On Wilson's Theorem and Polignac Conjecture

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(the twin prime conjecture) was developed, in 1922, by Godfrey Harold Hardy (1877-1947) and John Edensor Littlewood (1885-1977) to estimate the density of twin primes[2]. Twin prime characterization was discussed later by Clement in 1949 [3]. But an effective approach to the conjecture remains undeveloped. It is thus worthwhile to re-examine the derivation of Clement’s result from a more fundamental theorem in number theory, the Wilson’s Theorem. By generalizing and improving Wilson’s Theorem and Clement’s result, a superior approach to Polignac Conjecture can be adapted to obtain useful partial results to the mysterious problems.

II. WILSON’S THEOREM AND CLEMENT’S RESULT

Wilson derived his theorem on sufficient and necessary condition for a number p to be prime.

Theorem 1 (p − 1)! ≡ −1 (mod p) iff p is prime.

In 1949, Clement [3, 6] formulated another theorem based on Wilson’s Theorem. The following is our derivation of Clement’s result,

Theorem 2 4[(p − 1)! + 1] ≡ −p (mod p(p +2)) iff p, p + 2 are primes.

Proof. Obviously, p + 2 is not prime when p = 2. So we exclude p = 2 from our discussion, i.e. p and p + 2 are odd prime numbers. By Wilson’s Theorem, when p =6 2,

(p − 1)! ≡ −1 (mod p) iff p is odd prime. ⇒ (p − 1)!+1 ≡ 0 (mod p) iff p is odd prime.

⇒ 4[(p − 1)! + 1] ≡ 0 (mod p) and p =46 iff p is odd prime. ⇒ 4[(p − 1)! + 1] ≡ −p (mod p) and p =46 iff p is odd prime. (1)

Similarly,

(p +1)!+1 ≡ 0 (mod p +2) iff p + 2 is odd prime.

⇒ (p2 + p)(p − 1)!+1 ≡ 0 (mod p +2) iff p + 2 is odd prime. ⇒ [(p + 2)(p − 1) + 2](p − 1)!+1 ≡ 0 (mod p +2) iff p + 2 is odd prime. ⇒ 2(p − 1)!+1 ≡ 0 (mod p +2) iff p + 2 is odd prime. 3

⇒ 2[2(p − 1)! + 1] ≡ 0 (mod p +2) iff p + 2 is odd prime. ⇒ 2[2(p − 1)!+1]+(p + 2) ≡ 0 (mod p +2) iff p + 2 is odd prime.

⇒ 4[(p − 1)!+1]+ p ≡ 0 (mod p +2) iff p + 2 is odd prime. ⇒ 4[(p − 1)! + 1] ≡ −p (mod p +2) iff p + 2 is odd prime. (2)

From (1) and (2),

4[(p − 1)! + 1] ≡ −p (mod p(p +2)) iff p, p + 2 are odd prime numbers. 

III. DERIVATION FROM WILSON’S THEOREM FOR POLIGNAC CONJECTURE

In the derivation of Clement’s result, a method of determining unknown coefficients has been applied. This method is used to combine two congruence identities into one. Suppose we have

f1(p)+ C1 ≡ 0 (mod p)

and g(λ)f1(p)+ C2 ≡ 0 (mod p + λ), where (p,p + λ)=1and (g(λ),p(p + λ)) = 1, we establish two congruences:

Xg(λ)[f1(p)+ C1]+ Yp ≡ 0 (mod p) (3)

and X[g(λ)f1(p)+ C2]+ Y (p + λ) ≡ 0 (mod p + λ). (4)

If we choose X and Y in such a way that (X,p)=1and (X,p + λ) =1, and

Xg(λ)C1 = XC2 + Yλ the (3) and (4) is equivalent to

Xg(λ)f1(p)+ Yp + XC2 + Yλ ≡ 0 (mod p(p + λ)). (5)

Specifically, in deriving Clement’s result, we used X = 2 and Y = 1. This method can be employed again in attempting the Polignac Conjecture which states that for every positive number k, there are infinitely many pairs of primes in the form of p and p +2k. We propose a necessary and sufficient condition when p and p +2k are primes. 4

Theorem 3 Suppose p ∤ 2k(2k)! and p is odd. Then

2k(2k)![(p − 1)! + 1] ≡ [1 − (2k)!]p (mod p(p +2k)) iff p, p+2k are odd primes.

Proof. From Wilson’s theorem, we have

(p − 1)!+1 ≡ 0 (mod p) iff p is odd prime. (p +2k − 1)!+1 ≡ 0 (mod p +2k) iff p +2k is odd prime.

The LHS of the congruence can be rewritten as

(p +2k − 1)(P +2k − 2) ··· p(p − 1)!+1 ≡ (−1)(−2) ··· (−2k)(p − 1)!+1 ≡ (2k)!(p − 1)!+1 ≡ 0 (mod p +2k).

By setting λ =2k, g(λ)=(2k)!, f1(p)=(p − 1)!, C1 = C2 = 1, X =2k and Y = (2k)! − 1 in (5), we obtain

2k(2k)![(p − 1)! + 1] ≡ [1 − (2k)!]p (mod p(p +2k))

iff p, p+2k are odd primes, assuming p ∤ 2k(2k)! and p is odd. 

For example, when k = 2, we have,

for p ∤ 96, 96[(p − 1)! + 1] ≡ −23p (mod p(p +4)) iff p, p + 4 are odd primes.

IV. SIMIONOV’S GENERALIZATION OF WILSON’S THEOREM

By elementary mathematical manipulations, we can generalize Wilson’s Theorem. Simionov had found out the following generalization of Wilson’s Theorem [4]

Theorem 4 (k − 1)!(p − k)! ≡ (−1)k (mod p), iff p is prime ∀k ∈ Z+, k ≤ p (6)

Proof. By Wilson’s Theorem,

(p − 1)! ≡ −1 (mod p) iff p is prime. ⇔ (p − 1)(p − 2)(p − 3) ··· (p − k + 1)(p − k)! ≡ −1 (mod p) iff p is prime. ⇔ (−1)(−2)(−3) ··· (−k + 1)(p − k)! ≡ −1 (mod p) iff p is prime. 5

⇔ (−1)k−1(1)(2)(3) ··· (k − 1)(p − k)! ≡ −1 (mod p) iff p is prime.

− − ⇔ (−1)k 1(−1)k 1(1)(2)(3) ··· (k − 1)(p − k)! ≡ (−1)k (mod p) iff p is prime. ⇔ (−1)k−1(−1)k−1(k − 1)!(p − k)! ≡ (−1)k (mod p) iff p is prime. ⇔ (k − 1)!(p − k)! ≡ (−1)k (mod p) iff p is prime. 

Besides Simionov’s result, other generalizations include

Corollary 1 For two distinct odd prime numbers, p1,p2 and a positive odd integer p such that p +1= p1 + p2 (Partial consequence of Goldbach’s Conjecture [5])

(p − p1)!(p − p2)! ≡ −1 (mod p) iff p is prime

Corollary 2 For integers k1 and k2 such that 0 ≤ k1

k1!k2!(p − k1 − 1)!(p − k2 − 1)! ≡ −1 (mod p) if p is prime

However,these generalizations are less significant compared to Sinionov’s Theorem.

p+1 A direct consequence of Simionov’s Theorem can be obtained by substituting k = 2 into the equation (6),

p − 1 2 p+1 Theorem 5 [( )!] ≡ (−1) 2 iff p is odd prime (7) 2 Proof. This result can also be directly derived from Wilson’s Theorem as shown below, for an odd integer p, p +1 p − 1 (p − 1)(p − 2)(p − 3) ··· ( )( ) ··· 3 · 2 · 1 ≡ −1 (mod p) iff p is odd prime. 2 2 p +1 p − 1 ⇔ [(p − 1) · 1][(p − 2) · 2][(p − 3) · 3] ··· [( )( )] ≡ −1 (mod p) iff p is odd prime. 2 2 p − 1 ⇔ (−1)(−4)(−9) ··· (−( )2) ≡ −1 (mod p) iff p is odd prime. 2 p − 1 ⇔ (−12)(−22)(−32) ··· [−( )2] ≡ −1 (mod p) iff p is odd prime. 2 p−1 p − 1 2 ⇔ (−1) 2 [( )!] ≡ −1 (mod p) iff p is odd prime. 2 p−1 p−1 p − 1 2 p−1 ⇔ (−1) 2 (−1) 2 [( )!] ≡ (−1)(−1) 2 (mod p) iff p is odd prime. 2 p − 1 2 p+1 ⇔ [( )!] ≡ (−1) 2 (mod p) iff p is odd prime.  2 As the calculation for the factorial of a number can be tremendous, this result can reduce p−1 2 the computation for determining the primality of a number, as [( 2 )!] is much smaller than 6

(p − 1)!, especially for large p. Hence, Simionov’s finding can be viewed as an improvement of Wilson’s Theorem. We can take this approach when solving Twin Prime Conjecture and Polignac Conjecture.

V. DERIVATION FROM SIMIONOV’S THEOREM FOR POLIGNAC CONJECTURE

From Simionov’s result, we use the method of the mathematical manipulations mentioned in Section III to derive a sufficient and necessary condition for twin prime numbers to exist, i.e. for p and p + 2 to be primes

p − 1 2 p−1 Theorem 6 2[( )!] +(−1) 2 (5p + 2) ≡ 0 (mod p(p + 2)) 2 iff p and p + 2 are odd prime numbers.

Proof. From (7), for odd integer p,

p − 1 2 p−1 [( )!] +(−1) 2 ≡ 0 (mod p) iff p is odd prime. 2 p − 1 2 p−1 ⇒ 2[( )!] + 2(−1) 2 ≡ 0 (mod p) iff p is odd prime. 2 p − 1 2 p−1 ⇒ 2[( )!] +(−1) 2 (2+5p) ≡ 0 (mod p) iff p is odd prime. (8) 2 Similarly for p + 2,

p +1 2 p+1 [( )!] +(−1) 2 ≡ 0 (mod p +2) iff p + 2 is odd prime 2 p +1 2 p+1 ⇒ 8[( )!] + 8(−1) 2 ≡ 0 (mod p +2) iff p + 2 is odd prime. 2 p +1 2 p − 1 2 p−1 ⇒ 8( ) [( )!] +(−8)(−1) 2 ≡ 0 (mod p +2) iff p + 2 is odd prime. 2 2 2 p − 1 2 p−1 ⇒ 2(p +2p + 1)[( )!] + [5(p + 2) − 8] 2 ≡ 0 (mod p +2) iff p + 2 is odd prime. 2 p − 1 2 p−1 ⇒ 2[( )!] + (5p + 2)(−1) 2 ≡ 0 (mod p +2) iff p + 2 is odd prime. (9) 2 From (8) and (9),

p − 1 2 p−1 2[( )!] + (5p + 2)(−1) 2 ≡ 0 (mod p(p +2)) iff p, p + 2 are odd primes.  2 If we use the same method again on the pair of congruences

p − 1 2 p−1 [( )!] +(−1) 2 ≡ 0 (mod p) iff p is odd prime 2 p +2k − 1 2 p+2k−1 [( )!] +(−1) 2 ≡ 0 (mod p +2k) iff p +2k is odd prime 2 7

− − k 2 p 1 k p+2k 1 and setting X =4 (2k) and Y = (2k − 1)!! (−1) 2 − 4 (−1) 2 in (5) where (2n − 1)!! n denotes Qi=1(2i − 1), it easily follows that Theorem 7 Suppose p ∤ (2k − 1)!!2 and p is odd,

− 2 p − 1 2 p 1 2 k k+1 2k(2k − 1)!! [( )!] +(−1) 2 [(2k − 1)!! (p +2k)+4 (−1) p] ≡ 0 (mod p(p +2k)) 2 iff p, p +2k are primes.

This is a necessary and sufficient condition for p and p +2k to be primes assuming p ∤ (2k − 1)!!2 and p is odd. For example, when k = 2, we have, supposing p ∤ 9 and p is odd,

p − 1 2 p−1 36[( )!] +(−1) 2 [36 − 7p] ≡ 0 (mod p(p +4)) iff p, p + 4 are primes. 2

VI. CONCLUSION

In this paper, we have proposed necessary and sufficient conditions for p and p +2k to be primes where k ∈ Z+(Theorem 3 and Theorem 7). Unfortunately, because of the tedious calculation of the factorial of a large number, these theorems may not be very effective in searching for large twin primes. Though Theorem 7 reduces significantly the number of p−1 computational steps of Theorem 3, it is still a daunting task to calculate ( 2 )! when p becomes very large. Moreover, as we cannot prove that there are infinitely many numbers satisfying the condition yet, we cannot solve the Polignac Conjecture. Thus, we need to further improve these theorems or look for other alternative routes in attempting the Twin Prime Conjecture and the Polignac Conjecture.

ACKNOWLEDGEMENT We would like to express our gratitude to Assoc Prof Choi Kwok Pui of National University of Singapore in guiding us in the project on Benford’s Law which led us to examine the goodness-of-fit for prime number distribution to the Benford’s distribution. While tackling this intriguing problem, we extended our knowledge of primes and obtained the results stated in this paper along the way.

[1] Chen J. R., Shao P. C., Goldbach hypothesis, Shenyang: Liaoning Education Press, (1987). 8

[2] G.H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, Oxford Science Publications, (1979). [3] P.A. Clement, Congruences for sets of primes, American Mathematical Monthly, vol. 56, p. 23-25, (1949). [4] W.J. LeVeque, Fundamentals of Number Theory, New York, Dover, (1996). [5] Hale Unsal, Unsolved Problems about Prime Numbers, Cankaya University (2002). [6] Introduction to twin primes and Brun’s constant computation, http://numbers.computation.free.fr/Constants/Primes/twin.html (2002).