Pépin's test explained

In mathematics, Pépin's test is a primality test, which can be used to determine whether a Fermat number is prime. It is a variant of Proth's test. The test is named after a French mathematician, Théophile Pépin.

Description of the test

Let

2n
F
n=2

+1

be the nth Fermat number. Pépin's test states that for n > 0,

Fn

is prime if and only if
(Fn-1)/2
3

\equiv-1\pmod{Fn}.

The expression
(Fn-1)/2
3
can be evaluated modulo

Fn

by repeated squaring. This makes the test a fast polynomial-time algorithm. However, Fermat numbers grow so rapidly that only a handful of Fermat numbers can be tested in a reasonable amount of time and space.

Other bases may be used in place of 3. These bases are:

3, 5, 6, 7, 10, 12, 14, 20, 24, 27, 28, 39, 40, 41, 45, 48, 51, 54, 56, 63, 65, 75, 78, 80, 82, 85, 90, 91, 96, 102, 105, 108, 112, 119, 125, 126, 130, 147, 150, 156, 160, ... .

The primes in the above sequence are called Elite primes, they are:

3, 5, 7, 41, 15361, 23041, 26881, 61441, 87041, 163841, 544001, 604801, 6684673, 14172161, 159318017, 446960641, 1151139841, 3208642561, 38126223361, 108905103361, 171727482881, 318093312001, 443069456129, 912680550401, ...

For integer b > 1, base b may be used if and only if only a finite number of Fermat numbers Fn satisfies that

\left(b
Fn

\right)=1

, where
\left(b
Fn

\right)

is the Jacobi symbol.

In fact, Pépin's test is the same as the Euler-Jacobi test for Fermat numbers, since the Jacobi symbol

\left(b
Fn

\right)

is −1, i.e. there are no Fermat numbers which are Euler-Jacobi pseudoprimes to these bases listed above.

Proof of correctness

Sufficiency: assume that the congruence

(Fn-1)/2
3

\equiv-1\pmod{Fn}

holds. Then
Fn-1
3

\equiv1\pmod{Fn}

, thus the multiplicative order of 3 modulo

Fn

divides
2n
F
n-1=2
, which is a power of two. On the other hand, the order does not divide

(Fn-1)/2

, and therefore it must be equal to

Fn-1

. In particular, there are at least

Fn-1

numbers below

Fn

coprime to

Fn

, and this can happen only if

Fn

is prime.

Necessity: assume that

Fn

is prime. By Euler's criterion,
(Fn-1)/2
3\equiv\left(
3{F
n}\right)\pmod{F

n}

,where
\left(3{F
n}\right)
is the Legendre symbol. By repeated squaring, we find that
2n
2

\equiv1\pmod3

, thus

Fn\equiv2\pmod3

, and
\left(Fn
3\right)=-1
.As

Fn\equiv1\pmod4

, we conclude
\left(3{F
n}\right)=-1
from the law of quadratic reciprocity.

Historical Pépin tests

Because of the sparsity of the Fermat numbers, the Pépin test has only been run eight times (on Fermat numbers whose primality statuses were not already known).[1] [2] [3] Mayer, Papadopoulos and Crandall speculate that in fact, because of the size of the still undetermined Fermat numbers, it will take considerable advances in technology before any more Pépin tests can be run in a reasonable amount of time.[4]

YearProversFermat numberPépin test resultFactor found later?
1905Morehead & Western

F7

compositeYes (1970)
1909Morehead & Western

F8

compositeYes (1980)
1952Robinson

F10

compositeYes (1953)
1960Paxson

F13

compositeYes (1974)
1961Selfridge & Hurwitz

F14

compositeYes (2010)
1987Buell & Young

F20

compositeNo
1993Crandall, Doenias, Norrie & Young

F22

compositeYes (2010)
1999Mayer, Papadopoulos & Crandall

F24

compositeNo

References

2n
2

+1

, Comptes rendus de l'Académie des Sciences de Paris 85 (1877), pp. 329–333.

External links

Notes and References

  1. http://www.primepuzzles.net/conjectures/conj_004.htm Conjecture 4. Fermat primes are finite - Pepin tests story, according to Leonid Durman
  2. Wilfrid Keller: Fermat factoring status
  3. R. M. Robinson (1954): Mersenne and Fermat numbers,
  4. Richard E. Crandall, Ernst W. Mayer & Jason S. Papadopoulos (2003): The twenty-fourth Fermat number is composite,