Calculator for testing prime numbers

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Eric Martin

Created: 2015-09-06 22:27:48, Last updated: 2020-05-06 07:38:23

Based upon Professor Agrawal's paper.

PRIMES is in P
Manindra Agrawal Neeraj Kayal
Nitin Saxena∗
Department of Computer Science & Engineering
Indian Institute of Technology Kanpur
Kanpur-208016, INDIA
Email: {manindra,kayaln,nitinsa}
We present an unconditional deterministic polynomial-time algorithm that determines whether
an input number is prime or composite.
1 Introduction
Prime numbers are of fundamental importance in mathematics in general, and number theory in particular.
So it is of great interest to study different properties of prime numbers. Of special interest are
those properties that allow one to efficiently determine if a number is prime. Such efficient tests are also
useful in practice: a number of cryptographic protocols need large prime numbers.
Let PRIMES denote the set of all prime numbers. The definition of prime numbers already gives a
way of determining if a number n is in PRIMES: try dividing n by every number m ≤

n—if any m
divides n then it is composite, otherwise it is prime. This test was known since the time of the ancient
Greeks—it is a specialization of the Sieve of Eratosthenes (ca. 240 BC) that generates all primes less
then n. The test, however, is inefficient: it takes Ω(√
n) steps to determine if n is prime. An efficient
test should need only a polynomial (in the size of the input = dlog ne) number of steps. A property that
almost gives an efficient test is Fermat’s Little Theorem: for any prime number p, and any number a not
divisible by p, a
p−1 = 1 (mod p). Given an a and n it can be efficiently checked if a
n−1 = 1 (mod n) by
using repeated squaring to compute the (n − 1)th power of a. However, it is not a correct test since many
composites n also satisfy it for some a’s (all a’s in case of Carmichael numbers [Car10]). Nevertheless,
Fermat’s Little Theorem became the basis for many efficient primality tests.
Since the beginning of complexity theory in the 1960s—when the notions of complexity were formalized
and various complexity classes were defined—this problem (referred to as the primality testing problem)
has been investigated intensively. It is trivial to see that the problem is in the class co-NP: if n is not
prime it has an easily verifiable short certificate, viz., a non-trivial factor of n. In 1974, Pratt observed
that the problem is in the class NP too [Pra75] (thus putting it in NP ∩ co-NP).
In 1975, Miller [Mil76] used a property based on Fermat’s Little Theorem to obtain a deterministic
polynomial-time algorithm for primality testing assuming the Extended Riemann Hypothesis (ERH). Soon
afterwards, his test was modified by Rabin [Rab80] to yield an unconditional but randomized polynomialtime
algorithm. Independently, Solovay and Strassen [SS77] obtained, in 1974, a different randomized
polynomial-time algorithm using the property that for a prime n,


= a
2 (mod n) for every a (

the Jacobi symbol). Their algorithm can also be made deterministic under ERH. Since then, a number
of randomized polynomial-time algorithms have been proposed for primality testing, based on many
different properties.
In 1983, Adleman, Pomerance, and Rumely achieved a major breakthrough by giving a deterministic
algorithm for primality that runs in (log n)
O(log log log n)
time (all the previous deterministic algorithms
∗Last two authors were partially supported by MHRD grant MHRD-CSE-20010018.
required exponential time). Their algorithm was (in a sense) a generalization of Miller’s idea and used
higher reciprocity laws. In 1986, Goldwasser and Kilian [GK86] proposed a randomized algorithm based
on Elliptic Curves running in expected polynomial-time on almost all inputs (all inputs under a widely
believed hypothesis) that produces an easily verifiable short certificate for primality (until then, all
randomized algorithms produced certificates for compositeness only). Based on their ideas, a similar
algorithm was developed by Atkin [Atk86]. Adleman and Huang [AH92] modified the Goldwasser-Kilian
algorithm to obtain a randomized algorithm that runs in expected polynomial-time on all inputs.
The ultimate goal of this line of research has been, of course, to obtain an unconditional deterministic
polynomial-time algorithm for primality testing. Despite the impressive progress made so far, this goal
has remained elusive. In this paper, we achieve this. We give a deterministic, O∼(log15/2
n) time
algorithm for testing if a number is prime. Heuristically, our algorithm does better: under a widely
believed conjecture on the density of Sophie Germain primes (primes p such that 2p + 1 is also prime),
the algorithm takes only O∼(log6 n) steps. Our algorithm is based on a generalization of Fermat’s Little
Theorem to polynomial rings over finite fields. Notably, the correctness proof of our algorithm requires
only simple tools of algebra (except for appealing to a sieve theory result on the density of primes p
with p − 1 having a large prime factor—and even this is not needed for proving a weaker time bound of
n) for the algorithm). In contrast, the correctness proofs of earlier algorithms producing a
certificate for primality [APR83, GK86, Atk86] are much more complex.
In section 2, we summarize the basic idea behind our algorithm. In section 3, we fix the notation used.
In section 4, we state the algorithm and present its proof of correctness. In section 5, we obtain bounds
on the running time of the algorithm. Section 6 discusses some ways of improving the time complexity
of the algorithm.
2 The Idea
Our test is based on the following identity for prime numbers which is a generalization of Fermat’s Little
Theorem. This identity was the basis for a randomized polynomial-time algorithm in [AB03]:
Lemma 2.1. Let a ∈ Z, n ∈ N , n ≥ 2, and (a, n) = 1. Then n is prime if and only if
(X + a)
n = Xn + a (mod n). (1)
Proof. For 0 < i < n, the coefficient of x
in ((X + a)
n − (Xn + a)) is n

Suppose n is prime. Then n

= 0 (mod n) and hence all the coefficients are zero.
Suppose n is composite. Consider a prime q that is a factor of n and let q
||n. Then q
k does not divide

and is coprime to a
n−q and hence the coefficient of Xq
is not zero (mod n). Thus ((X +a)
is not identically zero over Zn.
The above identity suggests a simple test for primality: given an input n, choose an a and test whether
the congruence (1) is satisfied. However, this takes time Ω(n) because we need to evaluate n coefficients
in the LHS in the worst case. A simple way to reduce the number of coefficients is to evaluate both sides
of (1) modulo a polynomial of the form Xr − 1 for an appropriately chosen small r. In other words, test
if the following equation is satisfied:
(X + a)
n = Xn + a (mod Xr − 1, n). (2)
From Lemma 2.1 it is immediate that all primes n satisfy the equation (2) for all values of a and r. The
problem now is that some composites n may also satisfy the equation for a few values of a and r (and
indeed they do). However, we can almost restore the characterization: we show that for appropriately
chosen r if the equation (2) is satisfied for several a’s then n must be a prime power. The number of a’s
and the appropriate r are both bounded by a polynomial in log n and therefore, we get a deterministic
polynomial time algorithm for testing primality.
3 Notation and Preliminaries
The class P is the class of sets accepted by deterministic polynomial-time Turing machines [Lee90].
See [Lee90] for the definitions of classes NP, co-NP, etc.
Zn denotes the ring of numbers modulo n. Fp denotes the finite field with p elements, where p is prime.
Recall that if p is prime and h(X) is a polynomial of degree d and irreducible in Fp, then Fp[X]/(h(X)) is
a finite field of order p
. We will use the notation f(X) = g(X) (mod h(X), n) to represent the equation
f(X) = g(X) in the ring Zn[X]/(h(X)).
We use the symbol O∼(t(n)) for O(t(n)· poly(log t(n)), where t(n) is any function of n. For example,
O∼(logk n) = O(logk n · poly(log log n)) = O(logk+ n) for any > 0. We use log for base 2 logarithm,
and ln for natural logarithm.
N and Z denote the set of natural numbers and integers respectively. Given r ∈ N , a ∈ Z with
(a, r) = 1, the order of a modulo r is the smallest number k such that a
k = 1 (mod r). It is denoted
as or(a). For r ∈ N , φ(r) is Euler’s totient function giving the number of numbers less than r that are
relatively prime to r. It is easy to see that or(a) | φ(r) for any a, (a, r) = 1.
We will need the following simple fact about the lcm of first m numbers (see, e.g., [Nai82] for a proof).
Lemma 3.1. Let LCM(m) denote the lcm of first m numbers. For m ≥ 7:
LCM(m) ≥ 2
4 The Algorithm and Its Correctness
Input: integer n > 1.

  1. If (n = a
    b for a ∈ N and b > 1), output COMPOSITE.
  2. Find the smallest r such that or(n) > log2
  3. If 1 < (a, n) < n for some a ≤ r, output COMPOSITE.
  4. If n ≤ r, output PRIME.1
  5. For a = 1 to b
    φ(r) log nc do
    if ((X + a)
    n 6= Xn + a (mod Xr − 1, n)), output COMPOSITE;
  6. Output PRIME;
    Algorithm for Primality Testing
    Theorem 4.1. The algorithm above returns PRIME if and only if n is prime.
    In the remainder of the section, we establish this theorem through a sequence of lemmas. The following
    is trivial:
    Lemma 4.2. If n is prime, the algorithm returns PRIME.
    Proof. If n is prime then steps 1 and 3 can never return COMPOSITE. By Lemma 2.1, the for loop
    also cannot return COMPOSITE. Therefore the algorithm will identify n as PRIME either in step 4 or
    in step 6.
    The converse of the above lemma requires a little more work. If the algorithm returns PRIME in
    step 4 then n must be prime since otherwise step 3 would have found a non-trivial factor of n. So the only
    remaining case is when the algorithm returns PRIME in step 6. For the purpose of subsequent analysis
    we assume this to be the case.
    The algorithm has two main steps (2 and 5): step 2 finds an appropriate r and step 5 verifies the
    equation (2) for a number of a’s. We first bound the magnitude of the appropriate r.
    Lemma 4.3. There exist an r ≤ max{3, dlog5 ne} such that or(n) > log2
    1Lemma 4.3 shows that r ≤ dlog5 ne, so Step 4 is relevant only when n ≤ 5, 690, 034.
    Proof. This is trivially true when n = 2: r = 3 satisfies all conditions. So assume that n > 2. Then
    dlog5 ne > 10 and Lemma 3.1 applies. Observe that the largest value of k for any number of the form
    mk ≤ B = dlog5 ne, m ≥ 2, is blog Bc. Now consider the smallest number r that does not divide the
    blog Bc
    blog2 Ync
    i − 1).
    Note that (r, n) cannot be divisible by all the prime divisors of r since otherwise r will divide n
    blog Bc by
    the observation above. Therefore, r
    (r,n) will also not divide the above product. Using the fact that r is
    the smallest number not dividing the product, it follows that (r, n) = 1. Further, since r does not divide
    any of n
    i − 1 for 1 ≤ i ≤ blog2
    nc, or(n) > log2
    n. Finally,
    blog Bc
    blog2 Ync
    i − 1) < nblog Bc+ 1
    log2 n·(log2 n−1) ≤ n
    log4 n ≤ 2
    log5 n ≤ 2
    (The second inequality holds for all n ≥ 2). By Lemma 3.1, the lcm of first B numbers is at least 2B.
    Therefore, r ≤ B.
    Since or(n) > 1, there must exist a prime divisor p of n such that or(p) > 1. We have p > r since
    otherwise either step 3 or step 4 would decide about primality of n. Since (n, r) = 1 (otherwise either
    step 3 or step 4 will correctly identify n), p, n ∈ Z

    . Numbers p and r will be fixed in the remainder of
    this section. Also, let = b p φ(r) log nc. Step 5 of the algorithm verifies equations. Since the algorithm does not output COMPOSITE in
    this step, we have:
    (X + a)
    n = Xn + a (mod Xr − 1, n)
    for every a, 0 ≤ a ≤ (the equation for a = 0 is trivially satisfied). This implies: (X + a) n = Xn + a (mod Xr − 1, p) (3) for 0 ≤ a ≤. By Lemma 2.1, we have:
    (X + a)
    p = Xp + a (mod Xr − 1, p) (4)
    for 0 ≤ a ≤ . From equations 3 and 4 it follows that: (X + a) n p = X n p + a (mod Xr − 1, p) (5) for 0 ≤ a ≤. Thus both n and n
    behave like prime p in the above equation. We give a name to this
    Definition 4.4. For polynomial f(X) and number m ∈ N , we say that m is introspective for f(X) if
    [f(X)]m = f(Xm) (mod Xr − 1, p).
    It is clear from equations (5) and (4) that both n
    and p are introspective for X + a when 0 ≤ a ≤ . The following lemma shows that introspective numbers are closed under multiplication: Lemma 4.5. If m and m0 are introspective numbers for f(X) then so is m · m0 . Proof. Since m is introspective for f(X) we have: [f(X)]m·m0 = [f(Xm)]m0 (mod Xr − 1, p). Also, since m0 is introspective for f(X), we have (after replacing X by Xm in the introspection equation for m0 ): [f(Xm)]m0 = f(Xm·m0 ) (mod Xm·r − 1, p) = f(Xm·m0 ) (mod Xr − 1, p) (since Xr − 1 divides Xm·r − 1). Putting together the above two equations we get: [f(X)]m·m0 = f(Xm·m0 ) (mod Xr − 1, p). 4 For a number m, the set of polynomials for which m is introspective is also closed under multiplication: Lemma 4.6. If m is introspective for f(X) and g(X) then it is also introspective for f(X) · g(X). Proof. We have: [f(X) · g(X)]m = [f(X)]m · [g(X)]m = f(Xm) · g(Xm) (mod Xr − 1, p). The above two lemmas together imply that every number in the set I = {( n p ) i · p j | i, j ≥ 0} is introspective for every polynomial in the set P = { Q
    a=0(X + a)
    ea | ea ≥ 0}. We now define two groups
    based on these sets that will play a crucial role in the proof.
    The first group is the set of all residues of numbers in I modulo r. This is a subgroup of Z

    as already observed, (n, r) = (p, r) = 1. Let G be this group and |G| = t. G is generated by n and p
    modulo r and since or(n) > log2
    n, t > log2
    To define the second group, we need some basic facts about cyclotomic polynomials over finite fields.
    Let Qr(X) be r
    th cyclotomic polynomial over Fp. Polynomial Qr(X) divides Xr − 1 and factors into
    irreducible factors of degree or(p) [LN86]. Let h(X) be one such irreducible factor. Since or(p) > 1,
    degree of h(X) is greater than one. The second group is the set of all residues of polynomials in P
    modulo h(X) and p. Let G be this group. This group is generated by elements X, X + 1, X + 2, . . .,
    X + in the field F = Fp[X]/(h(X)) and is a subgroup of the multiplicative group of F. The following lemma proves a lower bound on the size of the group G. It is a slight improvement on a bound shown by Hendrik Lenstra Jr. [Len02], which, in turn, improved a bound shown in an earlier version of our paper [AKS03].2 Lemma 4.7 (Hendrik Lenstra Jr.). |G| ≥ t+

Proof. First note that since h(X) is a factor of the cyclotomic polynomial Qr(X), X is a primitive r
root of unity in F.
We now show that any two distinct polynomials of degree less than t in P will map to different
elements in G. Let f(X) and g(X) be two such polynomials in P. Suppose f(X) = g(X) in the field F.
Let m ∈ I. We also have [f(X)]m = [g(X)]m in F. Since m is introspective for both f and g, and h(X)
divides Xr − 1, we get:
f(Xm) = g(Xm)
in F. This implies that Xm is a root of the polynomial Q(Y ) = f(Y ) − g(Y ) for every m ∈ G. Since
(m, r) = 1 (G is a subgroup of Z

), each such Xm is a primitive r
th root of unity. Hence there will be
|G| = t distinct roots of Q(Y ) in F. However, the degree of Q(Y ) is less than t by the choice of f and g.
This is a contradiction and therefore, f(X) 6= g(X) in F.
Note that i 6= j in Fp for 1 ≤ i 6= j ≤ since = b
φ(r) log nc <

r log n < r and p > r. So the
elements X, X + 1, X + 2, . . ., X + are all distinct in F. Also, since degree of h is greater than one, X + a 6= 0 in F for every a, 0 ≤ a ≤. So there exist at least + 1 distinct polynomials of degree one in G. Therefore, there exist at least t+

distinct polynomials of degree < t in G.
In case n is not a power of p, size of G can also be upper bounded:
Lemma 4.8. If n is not a power of p then |G| ≤ n

Proof. Consider the following subset of I:
ˆI = {(
· p
| 0 ≤ i, j ≤ b√
If n is not a power of p then the set ˆI has (b

tc + 1)2 > t distinct numbers. Since |G| = t, at least two
numbers in ˆI must be equal modulo r. Let these be m1 and m2 with m1 > m2. So we have:
Xm1 = Xm2
(mod Xr − 1).
2Macaj [Mac02] also proved this lemma independently.
Let f(X) ∈ P. Then,
[f(X)]m1 = f(Xm1
) (mod Xr − 1, p)
= f(Xm2
) (mod Xr − 1, p)
= [f(X)]m2
(mod Xr − 1, p).
This implies
[f(X)]m1 = [f(X)]m2
in the field F. Therefore, f(X) ∈ G is a root of the polynomial Q0
(Y ) = Y
m1 − Y
in the field F.
3 As
f(X) is an arbitrary element of G, we have that the polynomial Q0
(Y ) has at least |G| distinct roots in
F. The degree of Q0
(Y ) is m1 ≤ (
· p)

tc ≤ n

. This shows |G| ≤ n

Armed with these estimates on the size of G, we are now ready to prove the correctness of the
Lemma 4.9. If the algorithm returns PRIME then n is prime.
Proof. Suppose that the algorithm returns PRIME. Lemma 4.7 implies that for t = |G| and = b p φ(r) log nc: |G| ≥ t +
t − 1

` + 1 + b

tlog nc

tlog nc

(since t > √
tlog n)


tlog nc + 1

tlog nc

(since ` = b
φ(r) log nc ≥ b√
tlog nc)


t log nc+1 (since b

tlog nc > blog2
nc ≥ 1)
≥ n

By Lemma 4.8, |G| ≤ n

if n is not a power of p. Therefore, n = p
for some k > 0. If k > 1 then the
algorithm will return COMPOSITE in step 1. Therefore, n = p.
This completes the proof of theorem.
5 Time Complexity Analysis and Improvements
It is straightforward to calculate the time complexity of the algorithm. In these calculations we use the
fact that addition, multiplication, and division operations between two m bits numbers can be performed
in time O∼(m) [vzGG99]. Similarly, these operations on two degree d polynomials with coefficients at
most m bits in size can be done in time O∼(d · m) steps [vzGG99].
Theorem 5.1. The asymptotic time complexity of the algorithm is O∼(log21/2
Proof. The first step of the algorithm takes asymptotic time O∼(log3 n) [vzGG99].
In step 2, we find an r with or(n) > log2
n. This can be done by trying out successive values of
r and testing if n
k 6= 1 (mod r) for every k ≤ log2
n. For a particular r, this will involve at most
n) multiplications modulo r and so will take time O∼(log2
n log r). By Lemma 4.3 we know that
only O(log5 n) different r’s need to be tried. Thus the total time complexity of step 2 is O∼(log7 n).
The third step involves computing gcd of r numbers. Each gcd computation takes time O(log n) [vzGG99],
and therefore, the time complexity of this step is O(r log n) = O(log6 n). The time complexity of step 4
is just O(log n).
In step 5, we need to verify b
φ(r) log nc equations. Each equation requires O(log n) multiplications
of degree r polynomials with coefficients of size O(log n). So each equation can be verified in
time O∼(r log2
n) steps. Thus the time complexity of step 5 is O∼(r
φ(r) log3 n) = O∼(r
2 log3 n) =
n). This time dominates all the other and is therefore the time complexity of the algorithm.
3This formulation of the argument is by Adam Kalai, Amit Sahai, and Madhu Sudan [KSS02].
The time complexity of the algorithm can be improved by improving the estimate for r (done in
Lemma 4.3). Of course the best possible scenario would be when r = O(log2
n) and in that case the time
complexity of the algorithm would be O∼(log6 n). In fact, there are two conjectures that support the
possibility of such an r (below ln is natural logarithm):
Artin’s Conjecture: Given any number n ∈ N that is not a perfect square, the number of primes
q ≤ m for which oq(n) = q − 1 is asymptotically A(n) ·
ln m where A(n) is Artin’s constant with
A(n) > 0.35.
Sophie-Germain Prime Density Conjecture: The number of primes q ≤ m such that 2q + 1 is also
a prime is asymptotically 2C2m
ln2 m
where C2 is the twin prime constant (estimated to be approximately
0.66). Primes q with this property are called Sophie-Germain primes.
Artin’s conjecture—if it becomes effective for m = O(log2
n)—immediately shows that there is an
r = O(log2
n) with the required properties. There has been some progress towards proving Artin’s
conjecture [GM84, GMM85, HB86], and it is also known that this conjecture holds under Generalized
Riemann Hypothesis.
If the second conjecture holds, we can conclude that r = O∼(log2
By density of Sophie-Germain primes, there must exist at least log2
n such primes between
8 log2
n and c log2
n(log log n)
for suitable constant c. For any such prime q, either oq(n) ≤ 2
or oq(n) ≥
. Any q for which oq(n) ≤ 2 must divide n
2 − 1 and so the number of such q
is bounded by O(log n). This implies that there must exist a prime r = O∼(log2
n) such that
or(n) > log2
n. Such an r will yield an algorithm with time complexity O∼(log6 n).
There has been progress towards proving this conjecture as well. Let P(m) denote the greatest prime
divisor of number m. Goldfeld [Gol69] showed that primes q with P(q − 1) > q
2 +c
, c ≈
12 , occur with
positive density. Improving upon this, Fouvry has shown:
Lemma 5.2. [Fou85] There exist constants c > 0 and n0 such that, for all x ≥ n0:
|{q | q is prime, q ≤ x and P(q − 1) > q
3 }| ≥ c
ln x
The above lemma is now known to hold for exponents up to 0.6683 [BH96]. Using the above lemma,
we can improve the analysis of our algorithm:
Theorem 5.3. The asymptotic time complexity of the algorithm is O∼(log15/2
Proof. As argued above, a high density of primes q with P(q−1) > q
3 implies that step 2 of the algorithm
will find an r = O(log3 n) with or(n) > log2
n. This brings the complexity of the algorithm down to
Recently, Hendrik Lenstra and Carl Pomerance [LP03a] have come up with a modified version of our
algorithm whose time complexity is provably O∼(log6 n).
6 Future Work
In our algorithm, the for loop in step 5 needs to run for b
φ(r) log nc times to ensure that the size of
the group G is large enough. The number of iterations of the loop could be reduced if we can show that
a still smaller set of (X + a)’s generates a group of the required size. This seems very likely.
One can further improve the complexity to O∼(log3 n) if the following conjecture—given in [BP01]
and verified for r ≤ 100 and n ≤ 1010 in [KS02]—is proved:
Conjecture. If r is a prime number that does not divide n and if
(X − 1)n = Xn − 1 (mod Xr − 1, n), (6)
then either n is prime or n
2 = 1 (mod r).
If this conjecture is true, we can modify the algorithm slightly to first search for an r which does
not divide n
2 − 1. Such an r can assuredly be found in the range [2, 4 log n]. This is because the
product of prime numbers less than x is at least e
(see [Apo97]). Thereafter we can test whether
the congruence (6) holds or not. Verifying the congruence takes time O∼(r log2
n). This gives a time
complexity of O∼(log3 n).
Recently, Hendrik Lenstra and Carl Pomerance [LP03b] have given a heuristic argument that suggests
that the above conjecture is false. However, some variant of the conjecture may still be true (for example,
if we force r > log n).
We wish to express our gratitude to Hendrik Lenstra Jr. for allowing us to use his observation on improving
the lower bound on the size of the group G. This has made the proof completely elementary (earlier version
required the density bound of Lemma 5.2), simplified the proof and improved the time complexity!
We are also thankful to Adam Kalai, Amit Sahai, and Madhu Sudan for allowing us to use their proof
of Lemma 4.8. This has made the proofs of both upper and lower bounds on size of G similar (both are
now based on the number of roots of a polynomial in a field).
We thank Somenath Biswas, Rajat Bhattacharjee, Jaikumar Radhakrishnan, and V. Vinay for many
useful discussions.
We thank Erich Bach, Abhijit Das, G. Harman, Roger Heath-Brown, Pieter Moree, Richard Pinch,
and Carl Pomerance for providing us with useful references.
Since our preprint appeared, a number of researchers took the trouble of pointing out various omissions
in our paper. We are thankful to all of them. We have tried to incorporate their suggestions in the paper
and our apologies if we missed out on some.
We thank W. Carlip, L. Somer, L. Rempre and students of Deutsche Schuleracademie for pointing
out an error in the proof of Lemma 4.3.
Finally, we thank the anonymous referee of the paper whose suggestions and observations have been
very useful.
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PLANETCALC, Calculator for testing prime numbers