Pépin's test

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 for a French mathematician, Théophile Pépin.

Description of the test

Let $F_n=2^{2^n}+1$ be the nth Fermat number. Pépin's test states that for n > 0,

$F_n$ is prime if and only if $3^{(F_n-1)/2}\equiv-1\pmod{F_n}.$

The expression $3^{(F_n-1)/2}$ can be evaluated modulo $F_n$ 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, for example 5, 6, 7, or 10 (sequence A129802 in OEIS).

Proof of correctness

For one direction, assume that the congruence

$3^{(F_n-1)/2}\equiv-1\pmod{F_n}$

holds. Then $3^{F_n-1}\equiv1\pmod{F_n}$, thus the multiplicative order of 3 modulo $F_n$ divides $F_n-1=2^{2^n}$, which is a power of two. On the other hand, the order does not divide $(F_n-1)/2$, and therefore it must be equal to $F_n-1$. In particular, there are at least $F_n-1$ numbers below $F_n$ coprime to $F_n$, and this can happen only if $F_n$ is prime.

For the other direction, assume that $F_n$ is prime. By Euler's criterion,

$3^{(F_n-1)/2}\equiv\left(\frac3{F_n}\right)\pmod{F_n}$,

where $\left(\frac3{F_n}\right)$ is the Legendre symbol. By repeated squaring, we find that $2^{2^n}\equiv1\pmod3$, thus $F_n\equiv2\pmod3$, and $\left(\frac{F_n}3\right)=-1$. As $F_n\equiv1\pmod4$, we conclude $\left(\frac3{F_n}\right)=-1$ from the law of quadratic reciprocity.

References

• P. Pépin, Sur la formule $2^{2^n}+1$, Comptes Rendus Acad. Sci. Paris 85 (1877), pp. 329–333.

External links

• The Prime Glossary: Pepin's test