A mod p 3 analogue of a theorem of Gauss on binomial coefficients
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1 A mod p 3 analogue of a theorem of Gauss on binomial coefficients Dalhousie University, Halifax, Canada ELAZ, Schloß Schney, August 16, 01
2 Joint work with John B. Cosgrave Dublin, Ireland
3 We begin with a table: Mod p3 analogue
4 We begin with a table: Mod p3 analogue
5 We begin with a table: p ( p 1 ) p 1 (mod p) p 1 (mod 4)
6 Add in sums of squares: p ( p 1 ) p 1 (mod p) a b p 1 (mod 4), p = a + b.
7 Reformulating the table: ( p ) p 1 (mod p) < p a b p 1 (mod 4), p = a + b.
8 1. Introduction The table is an illustration of the following celebrated result: Theorem 1 (Gauss, 188) Let p 1 (mod 4) be a prime and write p = a + b, a 1 (mod 4). Then ( p 1 ) p 1 a 4 (mod p).
9 1. Introduction The table is an illustration of the following celebrated result: Theorem 1 (Gauss, 188) Let p 1 (mod 4) be a prime and write p = a + b, a 1 (mod 4). Then ( p 1 ) p 1 a 4 (mod p). Several different proofs are known, some using Jacobsthal sums".
10 Goal: Extend this to a congruence mod p.
11 Goal: Extend this to a congruence mod p. Need Fermat quotients: For m Z, m, and p m, define q p (m) := mp 1 1. p
12 Goal: Extend this to a congruence mod p. Need Fermat quotients: For m Z, m, and p m, define q p (m) := mp 1 1. p Beukers (1984) conjectured, and Chowla, Dwork & Evans (1986) proved: Theorem (Chowla, Dwork, Evans) Let p and a be as before. Then ( p 1 ) p 1 ( a a)( p pq p() ) (mod p ). 4
13 Goal: Extend this to a congruence mod p. Need Fermat quotients: For m Z, m, and p m, define q p (m) := mp 1 1. p Beukers (1984) conjectured, and Chowla, Dwork & Evans (1986) proved: Theorem (Chowla, Dwork, Evans) Let p and a be as before. Then ( p 1 ) p 1 ( a a)( p pq p() ) (mod p ). 4 Application: Search for Wilson primes.
14 . Gauss Factorials Recall Wilson s Theorem: p is a prime if and only if (p 1)! 1 (mod p).
15 . Gauss Factorials Recall Wilson s Theorem: p is a prime if and only if Define the Gauss factorial (p 1)! 1 (mod p). N n! = 1 j N gcd(j,n)=1 j.
16 . Gauss Factorials Recall Wilson s Theorem: p is a prime if and only if Define the Gauss factorial (p 1)! 1 (mod p). N n! = 1 j N gcd(j,n)=1 j. Theorem 3 (Gauss) For any integer n, { 1 (mod n) for n =, 4, p α, or p α, (n 1) n! 1 (mod n) otherwise, where p is an odd prime and α is a positive integer.
17 Recall Gauss Theorem: ( ) p 1! ) ) a (mod p).! (( p 1 4
18 Recall Gauss Theorem: ( ) p 1! ) ) a (mod p).! (( p 1 4 Can we have something like this for p in place of p, using Gauss factorials?
19 Recall Gauss Theorem: ( ) p 1! ) ) a (mod p).! (( p 1 4 Can we have something like this for p in place of p, using Gauss factorials? Idea: Use the mod p extension by Chowla et al.
20 Recall Gauss Theorem: ( ) p 1! ) ) a (mod p).! (( p 1 4 Can we have something like this for p in place of p, using Gauss factorials? Idea: Use the mod p extension by Chowla et al. Main technical device: We can show that ( p ) ( ) 1! (p 1)! p 1 p 1! 1 + p 1 p p p 1 j=1 1 j (mod p ).
21 We can derive a similar congruence for ( p ) 1! (mod p ). 4 p
22 We can derive a similar congruence for ( p ) 1! (mod p ). 4 p Also used is the congruence p 1 j=1 1 j q p () (mod p), and other similar congruences due to Emma Lehmer (1938) and others before her.
23 We can derive a similar congruence for ( p ) 1! (mod p ). 4 p Also used is the congruence p 1 j=1 1 j q p () (mod p), and other similar congruences due to Emma Lehmer (1938) and others before her. Altogether we have, after simplifying, ( ) p 1! ( p 1 ) p 1 ( ( ) ) p 1 p pq p() 4 p! (mod p ).
24 Combining this with the theorem of Chowla, Dwork & Evans: Theorem 4 Let p and a be as before. Then ) ( p 1 ( ( p 1 4 ) p! p! ) a p a (mod p ).
25 Combining this with the theorem of Chowla, Dwork & Evans: Theorem 4 Let p and a be as before. Then ) ( p 1 ( ( p 1 4 ) p! p! ) a p a (mod p ). Hopeless to conjecture an extension of the theorem of Chowla et al., but easily possible for the theorem above.
26 3. Extensions modulo p 3 By numerical experimentation we first conjectured Theorem 5 Let p and a be as before. Then ) ( p 3 1 ( ( p ) p! p! ) a p a p 8a 3 (mod p 3 ). (Proof later).
27 3. Extensions modulo p 3 By numerical experimentation we first conjectured Theorem 5 Let p and a be as before. Then ) ( p 3 1 ( ( p ) p! p! ) a p a p 8a 3 (mod p 3 ). (Proof later). Using more complicated congruences than the ones leading to Theorem 4 (but the same ideas), and going backwards, we obtain
28 Theorem 6 (Main result) Let p and a be as before. Then ( p 1 ) p 1 (a 4 ) p pa 8a 3 ( ( pq p() p E p 3 q p () )) (mod p 3 ).
29 Theorem 6 (Main result) Let p and a be as before. Then ( p 1 ) p 1 (a 4 ) p pa 8a 3 ( ( pq p() p E p 3 q p () )) (mod p 3 ). Here E p 3 is the Euler number defined by e t + e t = n=0 E n n! tn ( t < π).
30 Theorem 6 (Main result) Let p and a be as before. Then ( p 1 ) p 1 (a 4 ) p pa 8a 3 ( ( pq p() p E p 3 q p () )) (mod p 3 ). Here E p 3 is the Euler number defined by e t + e t = n=0 How can we prove Theorem 5? E n n! tn ( t < π).
31 Theorem 6 (Main result) Let p and a be as before. Then ( p 1 ) p 1 (a 4 ) p pa 8a 3 ( ( pq p() p E p 3 q p () )) (mod p 3 ). Here E p 3 is the Euler number defined by e t + e t = n=0 E n n! tn ( t < π). How can we prove Theorem 5? By further experimentation we first conjectured, and then proved the following generalization.
32 Theorem 7 Let p and a be as before and let α be an integer. Then ) ( p α 1 ( ( p α 1 4 ) p! p! ) a 1 p a 1 p 8a 3
33 Theorem 7 Let p and a be as before and let α be an integer. Then ) ( p α 1 ( ( p α 1 4 ) p! p! ) a 1 p a 1 p 8a 3 p 3 (a) 5
34 Theorem 7 Let p and a be as before and let α be an integer. Then ) ( p α 1 ( ( p α 1 4 ) p! p! ) a 1 p a 1 p 8a 3 p 3 (a) 5 5 p 4 (a) 7
35 Theorem 7 Let p and a be as before and let α be an integer. Then ) ( p α 1 ( ( p α 1 4 ) p! p! ) a 1 14 p 5 (a) 9 p a 1 p 8a 3 p 3 (a) 5 5 p 4 (a) 7
36 Theorem 7 Let p and a be as before and let α be an integer. Then ) ( p α 1 ( ( p α 1 4 ) p! p! ) a 1 p 5 p a 1 p 8a 3 p 3 (a) 5 5 p 4 (a) 7 p α 1 14 (a) 9... C α (a) α 1 (mod p α ).
37 Theorem 7 Let p and a be as before and let α be an integer. Then ) ( p α 1 ( ( p α 1 4 ) p! p! ) a 1 p 5 p a 1 p 8a 3 p 3 (a) 5 5 p 4 (a) 7 p α 1 14 (a) 9... C α (a) α 1 (mod p α ). Here C n := 1 an integer. n+1 ( n n ) is the nth Catalan number which is always
38 Theorem 7 Let p and a be as before and let α be an integer. Then ) ( p α 1 ( ( p α 1 4 ) p! p! ) a 1 p 5 p a 1 p 8a 3 p 3 (a) 5 5 p 4 (a) 7 p α 1 14 (a) 9... C α (a) α 1 (mod p α ). Here C n := 1 an integer. n+1 ( n n ) is the nth Catalan number which is always Theorem 5 is obviously a special case of Theorem 7.
39 4. Main Ingredients in the Proof The Jacobi sum J(χ, ψ) = χ(j)ψ(1 j), j mod p where χ and ψ are characters modulo p.
40 4. Main Ingredients in the Proof The Jacobi sum J(χ, ψ) = j mod p χ(j)ψ(1 j), where χ and ψ are characters modulo p. Fix a primitive root g mod p; let χ be a character of order 4 such that χ(g) = i.
41 4. Main Ingredients in the Proof The Jacobi sum J(χ, ψ) = j mod p χ(j)ψ(1 j), where χ and ψ are characters modulo p. Fix a primitive root g mod p; let χ be a character of order 4 such that χ(g) = i. Define integers a, b by ( ) p = a +b, a (mod 4), b a g (p 1)/4 (mod p). p
42 4. Main Ingredients in the Proof The Jacobi sum J(χ, ψ) = j mod p χ(j)ψ(1 j), where χ and ψ are characters modulo p. Fix a primitive root g mod p; let χ be a character of order 4 such that χ(g) = i. Define integers a, b by ( ) p = a +b, a (mod 4), b a g (p 1)/4 (mod p). p These are uniquely defined, differ from a and b of Gauss theorem only (possibly) in sign.
43 Then J(χ, χ) = ( 1) p 1 4 (a + ib ), J(χ 3, χ 3 ) = ( 1) p 1 4 (a ib ),
44 Then J(χ, χ) = ( 1) p 1 4 (a + ib ), J(χ 3, χ 3 ) = ( 1) p 1 4 (a ib ), On the other hand, J(χ, χ) 0 (mod p), J(χ 3, χ 3 ) = Γ p(1 1 ) Γ p (1 1 4 ). These are deep results, related to the Gross-Koblitz formula" (see, e.g., Gauss and Jacobi Sums by B. Berndt, R. Evans and K. Williams).
45 Γ p (z) is the p-adic gamma function defined by F(n) := ( 1) n 0<j<n p j Γ p (z) = lim n z F(n) (z Z p ), where n runs through any sequence of positive integers p-adically approaching z. j,
46 In particular, ( 1) p 1 4 (a ib ) = J(χ 3, χ 3 ) = Γ p(1 1 ) Γ p (1 1 4 ) Γ p(1 + pα 1 ) Γ p (1 + pα 1 4 ) (mod pα ) = F(1 + pα 1 ) F(1 + pα 1 = 4 ) ( ) p α 1 ( ( p α 1 4 ) p! p! ).
47 Raise to the power α: ( 1) p 1 4 (a + ib ) = J(χ, χ) 0 (mod p)
48 Raise to the power α: ( 1) p 1 4 (a + ib ) = J(χ, χ) 0 (mod p) (a + ib ) α 0 (mod p α ).
49 Raise to the power α: ( 1) p 1 4 (a + ib ) = J(χ, χ) 0 (mod p) (a + ib ) α 0 (mod p α ). Expand the left-hand side; get binomial coefficients;
50 Raise to the power α: ( 1) p 1 4 (a + ib ) = J(χ, χ) 0 (mod p) (a + ib ) α 0 (mod p α ). Expand the left-hand side; get binomial coefficients; separate real and imaginary parts;
51 Raise to the power α: ( 1) p 1 4 (a + ib ) = J(χ, χ) 0 (mod p) (a + ib ) α 0 (mod p α ). Expand the left-hand side; get binomial coefficients; separate real and imaginary parts; use the combinatorial identity (k = 0, 1,..., n 1) k j=0 ( 1) j j + 1 ( j j )( ) ( n + j k n 1 k = k j k ) ;
52 Raise to the power α: ( 1) p 1 4 (a + ib ) = J(χ, χ) 0 (mod p) (a + ib ) α 0 (mod p α ). Expand the left-hand side; get binomial coefficients; separate real and imaginary parts; use the combinatorial identity (k = 0, 1,..., n 1) k j=0 ( 1) j j + 1 ( j j )( ) ( n + j k n 1 k = k j k putting everything together, we obtain Theorem 7. ) ;
53 5. A Jacobi Analogue Let p 1 (mod 6). Then we can write 4p = r + 3s, r 1 (mod 3), 3 s, which determines r uniquely.
54 5. A Jacobi Analogue Let p 1 (mod 6). Then we can write 4p = r + 3s, r 1 (mod 3), 3 s, which determines r uniquely. In analogy to Gauss Theorem 1 we have Theorem 8 (Jacobi, 1837) Let p and r be as above. Then ( (p 1) ) 3 p 1 r 3 (mod p).
55 5. A Jacobi Analogue Let p 1 (mod 6). Then we can write 4p = r + 3s, r 1 (mod 3), 3 s, which determines r uniquely. In analogy to Gauss Theorem 1 we have Theorem 8 (Jacobi, 1837) Let p and r be as above. Then ( (p 1) ) 3 p 1 r 3 (mod p). This was generalized to mod p independently by Evans (unpublished, 1985) and Yeung (1989):
56 Theorem 9 (Evans; Yeung) Let p and r be as above. Then ( (p 1) ) 3 p 1 r + p r 3 (mod p ).
57 Theorem 9 (Evans; Yeung) Let p and r be as above. Then ( (p 1) ) 3 p 1 r + p r 3 (mod p ). With methods similar to those in the first part of this talk, we proved Theorem 10 Let p and r be as above. Then ( (p 1) ) 3 p 1 ( r + 3 ) pr p + (1 r ) p B p ( 13 ) (mod p 3 ). Here B n (x) is the nth Bernoulli polynomial.
58 6. More on Gauss Factorials Write out the factorial (p 1)!, exploit symmetry mod p: ( 1... p 1 p+1... (p 1) p 1 )!( 1) p 1 ( ) p 1! (mod p).
59 6. More on Gauss Factorials Write out the factorial (p 1)!, exploit symmetry mod p: ( 1... p 1 p+1... (p 1) p 1 )!( 1) p 1 Thus, with Wilson s Theorem, ( p 1 )! ( 1) p+1 (mod p). ( ) p 1! (mod p).
60 6. More on Gauss Factorials Write out the factorial (p 1)!, exploit symmetry mod p: ( 1... p 1 p+1... (p 1) p 1 )!( 1) p 1 Thus, with Wilson s Theorem, ( p 1 )! ( 1) p+1 (mod p). ( ) p 1! (mod p). This was apparently first observed by Lagrange (1773).
61 6. More on Gauss Factorials Write out the factorial (p 1)!, exploit symmetry mod p: ( 1... p 1 p+1... (p 1) p 1 )!( 1) p 1 Thus, with Wilson s Theorem, ( p 1 )! ( 1) p+1 (mod p). ( ) p 1! (mod p). This was apparently first observed by Lagrange (1773). For p 1 (mod 4) the RHS is 1, so (( ) ) ord p 1 p! = 4 for p 1 (mod 4).
62 In the case p 3 (mod 4) we get ( p 1 )! ±1 (mod p).
63 In the case p 3 (mod 4) we get ( p 1 What is the sign on the right? )! ±1 (mod p).
64 In the case p 3 (mod 4) we get ( p 1 What is the sign on the right? Theorem 11 (Mordell, 1961) For a prime p 3 (mod 4), ( p 1 )! ±1 (mod p). )! 1 (mod p) h( p) 1 (mod 4), where h( p) is the class number of Q( p).
65 In the case p 3 (mod 4) we get ( p 1 What is the sign on the right? Theorem 11 (Mordell, 1961) For a prime p 3 (mod 4), ( p 1 )! ±1 (mod p). )! 1 (mod p) h( p) 1 (mod 4), where h( p) is the class number of Q( p). Discovered independently by Chowla. This completely determines the order mod p of ( ) p 1!.
66 Now consider the two halves of the product 1 p 1 p+1 (p 1) and denote them, respectively, by Π () 1, Π().
67 Now consider the two halves of the product 1 p 1 p+1 (p 1) and denote them, respectively, by By Wilson s theorem: Π () 1, Π(). Π () 1 Π() 1 (mod p),
68 Now consider the two halves of the product 1 p 1 p+1 (p 1) and denote them, respectively, by By Wilson s theorem: and by symmetry: Π () 1, Π(). Π () 1 Π() 1 (mod p), Π () ( 1) p 1 Π () 1 (mod p).
69 What can we say about the three partial products Π (3) 1, Π(3), Π(3) 3 obtained by dividing the entire product (p 1)! into three equal parts?
70 What can we say about the three partial products Π (3) 1, Π(3), Π(3) 3 obtained by dividing the entire product (p 1)! into three equal parts? We require p 1 (mod 3); in fact, p 1 (mod 6).
71 What can we say about the three partial products Π (3) 1, Π(3), Π(3) 3 obtained by dividing the entire product (p 1)! into three equal parts? We require p 1 (mod 3); in fact, p 1 (mod 6). Then Π (3) 1 = 1 p 1 3, Π(3) = p+ 3 p 3, Π (3) 3 = p+1 3 (p 1).
72 What can we say about the three partial products Π (3) 1, Π(3), Π(3) 3 obtained by dividing the entire product (p 1)! into three equal parts? We require p 1 (mod 3); in fact, p 1 (mod 6). Then Π (3) 1 = 1 p 1 3, Π(3) = p+ 3 p 3, Π (3) 3 = p+1 3 (p 1). Once again there is an obvious symmetry: Π (3) 3 Π (3) 1 (mod p), (without a power of 1 since p 1 3 is always even.)
73 What can we say about the three partial products Π (3) 1, Π(3), Π(3) 3 obtained by dividing the entire product (p 1)! into three equal parts? We require p 1 (mod 3); in fact, p 1 (mod 6). Then Π (3) 1 = 1 p 1 3, Π(3) = p+ 3 p 3, Π (3) 3 = p+1 3 (p 1). Once again there is an obvious symmetry: Π (3) 3 Π (3) 1 (mod p), (without a power of 1 since p 1 3 is always even.) No obvious relation between Π (3) 1 and the middle third" Π (3).
74 Natural extension: For any M and for a prime p 1 (mod M), divide (p 1)! into the products
75 Natural extension: For any M and for a prime p 1 (mod M), divide (p 1)! into the products Π (M) j = p 1 M i=1 ( (j 1) p 1 M + i ), (j = 1,,..., M).
76 Natural extension: For any M and for a prime p 1 (mod M), divide (p 1)! into the products Π (M) j = p 1 M i=1 Once again, clear that ( (j 1) p 1 M + i ), (j = 1,,..., M). Π (M) M j ±Π (M) j (mod p), j = 1,,..., M 1.
77 Natural extension: For any M and for a prime p 1 (mod M), divide (p 1)! into the products Π (M) j = p 1 M i=1 Once again, clear that Example: ( (j 1) p 1 M + i ), (j = 1,,..., M). Π (M) M j ±Π (M) j (mod p), j = 1,,..., M 1.
78 p Π (3) 1 Π (3) Π (3) 3 p Π (4) 1 Π (4) Π (4) 3 Π (4)
79 p Π (3) 1 Π (3) Π (3) 3 p Π (4) 1 Π (4) Π (4) 3 Π (4) We observe: The obvious symmetries.
80 p Π (3) 1 Π (3) Π (3) 3 p Π (4) 1 Π (4) Π (4) 3 Π (4) We observe: The obvious symmetries. Π (3) 1 Π (3) (mod p) for p = 7 and p = 61.
81 Are these last cases just coincidences?
82 Are these last cases just coincidences? It turns out: Π (3) 1 Π (3) (mod p) also for p = 331, p = 547, p = 1951, and further relatively rare primes (explained below).
83 Are these last cases just coincidences? It turns out: Π (3) 1 Π (3) (mod p) also for p = 331, p = 547, p = 1951, and further relatively rare primes (explained below). In contrast: No primes p for which Π (3) 1 Π (3) (mod p), p 1 (mod 6), or Π (4) 1 ±Π (4) (mod p), p 1 (mod 4).
84 Are these last cases just coincidences? It turns out: Π (3) 1 Π (3) (mod p) also for p = 331, p = 547, p = 1951, and further relatively rare primes (explained below). In contrast: No primes p for which Π (3) 1 Π (3) (mod p), p 1 (mod 6), or Π (4) 1 ±Π (4) (mod p), p 1 (mod 4). The theorems of Gauss and Jacobi can be used to explain this:
85 Consider the quotient Q 4 (p) := Π(4) Π (4) 1
86 Consider the quotient Q 4 (p) := Π(4) Π (4) 1 = Π(4) 1 Π(4) ( ) Π (4) 1
87 Consider the quotient Q 4 (p) := Π(4) Π (4) 1 = Π(4) 1 Π(4) ( Π (4) 1 ) = p 1 (! ) p 1 4!
88 Consider the quotient Q 4 (p) := Π(4) Π (4) 1 = Π(4) 1 Π(4) ( Π (4) 1 ) = p 1 (! ) = p 1 4! ( p 1 ) p 1. 4
89 Consider the quotient Q 4 (p) := Π(4) Π (4) 1 Recall Gauss theorem: = Π(4) 1 Π(4) ( Π (4) 1 ( p 1 p 1 4 ) a ) = p 1 (! ) = p 1 4! (mod p), ( p 1 ) p 1. 4 where p 1 (mod 4), p = a + b, a 1 (mod 4).
90 Consider the quotient Q 4 (p) := Π(4) Π (4) 1 Recall Gauss theorem: = Π(4) 1 Π(4) ( Π (4) 1 ( p 1 p 1 4 ) a ) = p 1 (! ) = p 1 4! (mod p), ( p 1 ) p 1. 4 where p 1 (mod 4), p = a + b, a 1 (mod 4). It follows easily that Π (4) ±Π (4) 1 (mod p) for all p 1 (mod 4).
91 Consider the quotient Q 4 (p) := Π(4) Π (4) 1 Recall Gauss theorem: = Π(4) 1 Π(4) ( Π (4) 1 ( p 1 p 1 4 ) a ) = p 1 (! ) = p 1 4! (mod p), ( p 1 ) p 1. 4 where p 1 (mod 4), p = a + b, a 1 (mod 4). It follows easily that Π (4) ±Π (4) 1 (mod p) for all p 1 (mod 4). Similarly, Jacobi s theorem implies that Π (3) Π (3) 1 (mod p) for all p 1 (mod 6).
92 A number of further consequences can be derived from these classical theorems.
93 A number of further consequences can be derived from these classical theorems. Two samples: Corollary 1 For a prime p 1 (mod 6) we have Π (3) Π (3) 1 (mod p) p = 7x + 7x + 7, x Z.
94 A number of further consequences can be derived from these classical theorems. Two samples: Corollary 1 For a prime p 1 (mod 6) we have Π (3) Π (3) 1 (mod p) p = 7x + 7x + 7, x Z. The first such primes are 7, 61 (seen earlier), 331, 547, 1951.
95 Corollary 13 Let p 1 (mod 4). Then (a) p 1 4! 1 (mod p) only if p = 5.
96 Corollary 13 Let p 1 (mod 4). Then (a) p 1 4 (b)! 1 (mod p) only if p = 5. ( ) k p 1 4! 1 (mod p) for k = 1,, 4.
97 Corollary 13 Let p 1 (mod 4). Then (a) p 1 4 ( p 1 (b) (c)! 1 (mod p) only if p = 5. ) k 4! 1 (mod p) for k = 1,, 4. ( ) 8 p 1 4! 1 (mod p) holds for p = 17, 41, 3361, 46817, 65081,...
98 Corollary 13 Let p 1 (mod 4). Then (a) p 1 4 ( p 1 (b) (c)! 1 (mod p) only if p = 5. ) k 4! 1 (mod p) for k = 1,, 4. ( ) 8 p 1 4! 1 (mod p) holds for p = 17, 41, 3361, 46817, 65081,... Part (c) is related to the solution of a certain Pell equation.
99 How are we doing with time? Salvador Dalí, The Persistence of Memory, 1931.
100 7. Composite Moduli Recall Gauss generalization of Wilson s theorem, the prototype of all composite modulus extensions: { 1 (mod n) for n =, 4, p α, or p α, (n 1) n! 1 (mod n) otherwise, where p is an odd prime and α is a positive integer.
101 7. Composite Moduli Recall Gauss generalization of Wilson s theorem, the prototype of all composite modulus extensions: { 1 (mod n) for n =, 4, p α, or p α, (n 1) n! 1 (mod n) otherwise, where p is an odd prime and α is a positive integer. Everything we did before can be extended to composite moduli.
102 7. Composite Moduli Recall Gauss generalization of Wilson s theorem, the prototype of all composite modulus extensions: { 1 (mod n) for n =, 4, p α, or p α, (n 1) n! 1 (mod n) otherwise, where p is an odd prime and α is a positive integer. Everything we did before can be extended to composite moduli. E.g., the multiplicative orders of ( n 1 ) n! modulo n were completely determined; only orders 1,, 4 occur. (JBC & KD, 008).
103 Next, divide the product (n 1) n! into M partial products:
104 Next, divide the product (n 1) n! into M partial products: For n 1 (mod M), set Π (M) j := i I (M) j where, for j = 1,,..., M, I (M) j := { i (j 1) n 1 M i, (j = 1,,..., M), + 1 i j n 1 M, gcd(i, n) = 1}.
105 Next, divide the product (n 1) n! into M partial products: For n 1 (mod M), set Π (M) j := i I (M) j where, for j = 1,,..., M, I (M) j := { i (j 1) n 1 M i, (j = 1,,..., M), + 1 i j n 1 M, gcd(i, n) = 1}. Dependence on n is implied in the notation;
106 Next, divide the product (n 1) n! into M partial products: For n 1 (mod M), set Π (M) j := i I (M) j where, for j = 1,,..., M, I (M) j := { i (j 1) n 1 M i, (j = 1,,..., M), + 1 i j n 1 M, gcd(i, n) = 1}. Dependence on n is implied in the notation; When n = p: reduces to previous case.
107 Example: n Π (3) 1 Π (3) Π (3) 3 n Π (4) 1 Π (4) Π (4) 3 Π (4)
108 Example: n Π (3) 1 Π (3) Π (3) 3 n Π (4) 1 Π (4) Π (4) 3 Π (4)
109 Example: n Π (3) 1 Π (3) Π (3) 3 n Π (4) 1 Π (4) Π (4) 3 Π (4) We see: In contrast to prime case, it can happen that all partial products are congruent to each other.
110 We pause to consider the number of elements in our subintervals: φ M,j (n) := #I (M) j. (Called totatives by J. J. Sylvester and later D. H. Lehmer).
111 We pause to consider the number of elements in our subintervals: φ M,j (n) := #I (M) j. (Called totatives by J. J. Sylvester and later D. H. Lehmer). When n = p is a prime, then φ M,j (n) = p 1 M for all j.
112 We pause to consider the number of elements in our subintervals: φ M,j (n) := #I (M) j. (Called totatives by J. J. Sylvester and later D. H. Lehmer). When n = p is a prime, then φ M,j (n) = p 1 M for all j. When M = 1, then φ 1,1 (n) = φ(n).
113 We pause to consider the number of elements in our subintervals: φ M,j (n) := #I (M) j. (Called totatives by J. J. Sylvester and later D. H. Lehmer). When n = p is a prime, then φ M,j (n) = p 1 M for all j. When M = 1, then φ 1,1 (n) = φ(n). When M =, then φ,1 (n) = φ, (n) = 1 φ(n).
114 We pause to consider the number of elements in our subintervals: φ M,j (n) := #I (M) j. (Called totatives by J. J. Sylvester and later D. H. Lehmer). When n = p is a prime, then φ M,j (n) = p 1 M for all j. When M = 1, then φ 1,1 (n) = φ(n). When M =, then φ,1 (n) = φ, (n) = 1 φ(n). In general, the situation is less straightforward.
115 We pause to consider the number of elements in our subintervals: φ M,j (n) := #I (M) j. (Called totatives by J. J. Sylvester and later D. H. Lehmer). When n = p is a prime, then φ M,j (n) = p 1 M for all j. When M = 1, then φ 1,1 (n) = φ(n). When M =, then φ,1 (n) = φ, (n) = 1 φ(n). In general, the situation is less straightforward. E.g., for n = 4: φ 3,1 (n) = φ 3,3 (n) = 1, but φ 3, (n) = 0.
116 When do we have equal distribution?
117 When do we have equal distribution? Theorem 14 (Lehmer, 1955) Let M and n 1 (mod M). If n has at least one prime factor p 1 (mod M), then φ M,j (n) = 1 M φ(n), (j = 1,,..., M).
118 When do we have equal distribution? Theorem 14 (Lehmer, 1955) Let M and n 1 (mod M). If n has at least one prime factor p 1 (mod M), then φ M,j (n) = 1 M φ(n), (j = 1,,..., M). Note: Condition is sufficient, but not necessary.
119 When Are the Partial Products Congruent to each other?
120 When Are the Partial Products Congruent to each other? Return to our table: n Π (3) 1 Π (3) Π (3) 3 n Π (4) 1 Π (4) Π (4) 3 Π (4)
121 When Are the Partial Products Congruent to each other? Return to our table: n Π (3) 1 Π (3) Π (3) 3 n Π (4) 1 Π (4) Π (4) 3 Π (4) Note: 91 = 7 13, 65 = 5 13, 85 = 5 17.
122 This observation holds in general:
123 This observation holds in general: Theorem 15 Let M and n 1 (mod M). If n has at least two distinct prime factors 1 (mod M), then Π (M) j ( ) n 1 M! (mod n), j = 1,,..., M. n
124 This observation holds in general: Theorem 15 Let M and n 1 (mod M). If n has at least two distinct prime factors 1 (mod M), then Π (M) j ( ) n 1 M! (mod n), j = 1,,..., M. n Result is best possible.
125 This observation holds in general: Theorem 15 Let M and n 1 (mod M). If n has at least two distinct prime factors 1 (mod M), then Π (M) j ( ) n 1 M! (mod n), j = 1,,..., M. n Result is best possible. On the other hand, condition is sufficient but not necessary.
126 Idea of proof: 1. Observe that Π (M) j = ( ) j n 1 M ( (j 1) n 1 M n! )n!, j = 1,,..., M,
127 Idea of proof: 1. Observe that Π (M) j = ( ) j n 1 M ( (j 1) n 1 M n! )n!, j = 1,,..., M,. Let n = p α q β w for p, q 1 (mod M) and gcd(pq, w) = 1.
128 Idea of proof: 1. Observe that Π (M) j = ( ) j n 1 M ( (j 1) n 1 M n! )n!, j = 1,,..., M,. Let n = p α q β w for p, q 1 (mod M) and gcd(pq, w) = Break the range of the product in ( j n 1 ) M n! into a number of products of approximately equal length, a shorter tail."
129 Idea of proof: 1. Observe that Π (M) j = ( ) j n 1 M ( (j 1) n 1 M n! )n!, j = 1,,..., M,. Let n = p α q β w for p, q 1 (mod M) and gcd(pq, w) = Break the range of the product in ( j n 1 ) M n! into a number of products of approximately equal length, a shorter tail." 4. Then evaluate the products of the first type using the Gauss-Wilson theorem (mod q β w).
130 Idea of proof: 1. Observe that Π (M) j = ( ) j n 1 M ( (j 1) n 1 M n! )n!, j = 1,,..., M,. Let n = p α q β w for p, q 1 (mod M) and gcd(pq, w) = Break the range of the product in ( j n 1 ) M n! into a number of products of approximately equal length, a shorter tail." 4. Then evaluate the products of the first type using the Gauss-Wilson theorem (mod q β w). 5. Carefully count which elements to include/exclude.
131 Idea of proof: 1. Observe that Π (M) j = ( ) j n 1 M ( (j 1) n 1 M n! )n!, j = 1,,..., M,. Let n = p α q β w for p, q 1 (mod M) and gcd(pq, w) = Break the range of the product in ( j n 1 ) M n! into a number of products of approximately equal length, a shorter tail." 4. Then evaluate the products of the first type using the Gauss-Wilson theorem (mod q β w). 5. Carefully count which elements to include/exclude. 6. Use Chinese Remainder Theorem modulo q β w and q α w.
132 In the case of at least three distinct prime factors 1 (mod M) we can say more:
133 In the case of at least three distinct prime factors 1 (mod M) we can say more: Theorem 16 Let M and n 1 (mod M). If n has at least three distinct prime factors 1 (mod M), then Π (M) j 1 (mod n), j = 1,,..., M.
134 In the case of at least three distinct prime factors 1 (mod M) we can say more: Theorem 16 Let M and n 1 (mod M). If n has at least three distinct prime factors 1 (mod M), then Π (M) j 1 (mod n), j = 1,,..., M. Method of proof is similar to that of previous result.
135 Summary: # of prime factors 1 (mod M) All Π (M) 1,..., Π (M) M : 1 have the same number of factors are congruent to each other (mod M) 3 are congruent to 1 (mod M)
136 Outlook Main task: Understand the multiplicative orders of ( ) n 1! (mod n). M n
137 Outlook Main task: Understand the multiplicative orders of ( ) n 1! (mod n). M n When M = : Completely determined.
138 Outlook Main task: Understand the multiplicative orders of ( ) n 1! (mod n). M n When M = : Completely determined. Let M 3 and n 1 (mod M).
139 Outlook Main task: Understand the multiplicative orders of ( ) n 1! (mod n). M n When M = : Completely determined. Let M 3 and n 1 (mod M). Then, when n has 3 prime factors 1 (mod M): order is always 1.
140 Outlook Main task: Understand the multiplicative orders of ( ) n 1! (mod n). M n When M = : Completely determined. Let M 3 and n 1 (mod M). Then, when n has 3 prime factors 1 (mod M): order is always 1. n has prime factors 1 (mod M): order is a divisor of M.
141 Outlook Main task: Understand the multiplicative orders of ( ) n 1! (mod n). M n When M = : Completely determined. Let M 3 and n 1 (mod M). Then, when n has 3 prime factors 1 (mod M): order is always 1. n has prime factors 1 (mod M): order is a divisor of M. n has 0 or 1 prime factor 1 (mod M): The most interesting and difficult case. Several results already published. Further work in progress.
142 Thank you Danke Mod p3 analogue
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