{2} p _{3}^{2} + p _{1}^{2} + p _{2}^{2} + p _{3}^{2} - 1} .

\frac{r _{1} r _{2} r _{3}}{j _{1} j _{2} j _{3}} = \frac{n ^{2} - p _{1}^{2} p _{2} p _{3} - p _{1} p _{2}^{2} p _{3} - p _{1} p _{2} p _{3}^{2} + p _{1} p _{2} + p _{2} p _{3} + p _{1} p _{3} - 1}{n ^{2} - p _{1}^{2} p _{2}^{2} - p _{1}^{2} p _{3}^{2} - p _{2}^{2} p _{3}^{2} + p _{1}^{2} + p _{2}^{2} + p _{3}^{2} - 1} .

\frac{r _{1} r _{2} r _{3}}{j _{1} j _{2} j _{3}} < \frac{n ^{2} + p _{1} p _{2} + p _{2} p _{3} + p _{1} p _{3}}{n ^{2} - p _{1}^{2} p _{2}^{2} - p _{1}^{2} p _{3}^{2} - p _{2}^{2} p _{3}^{2}} .

\frac{r _{1} r _{2} r _{3}}{j _{1} j _{2} j _{3}} < \frac{n ^{2} + p _{1} p _{2} + p _{2} p _{3} + p _{1} p _{3}}{n ^{2} - p _{1}^{2} p _{2}^{2} - p _{1}^{2} p _{3}^{2} - p _{2}^{2} p _{3}^{2}} .

\frac{1 + ( p _{1} p _{2} + p _{2} p _{3} + p _{1} p _{2} ) / n ^{2}}{1 - ( p _{1}^{2} p _{2}^{2} + p _{1}^{2} p _{3}^{2} + p _{2}^{2} p _{3}^{2} ) / n ^{2}} > 1 + \frac{1}{C} .

\frac{1 + ( p _{1} p _{2} + p _{2} p _{3} + p _{1} p _{2} ) / n ^{2}}{1 - ( p _{1}^{2} p _{2}^{2} + p _{1}^{2} p _{3}^{2} + p _{2}^{2} p _{3}^{2} ) / n ^{2}} > 1 + \frac{1}{C} .

\frac{r _{1} r _{2} r _{3}}{j _{1} j _{2} j _{3}} > \frac{n ^{2} - p _{1}^{2} p _{2} p _{3} - p _{1} p _{2}^{2} p _{3} - p _{1} p _{2} p _{3}^{2}}{n ^{2} + p _{1}^{2} + p _{2}^{2} + p _{3}^{2}} .

\frac{r _{1} r _{2} r _{3}}{j _{1} j _{2} j _{3}} > \frac{n ^{2} - p _{1}^{2} p _{2} p _{3} - p _{1} p _{2}^{2} p _{3} - p _{1} p _{2} p _{3}^{2}}{n ^{2} + p _{1}^{2} + p _{2}^{2} + p _{3}^{2}} .

\frac{1 - ( p _{1}^{2} p _{2} p _{3} + p _{1} p _{2}^{2} p _{3} + p _{1} p _{2} p _{3}^{2} ) / n ^{2}}{1 + ( p _{1}^{2} + p _{2}^{2} + p _{3}^{2} ) / n ^{2}} < 1 - \frac{1}{C} .

\frac{1 - ( p _{1}^{2} p _{2} p _{3} + p _{1} p _{2}^{2} p _{3} + p _{1} p _{2} p _{3}^{2} ) / n ^{2}}{1 + ( p _{1}^{2} + p _{2}^{2} + p _{3}^{2} ) / n ^{2}} < 1 - \frac{1}{C} .

(p_{1}^{2} - 1) (p_{2}^{2} - 1) (p_{3}^{2} - 1) = (p_{2} p_{3} - 1) (p_{1} p_{3} - 1) (p_{1} p_{2} - 1) .

(p_{1}^{2} - 1) (p_{2}^{2} - 1) (p_{3}^{2} - 1) = (p_{2} p_{3} - 1) (p_{1} p_{3} - 1) (p_{1} p_{2} - 1) .

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Full text

A Probable Prime Test With High Confidence

Jon Grantham

Institute for Defense Analyses, Center for Computing Sciences, 17100 Science Drive, Bowie, MD 20715

Monier and Rabin proved that an odd composite can pass the Strong Probable Prime Test for at most $\frac{1}{4}$ of the possible bases. In this paper, a probable prime test is developed using quadratic polynomials and the Frobenius automorphism. The test, along with a fixed number of trial divisions, ensures that a composite $n$ will pass for less than $\frac{1}{7710}$ of the polynomials $x^{2}-bx-c$ with $\left(b^{2}+4c\over n\right)=-1$ and $\left(-c\over n\right)=1$ . The running time of the test is asymptotically $3$ times that of the Strong Probable Prime Test.

:

11Y11

§1 Background

Perhaps the most common method for determining whether or not a number is prime is the Strong Probable Prime Test. Given an odd integer $n$ , let $n=2^{r}s+1$ with $s$ odd. Choose a random integer $a$ with $1\leq a\leq n-1$ . If $a^{s}\equiv 1\bmod n$ or $a^{2^{j}s}\equiv-1\bmod n$ for some $0\leq j\leq r-1$ , then $n$ passes the test. An odd prime will pass the test for all $a$ .

The test is very fast; it requires no more than $(1+o(1))\log_{2}n$ multiplications mod $n$ , where $\log_{2}n$ denotes the base $2$ logarithm. The catch is that a number which passes the test is not necessarily prime. Monier [9] and Rabin [13], however, showed that a composite $n$ passes for at most $\frac{1}{4}$ of the possible bases $a$ . Thus, if the bases $a$ are chosen at random, composite $n$ will pass $k$ iterations of the Strong Probable Prime Test with probability at most $\frac{1}{4^{k}}$ .

Recently, Arnault [2] has shown that any composite $n$ passes the Strong Lucas Probable Prime Test for at most $\frac{4}{15}$ of the bases $(b,c)$ , unless $n$ is the product of twin primes having certain properties (these composites are easy to detect). Also, Jones and Mo [8] have introduced an Extra Strong Lucas Probable Prime Test. Composite $n$ will pass this test with probability at most $\frac{1}{8}$ for a random choice of the parameter $b$ . These authors did not concern themselves with the issue of running time. The methods of [10], however, show each test can be performed in twice the time it takes to perform the Strong Probable Prime Test. By contrast, two iterations of the Strong Probable Prime Test take the same amount of time and are passed by a composite with probability at most $\frac{1}{16}$ .

Pomerance, Selfridge and Wagstaff have proposed a test, based on a combination of the Strong Probable Prime Test and the Lucas Probable Prime Test, that seems very powerful [12]. Indeed, nobody has yet claimed the $620 that they offer for a composite that passes it [6], even though they have relaxed the conditions of the offer so that the Probable Prime Test no longer need be “Strong”. Adams and Shanks proposed a test based on Perrin’s sequence [1]. The Q and I cases of their test also have no known pseudoprimes. The success of both of these tests suggests that it is possible to construct a test that is, in some sense, stronger than the Strong Probable Prime Test.

My goal is to provide a test which is always passed by primes, but passed by composites with probability less than $\frac{1}{7710}$ . This test has running time bounded by $(3+o(1))\log_{2}n$ multiplications mod $n$ . By comparison, $3$ iterations of the Strong Probable Prime Test have a running time bounded by $(3+o(1))\log_{2}n$ multiplications and have a probability of error at most $\frac{1}{64}$ .

The number $\frac{1}{7710}$ is slightly greater than $\frac{17}{2^{17}}+\epsilon$ , and $\frac{17}{2^{17}}$ is the limit of the analysis in this paper. The lack of counterexamples to the PSW and Perrin tests indicates that the true probability of error may be much lower. See Section 4 for possible improved versions of the test.

I would like to thank Carl Pomerance for his comments on this paper, including the suggestion of the technique described in Proposition 3.3.

§2 The Quadratic Frobenius Test

The following test is based on the concepts of Frobenius probable primes and strong Frobenius probable primes, which I introduced in another paper [5]. These tests involve computations in the ring $\mathbb{Z}[x]/(n,f(x))$ , where $f(x)\in\mathbb{Z}[x]$ and $n$ is an odd positive integer. We will be considering the special case $f(x)=x^{2}-bx-c$ . For convenience of notation, we introduce the integer $B$ , which we later show can be taken to be $50000$ .

Definition

Suppose $n>1$ is odd, $\left(b^{2}+4c\over n\right)=-1$ , and $\left(-c\over n\right)=1$ . The Quadratic Frobenius Test (QFT) with parameters $(b,c)$ consists of the following.

Test $n$ for divisibility by primes less than or equal to $\min\{B,\sqrt{n}\}$ . If it is divisible by one of these primes, declare $n$ to be composite and stop.
Test whether $\sqrt{n}\in\mathbb{Z}$ . If it is, declare $n$ to be composite and stop.
Compute $x^{\frac{n+1}{2}}\bmod(n,x^{2}-bx-c)$ . If $x^{\frac{n+1}{2}}\not\in\mathbb{Z}/n\mathbb{Z}$ , declare $n$ to be composite and stop.
Compute $\topsmash{x^{n+1}}\bmod(n,x^{2}-bx-c)$ . If $\topsmash{x^{n+1}}\not\equiv-c$ , declare $n$ to be composite and stop.
Let $\topsmash{n^{2}-1=2^{r}s}$ , where $s$ is odd. If $x^{s}\not\equiv 1\bmod(n,x^{2}-bx-c)$ , and $\topsmash{x^{2^{j}s}\not\equiv-1\bmod(n,x^{2}-bx-c)}$ for all $0\leq j\leq r-2$ , declare $n$ to be composite and stop.

If $n$ is not declared composite in Steps 1–5, declare $n$ to be a probable prime.

Definition

Suppose $n,b,c$ are as above. We say $n$ passes the QFT with parameters $(b,c)$ if the test declares $n$ to be a probable prime.

The requirement that $\left(-c\over n\right)=1$ complicates the test somewhat. Without it and Step 3, an accuracy of only $\frac{1}{495}$ can be proven.

Steps 4 and 5 of the QFT are equivalent to the Strong Frobenius Probable Prime Test described in [5] for the polynomial $x^{2}-bx-c$ . As is stated there, the test is not entirely new, but rather a combination of ideas contained in earlier probable prime tests. The stricter condition $j\leq r-2$ is possible because the restriction $\left(-c\over n\right)=1$ forces $x$ to be a square in the finite field $\mathbb{Z}[x]/(n,x^{2}-bx-c)$ , if $n$ is a prime.

Traditionalists might prefer to rephrase the test in terms of Lucas sequences. For example, Step 4 is equivalent to $U_{n+1}(b,c)\equiv 0$ and $V_{n+1}(b,c)\equiv-2c\bmod n$ , where $U_{k}$ and $V_{k}$ are the standard Lucas sequences. While this rephrasing might be useful to understand what is going on in the test, the proof of accuracy relies on properties of finite fields, so I do not feel it beneficial to use this notation.

Definition

We define one iteration of the Random Quadratic Frobenius Test (RQFT) for an odd integer $n>1$ to consist of the following:

Choose pairs $(b,c)$ at random with $1\leq b,c<n$ until one is found with $\left({b^{2}+4c}\over n\right)=-1$ and $\left(-c\over n\right)=1$ , or with $\gcd(b^{2}+4c,n)$ , $\gcd(b,n)$ or $\gcd(c,n)$ a nontrivial divisor of $n$ . However, if the latter case occurs before the former, declare $n$ to be composite and stop. If after $B$ pairs are tested, none is found satisfying the above conditions, declare $n$ to be a probable prime and stop.
Perform the QFT with parameters $(b,c)$ .

Of course, if more than one iteration of the RQFT is performed, Steps 1 and 2 of the QFT can be omitted in subsequent iterations.

Step 1 of the RQFT requires a declaration of probable primality if one is extremely unlucky in choosing pairs $(b,c)$ . An objection could be made that the declaration is not based on any actual evidence that $n$ is prime. This objection would be an accurate one, but the declaration is only done in the extremely unlikely case that no suitable pair is found. Since we want to be certain to declare primes to be probable primes, we must declare $n$ to be a probable prime.

Without this limit on the number of pairs, the running time of the test is not deterministic. Purists are welcome to delete this portion of the test, but they will be left with a probabilistic running time.

Proposition 2.1

If $p$ is an odd prime with $\topsmash{\left({b^{2}+4c}\over p\right)}=-1$ and $\topsmash{\left(-c\over p\right)}=1$ , then $p$ passes the Quadratic Frobenius Test with parameters $(b,c)$ .

Demonstration Proof

Proposition 2.1 follows from elementary properties of finite fields. Most of these can be found in [5]. The only differences here are the condition that $\left(-c\over p\right)=1$ , Step 3, and the elimination of the possibility that $x^{2^{r-1}s}\equiv-1$ .

Since $x^{p+1}\equiv-c$ and $\left(-c\over p\right)=1$ , we have $x^{\frac{p+1}{2}}\in\mathbb{Z}/p\mathbb{Z}$ , so $p$ passes Step 3.

It remains to show that $\left(-c\over p\right)=1$ implies that $x^{2^{r-1}s}\equiv 1$ . Note that $x^{2^{r-1}s}=x^{\frac{p^{2}-1}{2}}=(x^{(p+1)/2})^{p-1}\equiv 1$ , since $x^{\frac{p+1}{2}}\in\mathbb{Z}/p\mathbb{Z}$ .

Proposition 2.2

Let $p$ be an odd prime. Let $\epsilon_{1},\epsilon_{2}\in\{-1,1\}$ . If $\epsilon_{1}\neq\epsilon_{2}$ , there are $\frac{(p-1)^{2}}{4}$ pairs $(b,c)\bmod p$ such that $\botsmash{\left({b^{2}+4c}\over p\right)}=\epsilon_{1}$ and $\botsmash{\left(-c\over p\right)}=\epsilon_{2}$ , otherwise there are at most $\frac{(p-1)^{2}}{4}-\epsilon_{1}\left(\frac{p-1}{2}\right)$ such pairs.

Demonstration Proof

Pick any non-square $r\bmod p$ . Let $R=1$ if $\epsilon_{2}=1$ , and $R=r$ if $\epsilon_{2}=-1$ . Let $S=1$ if $\epsilon_{1}=1$ and $S=r$ if $\epsilon_{1}=-1$ . For each pair $(b,c)$ there exist four pairs $(x,y)$ with $-c\equiv Rx^{2}$ and $b^{2}+4c\equiv Sy^{2}$ . Substituting, $b^{2}=R(2x)^{2}+Sy^{2}$ .

We need to count

[TABLE]

If $\epsilon_{1}=\epsilon_{2}=1$ , then

[TABLE]

If $\epsilon_{1}=\epsilon_{2}=-1$ , then

[TABLE]

So $N_{2}\leq(p^{2}-1)/4+(p-1)/2$ .

If $\epsilon_{1}\neq\epsilon_{2}$ , then

[TABLE]

Proposition 2.3

Let $n$ be an odd squarefree number. Let $\epsilon_{1},\epsilon_{2}\in\{-1,1\}$ . If $\epsilon_{1}\neq\epsilon_{2}$ , there are at most $\frac{1}{4}\phi(n)^{2}$ pairs $(b,c)\bmod n$ such that $\left(b^{2}+4c\over n\right)=\epsilon_{1}$ and $\left(-c\over n\right)=\epsilon_{2}$ , otherwise there are at most $\frac{1}{4}\phi(n)^{2}+\frac{\epsilon_{1}}{2}\mu(n)\phi(n)$ such pairs.

Demonstration Proof

Let $n=p_{1}\dots p_{k}$ , where the $p_{i}$ are primes. The case $k=1$ follows from Proposition 2.2.

We proceed by induction on $k$ . Let $N_{\epsilon_{1},\epsilon_{2}}(n)$ be the number of pairs $(b,c)\bmod n$ such that $\left(b^{2}+4c\over n\right)=\epsilon_{1}$ and $\left(-c\over n\right)=\epsilon_{2}$ .

Let $n=mp$ . Then, by the Chinese Remainder Theorem,

[TABLE]

By the inductive hypothesis, if $\epsilon_{1}=\epsilon_{2}$ ,

[TABLE]

If $\epsilon_{1}\neq\epsilon_{2}$ ,

[TABLE]

Proposition 2.4

Let $n$ be an odd composite, not a perfect square. Let $M(n)$ be the number of pairs $(b,c)\bmod n$ such that $\left(b^{2}+4c\over n\right)=-1$ and $\left(-c\over n\right)=1$ , $n>(b^{2}+4c,n)>1$ , or $n>(c,n)>1$ . Then $M(n)>\frac{n^{2}}{4}$ .

Demonstration Proof

If $n$ is squarefree, exactly $\frac{3}{4}\phi(n)^{2}$ pairs have $\left(b^{2}+4c\over n\right)$ and $\left(-c\over n\right)$ non-zero, but not equal to the specified values. Note that the number of such pairs mod $np^{2}$ is $\frac{3}{4}\phi(np^{2})^{2}$ , since each such pair mod $n$ corresponds to $\frac{\phi(np^{2})^{2}}{\phi(n)^{2}}$ such pairs mod $np^{2}$ . Thus we can remove the restriction that $n$ be squarefree.

There are $n^{2}$ pairs $(b,c)\bmod n$ . Exactly $n$ pairs have $n|b^{2}+4c$ and exactly $n$ have $n|c$ . There is an overlap of one pair, $(0,0)$ .

So $M(n)>n^{2}-\frac{3}{4}\phi(n)^{2}-2n>n^{2}-\frac{3}{4}(n-2)^{2}-2n=\frac{n^{2}}{4}+n-3>\frac{n^{2}}{4}$ .

Corollary 2.5

The probability of failing to find a suitable pair $(b,c)$ in Step 1 of the RQFT is less than $(3/4)^{B}$ .

Definition

A positive integer $n$ is said to pass the Random Quadratic Frobenius Test with probability $\alpha$ if the number of $(b,c)$ with $0\leq b,c\leq n$ such that $n$ passes the Quadratic Frobenius Test with parameters $(b,c)$ is equal to $\alpha M(n)$ .

Theorem 2.6

An odd composite passes the RQFT with probability less than $\frac{1}{7710}$ .

The proof of Theorem 2.6 will be given in a sequence of lemmas.

Lemma 2.7

Let $n$ be an odd integer. If $p$ is a prime such that $p^{2}$ divides $n$ , then $n$ passes the RQFT with probability less than $\frac{4}{p}$ .

Demonstration Proof

Let $k$ be such that $\topsmash{p^{k}}|n$ , but $\topsmash{p^{k+1}}\nmid n$ . If $n$ passes the QFT with parameters $(b,c)$ , then $x^{n+1}\equiv-c\bmod(n,x^{2}-bx-c)$ . By (5) in the definition of the QFT, $x^{n^{2}-1}\equiv 1\bmod(n,x^{2}-bx-c)$ . So ${c}^{n-1}\equiv 1\bmod p^{k}$ . There can be at most $\gcd(n-1,p^{k}-p^{k-1})=\gcd(n-1,p-1)\leq{p-1}$ solutions to this congruence mod $p^{k}$ . Hence there are at most $p-1$ choices for $c\bmod p^{k}$ for which it is possible that $n$ passes the QFT with parameters $(b,c)$ , for some $b$ . Thus there are at most $p^{k+1}-p^{k}$ pairs $(b,c)\bmod p^{k}$ such that $n$ passes the QFT with parameters $(b,c)$ .

By the Chinese Remainder Theorem, each pair mod $p^{k}$ corresponds to $(n/p^{k})^{2}$ pairs mod $n$ . Thus $n$ passes for at most $\botsmash{(p^{k+1}-p^{k})\frac{n^{2}}{p^{2k}}=(1-\frac{1}{p})\frac{n^{2}}{p^{k-1}}<\frac{4}{p^{k-1}}M(n)}$ pairs by Proposition 2.4.

The lemma follows since $k\geq 2$ .

Lemma 2.8

Let $n$ be an odd composite with $p|n$ . We write that “ $n$ passes the QFT with parameters $(b,c)\bmod p$ ” if $n$ passes the QFT for some parameters $(b^{\prime},c^{\prime})$ , with $(b^{\prime},c^{\prime})\equiv(b,c)\bmod p$ . There are at most $\frac{p-1}{2}$ distinct pairs $(b,c)\bmod p$ with $\left({b^{2}+4c}\over p\right)=1$ such that $n$ passes the Quadratic Frobenius Test with parameters $(b,c)\bmod p$ .

Demonstration Proof

First, we count the number of pairs $(b,c)\bmod p$ with $\left({b^{2}+4c}\over p\right)=1$ such that $x^{n+1}\equiv-c\bmod(p,x^{2}-bx-c)$ . If $a_{1}$ and $a_{2}$ are the two roots of $x^{2}-bx-c\bmod p$ , then the above congruence gives $x^{n+1}\equiv-c\bmod(p,x-a_{1})$ . This congruence is equivalent to $a_{1}^{n+1}\equiv-c\equiv a_{1}a_{2}\bmod p$ , or $a_{1}^{n}\equiv a_{2}$ , since $p\nmid a_{1}$ . Similarly, $a_{2}^{n}\equiv a_{1}\bmod p$ .

Thus the number of polynomials $x^{2}-bx-c$ with $x^{n}\equiv b-x$ is equal to the number of sets of integers $\{a_{1},a_{2}\}\bmod p$ with $a_{1}^{n}\equiv a_{2}\bmod p$ and $a_{2}^{n}\equiv a_{1}\bmod p$ . This is no more than $(p-1)/2$ .

Lemma 2.9

The number of pairs $(b,c)\bmod n$ for which a squarefree integer $n$ with $k$ prime factors passes the QFT with parameters $(b,c)$ , such that $\left({b^{2}+4c}\over p\right)=1$ for some $p|n$ , is less than $\frac{n\phi(n)}{2B}$ if $k$ is even and less than $\frac{n\phi(n)}{B^{2}}$ if $k$ is odd.

Demonstration Proof

Write $n=p_{1}p_{2}\dots p_{k}$ . Note $k>1$ , since $k=1$ would imply that $\left({b^{2}+4c}\over n\right)=-1$ and $\left({b^{2}+4c}\over p_{1}\right)=1$ .

We consider separately the cases where $\left({b^{2}+4c}\over p_{i}\right)=1$ for exactly one $i$ , and for more than one $i$ .

**Case 1. ** Consider those pairs $(b,c)$ for which $n$ passes the QFT with parameters $(b,c)$ such that $\left({b^{2}+4c}\over p_{i}\right)=1$ for exactly one $p_{i}|n$ . Note that such an $n$ can only have $\left({b^{2}+4c}\over n\right)=-1$ when $k$ is even.

There are at most $\frac{p_{i}-1}{2}$ pairs $(b,c)\bmod p_{i}$ for which $n$ can pass the QFT, and for $j\neq i$ , at most $\frac{p_{j}^{2}-p_{j}}{2}$ pairs mod $p_{j}$ with $\left({b^{2}+4c}\over p_{j}\right)=-1$ .

So by the Chinese Remainder Theorem, there are at most $\sum_{1\leq i\leq k}\frac{n\phi(n)}{2^{k}p_{i}}$ pairs $(b,c)$ for which $n$ passes the QFT and $\left({b^{2}+4c}\over p_{i}\right)=1$ for exactly one $p_{i}|n$ . Since any prime $p_{i}|n$ must be at least $B$ , the total number of pairs is at most $\frac{kn\phi(n)}{2^{k}B}$ .

**Case 2. ** For each pair $(b,c)$ , let $L(b,c)$ be the $k$ -tuple whose $i$ th coordinate is $\left({b^{2}+4c}\over p_{i}\right)$ .

The $k$ -tuples in $\{1,-1\}^{k}$ with at least two $1$ s correspond to pairs $(b,c)$ with $\left({b^{2}+4c}\over p\right)=1$ for more than one prime $p|n$ . Let $V$ denote this set of $k$ -tuples.

Given an element of $L\in V$ , let $p_{i}<p_{j}$ be the largest two primes with $\left({b^{2}+4c}\over p\right)=1$ . By Lemma 2.8, the number of pairs $(b,c)$ for which $n$ passes the QFT mod $p_{i}$ is at most $\frac{p_{i}-1}{2}$ , and similarly mod $p_{j}$ . The number of pairs mod $p_{k}$ for $k\neq i,j$ is at most $p_{k}(p_{k}-1)/2$ . By the Chinese Remainder Theorem, the total number of pairs $(b,c)$ with $L(b,c)=L$ such that $n$ passes the QFT with parameters $(b,c)$ is at most $[(p_{i}-1)/2][(p_{j}-1)/2]\prod_{k\neq i,j}p_{k}(p_{k}-1)/2=\frac{n\phi(n)}{2^{k}p_{i}p_{j}}$ . The total number of $k$ -tuples $L\in V$ is less than $2^{k}$ .

So the number of pairs $(b,c)$ such that $\smash{\left({b^{2}+4c}\over p\right)=1}$ for more than one prime dividing $n$ is less than or equal to

[TABLE]

We now combine the analysis of the two cases to complete the proof of the Lemma.

For $k=2$ , we have $\left({b^{2}+4c}\over p_{1}\right)=1$ and $\left({b^{2}+4c}\over p_{2}\right)=-1$ , or vice versa, so Case 2 is impossible. The proof easily follows by taking $k=2$ in Case 1.

For $k>2$ even, we have shown that the total number of pairs $(b,c)$ mod $n$ for which $n$ can pass the QFT is less than $n\phi(n)\left(\frac{k}{2^{k}B}+\frac{1}{B^{2}}\right)$ . This is less than $\frac{n\phi(n)}{2B}$ .

For $k$ odd, Case 1 cannot occur, so the Lemma is proven.

Corollary 2.10

A squarefree integer $n$ with an even number of prime factors passes the RQFT with probability less than $\frac{2}{B}$ .

Demonstration Proof

Combine Proposition 2.4 with Lemma 2.9.

Lemma 2.11 is a more exact form of Proposition 6.2 of [5] for quadratic polynomials.

Lemma 2.11

If an odd squarefree number $n$ has $3$ prime factors, the probability that it passes the RQFT is less than $\frac{4}{B^{2}}+\frac{3(B^{2}+1)}{2(B^{4}-3B^{2})}$ .

Demonstration Proof

Write $n=p_{1}p_{2}p_{3}$ . By Lemma 2.9, we know that $n$ passes for at most $\frac{4M(n)}{B^{2}}$ pairs with $\left({b^{2}+4c}\over n\right)=-1$ and $\left({b^{2}+4c}\over p_{i}\right)=1$ for some $i$ . So it suffices to show that $n$ passes the QFT for at most $\frac{3(B^{2}+1)}{2(B^{4}-3B^{2})}\frac{n^{2}}{4}$ pairs with $\left({b^{2}+4c}\over p_{i}\right)=-1$ for all $i$ .

We know that $x^{n+1}\equiv-c\bmod(p_{i},x^{2}-bx-c)$ , and $x^{p_{i}+1}\equiv-c$ . Since $c$ is invertible mod $n$ , $x$ is invertible mod $(n,x^{2}-bx-c)$ , and $x^{n-p_{i}}\equiv 1$ . For each $p_{i}$ , we need to know how many solutions there are in $\mathbb{F}_{p_{i}^{2}}$ to $y^{n-p_{i}}\equiv 1$ . This congruence holds for exactly $\gcd(n-p_{i},p_{i}^{2}-1)$ elements. Each pair $(b,c)$ corresponds to two elements with minimal polynomial $x^{2}-bx-c$ , which either both do or both do not satisfy $y^{n-p_{i}}\equiv 1$ .

Let $\gcd(n-p_{i},p_{i}^{2}-1)=k_{i}$ . By the preceding paragraph, there are at most $k_{i}/2$ pairs $(b,c)\bmod p_{i}$ with $x^{n-p_{i}}\equiv 1\bmod x^{2}-bx-c\bmod p_{i}$ . Then by the Chinese Remainder Theorem, $n$ passes the QFT for at most $\botsmash{k_{1}k_{2}k_{3}/8}$ pairs with all Jacobi symbols equal to $-1$ .

Since $(p_{1},p_{1}^{2}-1)=1$ , $k_{1}=\gcd(p_{2}p_{3}-1,p_{1}^{2}-1)$ , and similarly for $k_{2}$ and $k_{3}$ . Let $j_{i}=(p_{i}^{2}-1)/k_{i}$ . Let $r_{1}=(p_{2}p_{3}-1)/k_{1}$ , and define $r_{2}$ and $r_{3}$ analogously.

Then $r_{1}(p_{1}^{2}-1)=j_{1}(p_{2}p_{3}-1)$ , $r_{2}(p_{2}^{2}-1)=j_{2}(p_{1}p_{3}-1)$ , and $r_{3}(p_{3}^{2}-1)=j_{3}(p_{1}p_{2}-1)$ .

Let $C=j_{1}j_{2}j_{3}$ . Assume that $r_{1}r_{2}r_{3}>C$ . Then $\frac{r_{1}r_{2}r_{3}}{j_{1}j_{2}j_{3}}\geq 1+\frac{1}{C}$ .

Multiplying the three equalities above, we have that

[TABLE]

Multiplying the right hand side out,

[TABLE]

Thus

[TABLE]

Therefore

[TABLE]

We now use the facts that $n=p_{1}p_{2}p_{3}$ and $p_{i}>B$ to obtain $\frac{1+3/B^{4}}{1-3/B^{2}}>1+\frac{1}{C}$ .

Thus $C>\frac{B^{4}-3B^{2}}{3B^{2}+3}$ .

If $r_{1}r_{2}r_{3}<C$ , then $\frac{r_{1}r_{2}r_{3}}{j_{1}j_{2}j_{3}}\leq 1-\frac{1}{C}$ . From above,

[TABLE]

Therefore

[TABLE]

We now use the facts that $n=p_{1}p_{2}p_{3}$ and $p_{i}>B$ to obtain $\frac{1-3/B^{2}}{1+3/B^{4}}<1-\frac{1}{C}$ . Thus $C>\frac{B^{4}+3}{3B^{2}+3}>\frac{B^{4}-3B^{2}}{3B^{2}+3}$ .

We have that $k_{1}k_{2}k_{3}/8<\frac{n^{2}}{8C}$ . The lemma is proven unless $j_{1}j_{2}j_{3}=r_{1}r_{2}r_{3}$ . We will now show that this condition is impossible.

Once again, we multiply out the three equalities involving the $r_{i}$ and $j_{i}$ terms, but this time we can cancel $r_{1}r_{2}r_{3}$ with $j_{1}j_{2}j_{3}$ . We get

[TABLE]

But $(p_{1}^{2}-1)(p_{2}^{2}-1)<(p_{1}p_{2}-1)^{2}$ . Multiplying by similar inequalities for $p_{1},p_{3}$ and $p_{2},p_{3}$ , and taking the square root, we have a contradiction.

Therefore, the only possibility is $j_{1}j_{2}j_{3}\neq r_{1}r_{2}r_{3}$ , and we have shown that this assumption gives the probability stated in the theorem.

Lemma 2.12

If $n$ is squarefree and has $k$ prime factors, where $k$ is odd, $n$ passes the RQFT with probability less than $\frac{1}{2^{3k-2}}+\frac{1}{2^{4k-3}}+\frac{4}{B^{2}}$ .

Demonstration Proof

By Lemma 2.9, the number of pairs $(b,c)$ such that $\left({b^{2}+4c}\over p\right)=1$ for some prime $p|n$ , and $n$ passes the QFT with parameters $(b,c)$ , is less than $\frac{n^{2}}{B^{2}}$ .

Now assume that $\left({b^{2}+4c}\over p\right)=-1$ for all $p|n$ .

Write $n=p_{1}p_{2}\dots p_{k}$ . Let $J$ be the largest integer such that $2^{J+1}|\gcd(p_{1}^{2}-1,\dots,p_{k}^{2}-1)$ . Then for each $i$ , $p_{i}^{2}\equiv 1\bmod 2^{J+1}$ , so $n^{2}\equiv 1\bmod 2^{J+1}$ . Thus $J<r$ .

The number of solutions that $y^{2^{j}s}\equiv-1$ can have in $\mathbb{F}_{p^{2}}$ is either [math] or $\gcd(p^{2}-1,2^{j}s)$ . Furthermore, it only has solutions if $-1$ is a perfect $2^{j}$ th power. This will only happen if $2^{j+1}|{p^{2}-1}$ .

So there are no pairs $(b,c)$ for which $\topsmash{x^{2^{j}s}\equiv-1\bmod(n,x^{2}-bx-c)}$ if $j>J$ .

For at least one prime $p|n$ , $2^{J+1}$ is the highest power of $2$ dividing $p^{2}-1$ . In Step 4 of the QFT, we show that $-c$ is a square mod $n$ , and therefore mod $p$ . By Proposition 2.1, it follows that $x^{\frac{p^{2}-1}{2}}\equiv 1\bmod(p,x^{2}-bx-c)$ . So $x$ has order dividing $2^{J}\ell$ , for some odd number $\ell$ . If $x^{2^{J}s}\equiv-1\bmod(p,x^{2}-bx-c)$ , then $2^{J+1}$ divides the order of $x$ , and we have a contradiction. Thus there are no pairs for which $x^{2^{J}s}\equiv-1\bmod(n,x^{2}-bx-c)$ .

We consider $0\leq j<J$ . For each $p_{i}$ , there are $\gcd(p_{i}^{2}-1,2^{j}s)$ solutions to $y^{2^{j}s}\equiv-1$ in $\mathbb{F}_{p_{i}^{2}}$ . Each solution $y\not\in\mathbb{F}_{p_{i}}$ produces a solution to $x^{2^{j}s}\equiv-1\bmod(p_{i},x^{2}-bx-c)$ , where $x^{2}-bx-c$ is the minimum polynomial of $y$ . But for each minimum polynomial, both of its roots in $\mathbb{F}_{p_{i}^{2}}$ will be solutions. (The roots are distinct by the restriction on $c$ .)

Therefore, there are at most $\gcd(p_{i}^{2}-1,2^{j}s)/2$ pairs $(b,c)$ mod $p_{i}$ for which $x^{2^{j}s}\equiv-1\bmod(p_{i},x^{2}-bx-c)$ . Let $G=\prod_{p_{i}}\gcd(p_{i}^{2}-1,s)$ . By the Chinese Remainder Theorem, there are at most $\prod_{1\leq i\leq k}\gcd(p_{i}^{2}-1,2^{j}s)/2$ pairs mod $n$ for which $x^{2^{j}s}\equiv-1$ mod $(n,x^{2}-bx-c)$ . But $\prod_{1\leq i\leq k}\gcd(p_{i}^{2}-1,2^{j}s)/2=\prod_{1\leq i\leq k}2^{j-1}\gcd(p_{i}^{2}-1,s)=2^{(j-1)k}G$ . Similarly, the number of pairs $(b,c)$ for which $x^{s}\equiv 1$ mod $(n,x^{2}-bx-c)$ is at most $G/2^{k}$ . Thus the total number of pairs for which $n$ passes the QFT is bounded by $G/2^{k}+\sum_{j=0}^{J-1}2^{(j-1)k}G=\left[1+\frac{2^{Jk}-1}{2^{k}-1}\right]\frac{G}{2^{k}}$ .

Since $\gcd(p_{i}^{2}-1,s)<p_{i}^{2}/2^{J+1}$ , the number of pairs for which $n$ passes the QFT is less than $\left[1+\frac{2^{Jk}-1}{2^{k}-1}\right]\frac{n^{2}}{2^{(J+1)k}2^{k}}$ . We have $J\geq 2$ , and this expression is maximized at $J=2$ , where it is equal to $(2^{k}+2)\frac{n^{2}}{2^{4k}}=n^{2}(\frac{1}{2^{3k}}+\frac{1}{2^{4k-1}})$ .

The lemma now follows from Proposition 2.4.

Demonstration Proof of Theorem 2.6

Choose $B=50000$ . If $n$ is not squarefree, or has an even number of prime factors, Lemma 2.7 and Corollary 2.10 prove Theorem 2.6. If $n$ is squarefree and has $3$ prime factors, we apply Lemma 2.11. If $n$ is squarefree and has $k$ prime factors, $k>3$ , we apply Lemma 2.12. The bound for the probability given by that lemma is largest when $k=5$ and is $1/2^{13}+1/2^{17}+1/25000^{2}$ .

Note that adding on the number $\topsmash{\left(\frac{3}{4}\right)^{B}}$ from Corollary 2.5 does not increase the probability above $\frac{1}{7710}$ .

§3 Running Time

Atkin [3] has suggested a unit of running time of probable prime tests based on the running time of the Strong Probable Prime Test. The unit is named the “selfridge” to honor John Selfridge for his discovery of the Strong Probable Prime Test. We give a slightly different formalization below.

Definition

An algorithm with input $n$ is said to have running time of $k$ selfridges if it can be completed in the time it takes to perform $(k+o(1))\log_{2}n$ multiplications mod $n$ . Calculation of an inverse will take time equal to $O(1)$ multiplications and computation of $\left({\ }\over n\right)$ will take $O(1)$ multiplications. Here, $O(1)$ is bounded as $n\rightarrow\infty$ . We will assume that addition takes $o(1)$ multiplications, where $o(1)\rightarrow 0$ as $n\rightarrow\infty$ .

If a multiplication algorithm that took $O(\log n)$ bit operations were discovered, the last assumption would be invalidated. The definition of “selfridge,” however, is meant to be a practical, if not completely precise, method of comparing different probable prime tests. Since the development of a multiplication algorithm that takes $O(\log n)$ bit operations appears to be highly improbable, I make the definition with a clear conscience.

Note that we use the term “multiplication” to refer both to the arithmetic operation and the time required to perform it. The above definition of the term “selfridge” is motivated by the following proposition.

Proposition 3.1

The Strong Probable Prime Test has a running time of at most $1$ selfridge.

Demonstration Proof

If we write $n=2^{r}s+1$ , with $s$ odd, then the Strong Probable Prime Test requires, at most, the raising of an integer to the $s$ power mod $n$ and the completion of $r$ squarings (which are multiplications). By [7], exponentiation to the $t$ th power can be done in $(1+o(1))\log_{2}t$ multiplications using easily constructed addition chains. Since $\log_{2}n>\log_{2}s+r$ , the test can be performed in $(1+o(1))\log_{2}n$ multiplications mod $n$ . Note that for most odd $n$ , the running time of the test is equal to $1$ selfridge.

It is the goal of this section to show that the Quadratic Frobenius Test has running time of at most $3$ selfridges. One “trick” used to attain this running time relies on the use of Lemma 4.8 of [5].

Lemma 4.8 of \cite{\me}

Let $m,n$ be positive integers, and let $f(x),g(x),r(x)\in\mathbb{Z}[x]$ . If $f(r(x))\equiv 0\bmod(n,f(x))$ and $x^{m}\equiv g(x)\bmod(n,f(x))$ , then $r(x)^{m}\equiv g(r(x))\bmod(n,f(x))$ .

Before analyzing the running time, we need to know how many multiplications mod $n$ it takes to perform a multiplication mod $(n,x^{2}-bx-c)$ .

Proposition 3.2

Multiplication $\bmod(n,x^{2}-bx-c)$ can be done in at most $5+o(1)$ multiplications $\bmod\ n$ .

Demonstration Proof

To multiply $(dx+e)(fx+g)\equiv(dg+ef+bdf)x+eg+cdf\bmod(n,x^{2}-bx-c)$ , we compute the $5$ products $df$ , $eg$ , $b(df)$ , $c(df)$ , and $(d+e)(f+g)$ . Then we can compute $dg+ef+bdf=(d+e)(f+g)+bdf-df-eg$ and $eg+cdf$ by addition and subtraction.

Proposition 3.3

Let $A_{j}=x^{j}+(b-x)^{j}\bmod(n,x^{2}-bx-c)$ , and let $B_{j}=\frac{x^{j}-(b-x)^{j}}{2x-b}$ $\bmod(n,x^{2}-bx-c)$ . Let $C_{j}=c^{j}\bmod n$ . Given values of $2^{-1}$ and $(b^{2}+4c)$ , $\bmod(n,x^{2}-bx-c)$ , then $(A_{j+k},B_{j+k},C_{j+k})$ can be computed from $(A_{j},B_{j},C_{j})$ and $(A_{k},B_{k},C_{k})$ in $8+o(1)$ multiplications $\bmod\ n$ . (We call this type of computation a chain addition.) $(A_{2j},B_{2j},C_{2j})$ can be computed from $(A_{j},B_{j},C_{j})$ in $3+o(1)$ multiplications $\bmod\ n$ . (We call this type of computation a doubling.) $(A_{j},B_{j},C_{j})$ can be computed in $(3+o(1))\log_{2}j$ multiplications $\bmod n$ . $x^{j}$ can be computed from $(A_{j},B_{j})$ in $2+o(1)$ multiplications mod $n$ .

Demonstration Proof

We have $A_{j},B_{j}\in\mathbb{Z}/n\mathbb{Z}$ . Note the identities $A_{j+k}=2^{-1}(A_{j}A_{k}+(b^{2}+4c)B_{j}B_{k})$ , $B_{j+k}=2^{-1}(A_{j}B_{k}+A_{k}B_{j})$ , and $C_{j+k}=C_{j}C_{k}$ .

This shows a chain addition takes $8+o(1)$ multiplications mod $n$ .

Also note that $A_{2j}=A_{j}^{2}-2(-1)^{j}C_{j}$ , $B_{2j}=A_{j}B_{j}$ , and $C_{2j}=C_{j}^{2}$ . $2(-1)^{j}C_{j}=C_{j}+C_{j}$ or $-C_{j}-C_{j}$ (depending on the parity of $j$ ), so this part of the computation is $o(1)$ multiplications. So a doubling can be achieved in $3+o(1)$ multiplications mod $n$ .

Once we have $A_{j}$ and $B_{j}$ , we can compute $x^{j}=B_{j}x+2^{-1}(A_{j}-bB_{j})$ with one multiplication of $b$ and $B_{j}$ , one subtraction, and one multiplication by $2^{-1}$ . Thus finding $x^{j}$ from the pair $(A_{j},B_{j})$ costs $2+o(1)$ multiplications mod $n$ .

We can compute $(A_{j},B_{j},C_{j})$ using $(1+o(1))\log_{2}j$ steps (doublings or chain additions) by the addition chain methods described in [7]. Since $o(\log n)$ of these steps will not be doublings, we can compute $(A_{j},B_{j},C_{j})$ in $(3+o(1))\log_{2}j$ multiplications mod $n$ .

Theorem 3.4

The Random Quadratic Frobenius Test has running time of at most $3$ selfridges.

Demonstration Proof

For a given $n$ , Step 1 of the RQFT will take at most $B$ tries in searching for a suitable pair $(b,c)$ . Thus it takes $O(1)$ multiplications mod $n$ . It remains to show that the QFT has running time of $3$ selfridges.

Steps 1 and 2 of the QFT have running time bounded by a fixed number of multiplications mod $n$ .

Assume $n\equiv 1\bmod 4$ . Write $n-1=2^{r^{\prime}}s^{\prime}$ , with $s^{\prime}$ odd. Then $n^{2}-1=2^{r^{\prime}+1}(2^{r^{\prime}-1}{s^{\prime}}^{2}+s^{\prime})$ . So $r=r^{\prime}+1$ and $s=2^{r^{\prime}-1}{s^{\prime}}^{2}+s^{\prime}$ . Write $t=\frac{s^{\prime}-1}{2}$ .

Computing $(A_{t},B_{t},C_{t})$ takes $(3+o(1))\log_{2}t$ steps, by Proposition 3.3. Computing $x^{n}$ from $(A_{t},B_{t},C_{t})$ can be accomplished by first computing $(A_{s^{\prime}},B_{s^{\prime}},C_{s^{\prime}})$ , which requires $11$ multiplications mod $n$ . We then perform $r^{\prime}-1$ doublings to get $(A_{\frac{n-1}{2}},B_{\frac{n-1}{2}},C_{\frac{n-1}{2}})$ . We can compute $x^{\frac{n-1}{2}}$ in $2+o(1)$ multiplications and then $x^{\frac{n+1}{2}}$ in $5+o(1)$ more. This calculation completes Step 3. Squaring this result, we complete Step 4 with at most $5+o(1)$ additional multiplications.

So the total number of multiplications mod $n$ is $(3+o(1))\log_{2}t+3r^{\prime}+15+o(1)=(3+o(1))\log_{2}n$ . So Steps 3 and 4 take $3$ selfridges when $n\equiv 1\bmod 4$ .

Note that Step 5 only needs to be performed if $x^{n+1}\equiv-c\bmod(n,x^{2}-bx-c)$ , which implies $x^{n}\equiv b-x$ .

For Step 5, note that $s=nt+t+\frac{n-1}{2}+1$ and, for $0\leq e<r$ , $2^{e+1}s=n2^{e}s^{\prime}+2^{e}s^{\prime}$ .

Let $\sigma$ be the map from $(\mathbb{Z}/n\mathbb{Z})[x]/(n,x^{2}-bx-c)$ to itself that sends $x$ to $b-x$ . We have $x^{nt}\equiv(b-x)^{t}\equiv\sigma(x^{t})$ , by Lemma 4.8 of [5].

So $x^{s}\equiv x^{nt}x^{t}x^{\frac{n-1}{2}}x\equiv\sigma(x^{t})x^{t}x^{\frac{n-1}{2}}x$ . We have already computed $(A_{t},B_{t})$ and $x^{\frac{n-1}{2}}$ in the process of computing $x^{n}$ , so it takes $2+o(1)$ multiplications to find $x^{t}$ and $x^{\frac{n-1}{2}}$ . Given $x^{t}$ , we require at most $1$ multiplication mod $n$ and an addition to compute $\sigma(x^{t})$ . It takes $3+o(1)$ multiplications mod $(n,x^{2}-bx-c)$ , or $15+o(1)\bmod n$ to multiply the results together to get $x^{s}$ . So we can compute $x^{s}$ with at most $18+o(1)$ additional multiplications mod $n$ .

We have $x^{2^{e+1}s}\equiv x^{n2^{e}s^{\prime}}x^{2^{e}s^{\prime}}$ . By Lemma 4.8 of [5], $x^{n2^{e}s^{\prime}}\equiv\sigma(x^{2^{e}s^{\prime}})$ . We have already computed $(A_{2^{e}s^{\prime}},B_{2^{e}s^{\prime}})$ in the process of computing $x^{n+1}$ . It takes $2+o(1)$ multiplications to compute $x^{2^{e}s^{\prime}}$ from these. We require $1$ multiplication mod $n$ and one addition to apply $\sigma$ , and $5+o(1)$ more to find the product $x^{n2^{e}s^{\prime}}x^{2^{e}s^{\prime}}$ . So we need $8+o(1)$ multiplications mod $n$ to compute $x^{2^{e+1}s}$ .

We want to know if $x^{s}\equiv\pm 1$ or if $x^{2^{e+1}s}\equiv-1$ for some $e<r$ . This can be accomplished by a binary search.

We first check whether or not $\topsmash{x^{s}}\equiv\pm 1$ . This step takes $22+o(1)$ multiplications mod $n$ .

We begin by computing $\topsmash{x^{2^{j}s}}\bmod(n,x^{2}-bx-c)$ for $j=[\frac{r}{2}]$ . If $x^{2^{j}s}\equiv-1$ , we can stop. If $x^{2^{j}s}\equiv 1$ , we know that $x^{2^{e+1}s}\equiv 1$ for $e\geq j$ , so we only need to check those $e$ less than $j$ . If $x^{2^{j}s}\not\equiv 1$ , then we know that $x^{2^{e+1}s}\not\equiv-1$ for $e<j$ , so we only need to check those $e$ greater than or equal to $j$ . We continue in a similar manner until we either find $e$ with $x^{2^{e+1}s}\equiv-1$ or prove that none exists. This takes at most $\log_{2}r+1$ steps, each of which requires at most $8+o(1)$ multiplications. $r<\log_{2}n+1$ , so the total number of multiplications mod $n$ in Step 5 is $O(\log\log n)$ when $n\equiv 1\bmod 4$ .

If $n\equiv-1\bmod 4$ , write $\smash{n+1=2^{r^{\prime}}s^{\prime}}$ . Then $\smash{n^{2}-1=2^{r^{\prime}+1}(2^{r^{\prime}-1}{s^{\prime}}^{2}-s^{\prime})}$ . Write $t=\frac{s^{\prime}-1}{2}$ .

In Step 3, we compute $(A_{t},B_{t},C_{t})$ , and then $(A_{s^{\prime}},B_{s^{\prime}},C_{s^{\prime}})$ . We double $r^{\prime}-1$ times, and then compute $x^{\frac{n+1}{2}}$ . We then square to get $x^{n+1}$ . By a similar analysis to the case when $n\equiv 1\bmod 4$ , Steps 3 and 4 require $3$ selfridges when $n\equiv-1\bmod 4$ .

For Step 5, observe that $s=nt-t+\frac{n+1}{2}-1$ and $2^{e}s=n2^{e}s^{\prime}-{2^{e}}s^{\prime}$ . We can calculate $x^{t}$ and $x^{\frac{n+1}{2}}$ , and we can use the fact that $x^{nt}\equiv\sigma(x^{t})$ .

We can then proceed with Step 5 via the binary search described in the case where $n\equiv 1\bmod 4$ . Again, Step 5 takes $O(\log\log n)$ multiplications mod $n$ .

§4 Unanswered Questions

The most obvious question that arises is, “Can we do significantly better than $7710$ ?” Increasing $B$ will significantly improve each lemma except for Lemma 2.12. In order to do significantly better than $7710$ , then, it seems that we need to develop a better analysis of the case where $n$ has $5$ prime factors. An improved analysis of the case where $n$ has $k$ prime factors, for any odd $k$ , would be even better.

The answer to this question is probably “yes”. Part of the motivation for this test was the combined Strong Probable Prime and Lucas Probable Prime tests of Pomerance, Selfridge, and Wagstaff [12]. Integers that have $\frac{p^{2}-1}{k}|n^{2}-1$ , with $k$ small, for each $p|n$ seem to offer the best chances of “fooling” either test, but such numbers have proven difficult to construct in practice. This statement does not preclude the existence of many such numbers, but it is encouraging to note that the best heuristics for constructing such numbers [11] would produce numbers with many prime factors.

The following theorem shows that composites that pass the test with high probability have special form.

Theorem 4.1

Let $n$ be an odd composite with $k$ prime factors. Let $r,s$ be such that $n^{2}-1=2^{r}s$ with $s$ odd. Let $G=\prod_{p|n}\gcd(p^{2}-1,s)$ . Let $J$ be the largest integer such that $2^{J+1}|p^{2}-1$ for every $p|n$ . If $\frac{n^{2}}{2^{k(J+1)}G}>\frac{B}{2^{3k+1}}$ , then $n$ passes the RQFT with probability less than $\frac{4}{B}$ , for any $B\geq 50000$ .

Demonstration Proof

The theorem follows immediately unless $k$ is odd and greater than $3$ . In the proof of Lemma 2.12 we used the bound $G<\frac{n^{2}}{2^{k(J+1)}}$ . Here we have the bound $G<\frac{2^{3k+1}}{B}\frac{n^{2}}{2^{k(J+1)}}$ . The multiple of $\frac{2^{3k+1}}{B}$ carries through to the final probability bound, giving a probability of error less than $\frac{4}{B}$ .

We showed that $\frac{n^{2}}{2^{k(J+1)}G}>1$ . It would be interesting to see computational results for small $n$ , and further theoretical results that might improve Theorem 2.6.

Scott Contini asks if a result analogous to that in [4] can be proven about the reliability of the RQFT in generating primes.

Another question is whether it would be useful to construct cubic and higher order versions of the Random Quadratic Frobenius Test. A basis for doing so can be found in [5]. The fraction $\frac{1}{7710}$ would have to be improved considerably to make the increased running time in larger finite fields worthwhile. One technique for doing so would be to exploit the fact that if $(n,m)=1$ , then for $m$ with $\phi(m)|d$ , $m|n^{d}-1$ . We could actually improve the Quadratic Frobenius Test by observing that $3|n^{2}-1$ for all $n$ coprime to $3$ . Checking $\operatorname{gcmd}(x^{(n^{2}-1)/3}-1,x^{2}-bx-c)$ may expose $n$ as composite. This modification would, however, further complicate the analysis of the running time, so it will be taken up in a future paper.

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