A mod p 3 analogue of a theorem of Gauss on binomial coefficients

A mod p 3 analogue of a theorem of Gauss on binomial coefficients Dalhousie University, Halifax, Canada ELAZ, Schloß Schney, August 16, 01

Joint work with John B. Cosgrave Dublin, Ireland

We begin with a table: Mod p3 analogue

We begin with a table: p ( p 1 ) p 1 (mod p) 5 4 13 0 7 17 70 9 343 10 37 4860 41 184756 10 53 10400600 39 61 10 73 67 89 10 97 18 p 1 (mod 4)

Add in sums of squares: p ( p 1 ) p 1 (mod p) a b 5 4 1 13 0 7 3 17 70 1 4 9 343 10 5 37 4860 1 6 41 184756 10 5 4 53 10400600 39 7 61 10 5 6 73 67 3 8 89 10 5 8 97 18 9 4 p 1 (mod 4), p = a + b.

Reformulating the table: ( p ) p 1 (mod p) < p a b 4 5 1 13 0 7 6 3 17 70 1 4 9 343 10 10 5 37 4860 1 6 41 184756 10 10 5 4 53 10400600 39 14 7 61 10 10 5 6 73 67 6 3 8 89 10 10 5 8 97 18 18 9 4 p 1 (mod 4), p = a + b.

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).

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".

Goal: Extend this to a congruence mod p.

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

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 1 + 1 pq p() ) (mod p ). 4

. Gauss Factorials Recall Wilson s Theorem: p is a prime if and only if (p 1)! 1 (mod p).

. 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.

. 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.

Recall Gauss Theorem: ( ) p 1! ) ) a (mod p).! (( p 1 4

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?

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.

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 ).

We can derive a similar congruence for ( p ) 1! (mod p ). 4 p

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.

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 1 4 1 + 1 pq p() 4 p! (mod p ).

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 ).

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.

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 3 1 4 ) p! p! ) a p a p 8a 3 (mod p 3 ). (Proof later).

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 3 1 4 ) 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

Theorem 6 (Main result) Let p and a be as before. Then ( p 1 ) p 1 (a 4 ) p pa 8a 3 ( ( 1 + 1 pq p() + 1 8 p E p 3 q p () )) (mod p 3 ).

Theorem 6 (Main result) Let p and a be as before. Then ( p 1 ) p 1 (a 4 ) p pa 8a 3 ( ( 1 + 1 pq p() + 1 8 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).

Theorem 6 (Main result) Let p and a be as before. Then ( p 1 ) p 1 (a 4 ) p pa 8a 3 ( ( 1 + 1 pq p() + 1 8 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).

Theorem 6 (Main result) Let p and a be as before. Then ( p 1 ) p 1 (a 4 ) p pa 8a 3 ( ( 1 + 1 pq p() + 1 8 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.

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

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

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

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

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 ).

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

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.

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

Then J(χ, χ) = ( 1) p 1 4 (a + ib ), J(χ 3, χ 3 ) = ( 1) p 1 4 (a ib ),

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).

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,

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! ).

Raise to the power α: ( 1) p 1 4 (a + ib ) = J(χ, χ) 0 (mod p)

Raise to the power α: ( 1) p 1 4 (a + ib ) = J(χ, χ) 0 (mod p) (a + ib ) α 0 (mod p α ).

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;

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;

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 ) ;

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.

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).

Theorem 9 (Evans; Yeung) Let p and r be as above. Then ( (p 1) ) 3 p 1 r + p r 3 (mod p ).

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 3 + 16 ) p B p ( 13 ) (mod p 3 ). Here B n (x) is the nth Bernoulli polynomial.

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).

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).

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).

In the case p 3 (mod 4) we get ( p 1 )! ±1 (mod p).

In the case p 3 (mod 4) we get ( p 1 What is the sign on the right? )! ±1 (mod p).

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!.

Now consider the two halves of the product 1 p 1 p+1 (p 1) and denote them, respectively, by Π () 1, Π().

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),

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).

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?

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).

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.)

Natural extension: For any M and for a prime p 1 (mod M), divide (p 1)! into the products

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).

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.

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.

p Π (3) 1 Π (3) Π (3) 3 p Π (4) 1 Π (4) Π (4) 3 Π (4) 4 7 5 1 1 13 3 13 6 3 3 6 19 5 17 7 3 3 7 31 8 9 6 6 37 7 3 7 37 16 5 5 16 43 3 19 3 41 13 7 7 13 61 14 14 14 53 6 7 7 6 67 0 33 0 61 19 7 7 19 73 33 1 33 73 18 35 35 18 79 37 3 37 89 4 4 97 1 11 1 97 0 8 8 0

Are these last cases just coincidences?

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).

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).

Consider the quotient Q 4 (p) := Π(4) Π (4) 1

Consider the quotient Q 4 (p) := Π(4) Π (4) 1 = Π(4) 1 Π(4) ( ) Π (4) 1

Consider the quotient Q 4 (p) := Π(4) Π (4) 1 = Π(4) 1 Π(4) ( Π (4) 1 ) = p 1 (! ) p 1 4!

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

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).

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).

A number of further consequences can be derived from these classical theorems.

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.

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.

Corollary 13 Let p 1 (mod 4). Then (a) p 1 4! 1 (mod p) only if p = 5.

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.

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,...

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.

How are we doing with time? Salvador Dalí, The Persistence of Memory, 1931.

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.

Next, divide the product (n 1) n! into M partial products:

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}.

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;

Example: n Π (3) 1 Π (3) Π (3) 3 n Π (4) 1 Π (4) Π (4) 3 Π (4) 4 70 9 1 9 61 19 7 7 19 73 33 1 33 65 8 8 8 8 76 9 15 9 69 31 6 6 31 79 37 3 37 73 18 35 35 18 8 33 5 33 77 16 31 31 16 85 8 9 8 81 40 40 88 5 7 5 85 13 13 13 13 91 9 9 9 89 4 4 94 3 43 3 93 34 10 10 34 97 1 11 1 97 0 8 8 0

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).

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).

When do we have equal distribution?

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).

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.

When Are the Partial Products Congruent to each other?

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) 4 70 9 1 9 61 19 7 7 19 73 33 1 33 65 8 8 8 8 76 9 15 9 69 31 6 6 31 79 37 3 37 73 18 35 35 18 8 33 5 33 77 16 31 31 16 85 8 9 8 81 40 40 88 5 7 5 85 13 13 13 13 91 9 9 9 89 4 4 94 3 43 3 93 34 10 10 34 97 1 11 1 97 0 8 8 0

This observation holds in general:

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

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.

Idea of proof: 1. Observe that Π (M) j = ( ) j n 1 M ( (j 1) n 1 M n! )n!, j = 1,,..., M,

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.

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. 3. 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).

In the case of at least three distinct prime factors 1 (mod M) we can say more:

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.

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)

Outlook Main task: Understand the multiplicative orders of ( ) n 1! (mod n). M n

Outlook Main task: Understand the multiplicative orders of ( ) n 1! (mod n). M n When M = : Completely determined.

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).

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.

Thank you Danke Mod p3 analogue