Relaxing the size constraints on the criterion of Proth

TL;DR

This paper extends Proth's primality test to larger numbers by adding a simple, efficient condition that preserves the test's biconditionality and can be computed concurrently, broadening its applicability.

Contribution

The authors introduce a new condition to Proth's theorem, allowing it to test a wider class of numbers without added complexity, and discuss an extension to an existing primality test.

Findings

Extended Proth's theorem to larger numbers

Maintained the theorem's biconditionality

Added minimal computational complexity

Abstract

We add one condition to the theorem of Proth to extend its applicability to $N=k2^n+1$ where $2^n>N^{1/3}$ as opposed to the former constraint of $2^n>k$. This additional condition adds barely any complexity or time to the test and can furthermore be calculated concurrently. Furthermore, it maintains the biconditionality of the theorem and thus makes it readily applicable. A note on an extension of the primality test of Brillhart, Lehmer, and Selfridge is also made.

Equations

$$2p+1$$

$$a^{N-1}$$

$$a^m$$

$$2p^z+1$$

$$a^{N-1}$$

$$a^{m{p}^{z-1}}$$

$$a^{\frac{N-1}{2}}$$

$$u$$

Full text

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then $N$ is prime.

If $a^{mp^{z}}\equiv 1\mod N$ but $a^{mp^{z-1}}\not\equiv 1\mod N$, then $p^{z}|ord_{N}(a)$. If $p^{z}|ord_{N}(a)$, and since $ord_{N}(a)=lcm[ord_{n_{i}^{z_{i}}}(a)]_{i}$, where $n_{i}^{z_{i}}$ represent the highest powers of each prime factor of $N$, at least one prime factor $n_{i}$ must satisfy $p^{z}|ord_{n_{i}^{z_{i}}}(a)$. But since, from (3), $ord_{n_{i}^{z_{i}}}(a)=n_{i}^{q}ord_{n_{i}}(a)$ for some $q$, and since $q\nmid N$, we know $p^{z}|ord_{n_{i}}(a)$. Therefore, since $n_{i}$ is prime, we can write

$p^{z}|ord_{n_{i}}(a)|\phi(n_{i})=n_{i}-1$.

Alternatively, $n_{i}\equiv 1\mod p^{z}$. So, $N/n_{i}\equiv 1\mod p^{z}$. But since $n_{i}\geq 2p^{z}+1>\sqrt{N}$, we know $N/n_{i}<\sqrt{N}<2p^{z}+1$, and thus $N/n_{i}$ is precisely $1$ and $N=n_{i}$ is prime.

This extension is valid, but not as strong as that of Proth because, again, it is not biconditional. However, utilizing quadratic reciprocity and techniques from (1), we can relax the size criterion on Proth’s primality test. Recall that for all primes,

a^{\frac{N-1}{2}}\equiv\mathchoice{\mathopen{}\left(\kern-1.0pt\vbox{\hbox{\set@color\genfrac{}{}{0.0pt}{0}{a}{N}}}\kern-9.12497pt\vbox{\hbox to9.12497pt{\kern-1.0pt\cleaders\hbox{\kern 0.5pt\vrule height=0.2pt,depth=0.2pt,width=1.0pt\kern 0.5pt}\hfil\kern-1.0pt}}\kern-1.0pt\right)\mathclose{}}{\mathopen{}\left(\kern-1.0pt\vbox{\hbox{\set@color\genfrac{}{}{0.0pt}{1}{\vphantom{1}a}{\vphantom{1}N}}}\kern-9.88748pt\vbox{\hbox to9.88748pt{\kern-0.5pt\cleaders\hbox{\kern 0.5pt\vrule height=0.2pt,depth=0.2pt,width=1.0pt\kern 0.5pt}\hfil\kern-0.5pt}}\kern-1.0pt\right)\mathclose{}}{\mathopen{}\left(\kern-1.0pt\vbox{\hbox{\set@color\genfrac{}{}{0.0pt}{2}{\vphantom{1}a}{\vphantom{1}N}}}\kern-7.06247pt\vbox{\hbox to7.06247pt{\kern 0.0pt\cleaders\hbox{\kern 0.5pt\vrule height=0.2pt,depth=0.2pt,width=1.0pt\kern 0.5pt}\hfil\kern 0.0pt}}\kern-1.0pt\right)\mathclose{}}{\mathopen{}\left(\kern-1.0pt\vbox{\hbox{\set@color\genfrac{}{}{0.0pt}{3}{\vphantom{1}a}{\vphantom{1}N}}}\kern-7.06247pt\vbox{\hbox to7.06247pt{\kern 0.0pt\cleaders\hbox{\kern 0.5pt\vrule height=0.2pt,depth=0.2pt,width=1.0pt\kern 0.5pt}\hfil\kern 0.0pt}}\kern-1.0pt\right)\mathclose{}}=-1\mod N

for all $a$ where \mathchoice{\mathopen{}\left(\kern-1.0pt\vbox{\hbox{\set@color\genfrac{}{}{0.0pt}{0}{a}{N}}}\kern-9.12497pt\vbox{\hbox to9.12497pt{\kern-1.0pt\cleaders\hbox{\kern 0.5pt\vrule height=0.2pt,depth=0.2pt,width=1.0pt\kern 0.5pt}\hfil\kern-1.0pt}}\kern-1.0pt\right)\mathclose{}}{\mathopen{}\left(\kern-1.0pt\vbox{\hbox{\set@color\genfrac{}{}{0.0pt}{1}{\vphantom{1}a}{\vphantom{1}N}}}\kern-9.88748pt\vbox{\hbox to9.88748pt{\kern-0.5pt\cleaders\hbox{\kern 0.5pt\vrule height=0.2pt,depth=0.2pt,width=1.0pt\kern 0.5pt}\hfil\kern-0.5pt}}\kern-1.0pt\right)\mathclose{}}{\mathopen{}\left(\kern-1.0pt\vbox{\hbox{\set@color\genfrac{}{}{0.0pt}{2}{\vphantom{1}a}{\vphantom{1}N}}}\kern-7.06247pt\vbox{\hbox to7.06247pt{\kern 0.0pt\cleaders\hbox{\kern 0.5pt\vrule height=0.2pt,depth=0.2pt,width=1.0pt\kern 0.5pt}\hfil\kern 0.0pt}}\kern-1.0pt\right)\mathclose{}}{\mathopen{}\left(\kern-1.0pt\vbox{\hbox{\set@color\genfrac{}{}{0.0pt}{3}{\vphantom{1}a}{\vphantom{1}N}}}\kern-7.06247pt\vbox{\hbox to7.06247pt{\kern 0.0pt\cleaders\hbox{\kern 0.5pt\vrule height=0.2pt,depth=0.2pt,width=1.0pt\kern 0.5pt}\hfil\kern 0.0pt}}\kern-1.0pt\right)\mathclose{}}=-1. Combine this with the following theorem.

$a^{\frac{N-1}{2}}\equiv-1\mod N$.

By the conditions, $a^{N-1}\equiv 1\mod N$, and thus $2^{n}|ord_{N}(a)$. Additionally, since it is congruent to $-1$, we know that, for each prime factor $n_{i}$ of $N$, $a^{\frac{N-1}{2}}-1\equiv-1-1=-2\mod n_{i}$. Since $N$ is odd, all prime factors are greater than $2$ and thus

$gcd(a^{\frac{N-1}{2}}-1,N)=1$.

Because of this we know that $2^{n}|ord_{n_{i}}(a)$ for all prime factors $n_{i}$ of $N$. Thus,

$2^{n}|ord_{n_{i}}(a)|\phi(n_{i})=n_{i}-1$.

Therefore, $n_{i}\equiv 1\mod 2^{n}$ for all prime factors $n_{i}$. But also $n_{i}\geq 2^{n}+1>\sqrt[3]{N}$. So $N/n_{1}n_{2}<N^{1/3}<2^{n}+1$. But since $N/n_{1}n_{2}\equiv 1\mod 2^{n}$, so $N/n_{1}n_{2}$ is precisely $1$ and $N$ is thus either prime or semiprime.

Combining the two aforementioned statements, we know that for a prime $N=k2^{n}+1$ with $2^{n}+1>\sqrt[3]{N}$, $a^{\frac{N-1}{2}}\equiv-1\mod N$. In the other direction, if $a^{\frac{N-1}{2}}\equiv-1\mod N$, then $N$ is either prime or semiprime. We can easily determine whether $N$ is semiprime given the constraints. Specifically, we utilize the fact that if $q<w$, then $q\mod w=q$. If $N$ is semiprime and satisfies the aforementioned conditions, then we may write $N=(2^{n}u+1)(2^{n}v+1)=2^{n}(2^{n}uv+u+v)+1$. Since $2^{n}2^{n}uv<N$, we know $u+v<uv<N^{1/3}<2^{n}$ (if $u+v\geq uv$, then because of the small values of $u,v$, the inequality less than $2^{n}$ will still hold). Therefore, we can find $u+v$ by calculating $k\mod 2^{n}$ and $uv$ by calculating $\frac{k-k\mod 2^{n}}{2^{n}}$. Solving this system of equations, we can find $u,v$ and thus the prime factors $2^{n}u+1$ and $2^{n}v+1$. Thus, if and only if the solution is not an integer, $N$ is prime. This means we have the following conditions after solving the system.

[TABLE]

where

$u=\frac{1}{2}(k\mod 2^{n}-\sqrt{(k\mod 2^{n})^{2}-4(\frac{k-k\mod 2^{n}}{2^{n}})}).$**

Furthermore, if $u\in\mathbb{N}$, then $N$ is semiprime and its factors are given by $2^{n}u+1$ and $2^{n}v+1$ as defined above. The second condition only requires a few modular multiplications to calculate and thus does not add significant complexity to the test. Additionally, regardless of the outcome of the second condition, at least one and up to two primes will be discerned ($N$ or its factors). If $2^{n}>N^{1/2}$, the second condition may be removed.