q-binomial coefficient congruences

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1 CARMA Analysis and Number Theory Seminar University of Newcastle August 23, 2011 Tulane University, New Orleans 1 / 35

2 Our first -analogs The natural number n has the -analog: [n] = n 1 1 = n 1 In the limit 1 a -analog reduces to the classical object. 2 / 35

3 Our first -analogs The natural number n has the -analog: [n] = n 1 1 = n 1 In the limit 1 a -analog reduces to the classical object. The -factorial: [n]! = [n] [n 1] [1] The -binomial coefficient: ( ) n = k ( ) [n]! n [k]! [n k]! = n k D1 2 / 35

4 A -binomial coefficient Example (6 ) 2 = ( )( ) 1 + = (1 + 2 ) ( ) ( ) }{{}}{{} =[3] =[5] Let us understand the first term a bit better! 3 / 35

5 Cyclotomic polynomials The nth cyclotomic polynomial: Φ n () = ( ζ k ) where ζ = e 2πi/n 1 k<n (k,n)=1 This is an irreducible polynomial with integer coefficients. irreducibility due to Gauss nontrivial 4 / 35

6 Cyclotomic polynomials The nth cyclotomic polynomial: Φ n () = ( ζ k ) where ζ = e 2πi/n 1 k<n (k,n)=1 This is an irreducible polynomial with integer coefficients. irreducibility due to Gauss nontrivial [n] = n 1 1 = Φ d () 1<d n d n For primes: [p] = Φ p() 4 / 35

7 Cyclotomic polynomials The nth cyclotomic polynomial: Φ n () = ( ζ k ) where ζ = e 2πi/n 1 k<n (k,n)=1 This is an irreducible polynomial with integer coefficients. irreducibility due to Gauss nontrivial [n] = n 1 1 = Φ d () 1<d n d n For primes: [p] = Φ p() An irreducible monic integral polynomial is cyclotomic if and only if its Mahler measure max( α, 1) is 1. 4 / 35

8 Some cyclotomic polynomials exhibited Example Φ mn () = Φ n ( m ) if m n Φ 2n () = Φ n ( ) for odd n > 1 Φ 2 () = + 1 Φ n () is palindromic Φ 3 () = Φ 6 () = Φ 9 () = Φ 21 () = Φ 102 () = / 35

9 Some cyclotomic polynomials exhibited Example Φ mn () = Φ n ( m ) if m n Φ 2n () = Φ n ( ) for odd n > 1 Φ 2 () = + 1 Φ n () is palindromic Φ 3 () = Φ 6 () = Φ 9 () = Φ 21 () = Φ 105 () = / 35

10 Back to -binomials [n] = n 1 1 = 1<d n d n Φ d () ( ) n = [n] [n 1] [n k + 1] k [k] [k 1] [1] How often does Φ d () appear in this? It appears n d n k d k d times 6 / 35

11 Back to -binomials [n] = n 1 1 = 1<d n d n Φ d () ( ) n = [n] [n 1] [n k + 1] k [k] [k 1] [1] How often does Φ d () appear in this? n n k k It appears times d d d Obviously nonnegative: the -binomials are indeed polynomials Also ( ) at most one: suare-free n always contains Φ n () if 0 < k < n. k Good way to compute -binomials and even get them factorized for free 6 / 35

12 The coefficients of -binomial coefficients Here s some -binomials in expanded form: Example ( ) 6 = ( ) 9 = All coefficients are positive! What is the degree of the -binomial? 7 / 35

13 The coefficients of -binomial coefficients Here s some -binomials in expanded form: Example ( ) 6 = ( ) 9 = All coefficients are positive! What is the degree of the -binomial? It is (n k)k. 7 / 35

14 The coefficients of -binomial coefficients Here s some -binomials in expanded form: Example ( ) 6 = ( ) 9 = All coefficients are positive! What is the degree of the -binomial? It is (n k)k. In fact, the coefficients are unimodal. Sylvester, / 35

15 -binomials: Pascal s triangle Define the -binomials via the -Pascal rule: ( ) ( ) ( ) n n 1 n 1 = + k k k 1 k D2 8 / 35

16 -binomials: Pascal s triangle Define the -binomials via the -Pascal rule: ( ) ( ) ( ) n n 1 n 1 = + k k k 1 k D (1 + ) (1 + ) / 35

17 -binomials: Pascal s triangle Define the -binomials via the -Pascal rule: ( ) ( ) ( ) n n 1 n 1 = + k k k 1 k D (1 + ) (1 + ) Example ( ) 4 = ( ) = / 35

18 -binomials: combinatorial ( ) n = w(s) where w(s) = k S ( n j k) s j j w(s) = normalized sum of S D3 Example {1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4} ( ) 4 = / 35

19 -binomials: combinatorial ( ) n = w(s) where w(s) = k S ( n j k) s j j w(s) = normalized sum of S D3 Example {1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4} ( ) 4 = ( ) n The coefficient of m in counts the number of k k-element subsets of n whose normalized sum is m 9 / 35

20 -binomials: combinatorial ( ) n = w(s) where w(s) = k S ( n j k) s j j w(s) = normalized sum of S D3 Example {1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4} ( ) 4 = ( ) n The coefficient of m in counts the number of k k-element subsets of n whose normalized sum is m words made from k ones and n k twos which have m inversions 9 / 35

21 -binomials: combinatorial ( ) n = w(s) where w(s) = k S ( n j k) s j j w(s) = normalized sum of S D3 Example {1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4} ( ) 4 = ( ) n The coefficient of m in counts the number of k k-element subsets of n whose normalized sum is m words made from k ones and n k twos which have m inversions partitions λ of m whose Ferrer s diagram fits in a k (n k) box 9 / 35

22 -Chu-Vandermonde Different representations make different properties apparent! Chu-Vandermonde: ( ) m + n = k j ( )( m n ) j k j 10 / 35

23 -Chu-Vandermonde Different representations make different properties apparent! ( ) m + n Chu-Vandermonde: = ( )( ) m n k j k j j Purely from the combinatorial representation: ( ) m + n = S k(k+1)/2 k S ( m+n k ) 10 / 35

24 -Chu-Vandermonde Different representations make different properties apparent! ( ) m + n Chu-Vandermonde: = ( )( ) m n k j k j j Purely from the combinatorial representation: ( ) m + n = S k(k+1)/2 k S ( m+n k ) = j S 1 ( m j ) S 2 ( n k j) S 1 + S 2 +(k j)m k(k+1)/2 10 / 35

25 -Chu-Vandermonde Different representations make different properties apparent! ( ) m + n Chu-Vandermonde: = ( )( ) m n k j k j j Purely from the combinatorial representation: ( ) m + n = S k(k+1)/2 k S ( m+n k ) = j = j S 1 ( m j ) S 2 ( k j) n ( ) ( ) m n j k j S 1 + S 2 +(k j)m k(k+1)/2 (m j)(k j) 10 / 35

26 Automatic proving of -identities In[1]:= " docs math mathematica packages Zeil.m"; -Zeilberger Package by Axel Riese RISC Linz V 2.42 (02/18/05) In[2]:= Out[2]= Zeil Binomial m, j, Binomial n, k j, ^ m j k j, j, 0, m n, k, k m n SUM 1 k SUM k 1 k P. Paule and A. Riese A Mathematica -Analogue of Zeilberger s Algorithm Based on an Algebraically Motivated Approach to -Hypergeometric Telescoping Fields Inst. Commun., Vol. 14, / 35

27 Automatic proving of -identities In[1]:= " docs math mathematica packages Zeil.m"; -Zeilberger Package by Axel Riese RISC Linz V 2.42 (02/18/05) In[2]:= Out[2]= Zeil Binomial m, j, Binomial n, k j, ^ m j k j, j, 0, m n, k, k m n SUM 1 k SUM k 1 k Encoded implementation in Mathematica at risk of bit rot? last version of Zeil by Alex Riese from 2005 many examples don t work in MMA7 anymore... Sage as a solution? P. Paule and A. Riese A Mathematica -Analogue of Zeilberger s Algorithm Based on an Algebraically Motivated Approach to -Hypergeometric Telescoping Fields Inst. Commun., Vol. 14, / 35

28 -binomials: algebraic Let be a prime power. ( ) n k = number of k-dim. subspaces of F n D4 12 / 35

29 -binomials: algebraic Let be a prime power. ( ) n = number of k-dim. subspaces of F n k D4 Number of ways to choose k linearly independent vectors in F n : ( n 1)( n ) ( n k 1 ) 12 / 35

30 -binomials: algebraic Let be a prime power. ( ) n = number of k-dim. subspaces of F n k D4 Number of ways to choose k linearly independent vectors in F n : ( n 1)( n ) ( n k 1 ) Hence the number of k-dim. subspaces of F n is: ( n 1)( n ) ( n k 1 ) ( k 1)( k ) ( k k 1 ) = ( ) n k 12 / 35

31 -binomials: noncommuting variables Suppose yx = xy where commutes with x, y. Then: (x + y) n = n j=0 ( ) n x j y n j j D5 13 / 35

32 -binomials: noncommuting variables Suppose yx = xy where commutes with x, y. Then: (x + y) n = n j=0 ( ) n x j y n j j D5 Example ( ) 4 x 2 y 2 = xxyy + xyxy + xyyx + yxxy + yxyx + yyxx 2 = ( )x 2 y 2 13 / 35

33 -binomials: noncommuting variables Suppose yx = xy where commutes with x, y. Then: (x + y) n = n j=0 ( ) n x j y n j j D5 Example ( ) 4 x 2 y 2 = xxyy + xyxy + xyyx + yxxy + yxyx + yyxx 2 = ( )x 2 y 2 Let X f(x) = xf(x) and Q f(x) = f(x). Then: QX f(x) = xf(x) = XQ f(x) 13 / 35

34 -calculus It all starts with the -derivative: D f(x) = f(x) f(x) x x 14 / 35

35 -calculus It all starts with the -derivative: D f(x) = f(x) f(x) x x Example D x s = (x)s x s x x = s 1 1 xs 1 = [s] x s 1 14 / 35

36 -calculus It all starts with the -derivative: D f(x) = f(x) f(x) x x Example D x s = (x)s x s x x = s 1 1 xs 1 = [s] x s 1 Define e x = n=0 x n [n]! D e x = e x e x e y e x+y unless yx = xy e x e x 1/ = 1 14 / 35

37 -calculus It all starts with the -derivative: D f(x) = f(x) f(x) x x Example D x s = (x)s x s x x = s 1 1 xs 1 = [s] x s 1 Define e x = n=0 x n [n]! Homework: Define cos (x), sin (x),... and develop some -trigonometry. D e x = e x e x e y e x+y unless yx = xy e x e x 1/ = 1 14 / 35

38 -calculus: the -integral Formally inverting D F (x) = f(x) gives: F (x) = x 0 f(x)d x := (1 ) n xf( n x) n=0 15 / 35

39 -calculus: the -integral Formally inverting D F (x) = f(x) gives: F (x) = x 0 f(x)d x := (1 ) n xf( n x) n=0 Theorem (Fundamental theorem of -calculus) Let 0 < < 1. Then D F (x) = f(x). F (x) is the uniue such function continuous at 0 with F (0) = 0. Fineprint: one needs for instance that f(x)x α is bounded on some (0, a]. 15 / 35

40 -calculus: special functions Define the -gamma function as Γ (s) = 0 x s 1 e x 1/ d x Γ (s + 1) = [s] Γ (s) Γ (n + 1) = [n]! 16 / 35

41 -calculus: special functions Define the -gamma function as -beta function: B (t, s) = Γ (s) = x s 1 e x 1/ d x x t 1 (1 x) s 1 d x D6 Γ (s + 1) = [s] Γ (s) Γ (n + 1) = [n]! B (t, s) = Γ (t)γ (s) Γ (t + s) B (t, s) = B (s, t) Here, (x a) n is defined by: f(x) = (D n f)(a) (x a)n [n] n 0! Explicitly: (x a) n = (x a)(x a) (x n 1 a) 16 / 35

42 Summary: the -binomial coefficient The -binomial coefficient: ( ) n = k Via a -version of Pascal s rule [n]! [k]! [n k]! Combinatorially, as the generating function of the element sums of k-subsets of an n-set Algebraically, as the number of k-dimensional subspaces of F n Via a binomial theorem for noncommuting variables Analytically, via -integral representations We have not touched: uantum groups arising in representation theory and physics 17 / 35

43 Classical binomial congruences John Wilson (1773, Lagrange): (p 1)! 1 mod p Charles Babbage (1819): Joseph Wolstenholme (1862): James W.L. Glaisher (1900): Wilhelm Ljunggren (1952): ( ) 2p 1 1 p 1 mod p 2 ( ) 2p 1 1 p 1 mod p 3 ( ) mp 1 1 p 1 mod p 3 ( ) ( ) ap a bp b mod p 3 18 / 35

44 Wilson s congruence Wilson s congruence (Lagrange, 1773) (p 1)! 1 mod p known to Ibn al-haytham, ca AD This congruence holds if and only if p is a prime. Not great as a practical primality test though... The problem of distinguishing prime numbers from composite numbers... is known to be one of the most important and useful in arithmetic.... The dignity of the science itself seems to reuire that every possible means be explored for the solution of a problem so elegant and so celebrated. C. F. Gauss, Disuisitiones Arithmeticae, / 35

45 Babbage s congruence (n 1)! + 1 is divisible by n if and only if n is a prime number In attempting to discover some analogous expression which should be divisible by n 2, whenever n is a prime, but not divisible if n is a composite number... Charles Babbage is led to: Theorem (Babbage, 1819) For primes p 3: ( ) 2p 1 1 mod p 2 p 1 20 / 35

46 Babbage s congruence (n 1)! + 1 is divisible by n if and only if n is a prime number In attempting to discover some analogous expression which should be divisible by n 2, whenever n is a prime, but not divisible if n is a composite number... Charles Babbage is led to: Theorem (Babbage, 1819) For primes p 3: ( ) 2p 1 1 mod p 2 p 1 ( ) 2n 1 (n + 1)(n + 2) (2n 1) = n (n 1) 20 / 35

47 Babbage s congruence (n 1)! + 1 is divisible by n if and only if n is a prime number In attempting to discover some analogous expression which should be divisible by n 2, whenever n is a prime, but not divisible if n is a composite number... Charles Babbage is led to: Theorem (Babbage, 1819) For primes p 3: ( ) 2p 1 1 mod p 2 p 1 ( ) 2n 1 (n + 1)(n + 2) (2n 1) = n (n 1) Does not uite characterize primes! n = / 35

48 A simple combinatorial proof We have ( ) 2p = p k ( )( p p ) k p k Note that p divides ( ) p unless k = 0 or k = p. k 21 / 35

49 A simple combinatorial proof We have Note that p divides ( ) 2p = p k ( )( p p ) k p k mod p 2 ( ) p unless k = 0 or k = p. k 21 / 35

50 A simple combinatorial proof We have ( ) 2p = p k ( )( p p ) k p k mod p 2 ( ) p Note that p divides unless k = 0 or k = p. k ( ) 2p 1 = 1 ( ) 2p which is only trouble when p = 2 p 1 2 p 21 / 35

51 A -analog of Babbage s congruence Using -Chu-Vandermonde ( ) 2p = ( ) ( ) p p (p k)2 p k k p k Again, [p] divides p2 + 1 ( ) p unless k = 0 or k = p. k mod [p] 2 22 / 35

52 A -analog of Babbage s congruence Using -Chu-Vandermonde ( ) 2p = ( ) ( ) p p (p k)2 p k k p k Again, [p] divides p2 + 1 ( ) p unless k = 0 or k = p. k mod [p] 2 Theorem ( ) 2p [2] p p 2 mod [p]2 22 / 35

53 Extending the -analog Actually, the same argument shows: Theorem (W. Edwin Clark, 1995) ( ) ( ) ap a mod [p] 2 bp b p2 23 / 35

54 Extending the -analog Actually, the same argument shows: Theorem (W. Edwin Clark, 1995) ( ) ( ) ap a mod [p] 2 bp b p2 Sketch of the corresponding classical congruence: ( ) ap ( ) ( ) p p = bp k 1 k a k k a=bp ( ) a b mod p 2 We get a contribution whenever b of the a many k s are p. 23 / 35

55 Extending the -analog Actually, the same argument shows: Theorem (W. Edwin Clark, 1995) ( ) ( ) ap a mod [p] 2 bp b p2 No restriction on p the argument is combinatorial. Sketch of the corresponding classical congruence: ( ) ap ( ) ( ) p p = bp k 1 k a k k a=bp ( ) a b mod p 2 We get a contribution whenever b of the a many k s are p. 23 / 35

56 Extending the -analog Actually, the same argument shows: Theorem (W. Edwin Clark, 1995) ( ) ( ) ap a mod [p] 2 bp b p2 No restriction on p the argument is combinatorial. Similar Sketchresults of the by corresponding Andrews; e.g.: classical congruence: (( )) ap ( ) ( ) ap ap p = (a b)b(p 2) mod [p] 2 bp bp k 1 k b a k k a=bp p ( ) George Andrews a b -analogs of the binomial coefficient congruences of Babbage, Wolstenholme and Glaisher Discrete Mathematics 204, 1999 mod p 2 We get a contribution whenever b of the a many k s are p. 23 / 35

57 Wolstenholme and Ljunggren Amazingly, the congruences hold modulo p 3! Theorem (Wolstenholme, 1862) ( ) 2p 1 For primes p 5: 1 mod p 3 p 1... for several cases, in testing numerically a result of certain investigations, and after some trouble succeeded in proving it to hold universally / 35

58 Wolstenholme and Ljunggren Amazingly, the congruences hold modulo p 3! Theorem (Wolstenholme, 1862) ( ) 2p 1 For primes p 5: 1 mod p 3 p 1... for several cases, in testing numerically a result of certain investigations, and after some trouble succeeded in proving it to hold universally... Theorem (Ljunggren, 1952) ( ) ap For primes p 5: bp ( ) a b mod p 3 Note the restriction on p proofs are algebraic. 24 / 35

59 Proof of Wolstenholme s congruence ( ) 2p 1 (2p 1)(2p 2) (p + 1) = p (p 1) ) p 1 = ( 1) p 1 k=1 ( 1 2p k 25 / 35

60 Proof of Wolstenholme s congruence ( ) 2p 1 (2p 1)(2p 2) (p + 1) = p (p 1) ) p 1 = ( 1) p 1 k=1 1 2p 0<i<p ( 1 2p k 1 i + 4p2 0<i<j<p 1 ij mod p 3 25 / 35

61 Proof of Wolstenholme s congruence ( ) 2p 1 (2p 1)(2p 2) (p + 1) = p (p 1) ) p 1 = ( 1) p 1 k=1 1 2p 0<i<p = 1 2p 0<i<p ( 1 2p k 1 i + 4p2 1 i + 2p2 0<i<j<p 0<i<p 1 ij 1 i 2 mod p 3 2p 2 0<i<p 1 i 2 25 / 35

62 Proof of Wolstenholme s congruence ( ) 2p 1 (2p 1)(2p 2) (p + 1) = p (p 1) ) p 1 = ( 1) p 1 k=1 1 2p 0<i<p = 1 2p 0<i<p ( 1 2p k 1 i + 4p2 1 i + 2p2 0<i<j<p 0<i<p 1 ij 1 i 2 = = mod p 3 2p 2 p 1 k=1 p 1 k=1 0<i<p p + (p k) p k ( 1 + p ) k 1 i 2 25 / 35

63 Proof of Wolstenholme s congruence II Wolstenholme s congruence therefore follows from the fractional congruences p 1 i=1 p 1 i=1 1 i 0 mod p2, (1) 1 i 2 0 mod p 26 / 35

64 Proof of Wolstenholme s congruence II Wolstenholme s congruence therefore follows from the fractional congruences p 1 i=1 p 1 i=1 1 i 0 mod p2, (1) 1 i 2 0 mod p If n is not a multiple of p 1 then, using a primitive root g, i n (gi) n g n i n 0 mod p 0<i<p 0<i<p 0<i<p 26 / 35

65 Proof of Wolstenholme s congruence II Wolstenholme s congruence therefore follows from the fractional congruences p 1 i=1 p 1 i=1 1 i 0 mod p2, (1) 1 i 2 0 mod p If n is not a multiple of p 1 then, using a primitive root g, i n (gi) n g n i n 0 mod p 0<i<p 0<i<p 0<i<p Comparing the p 3 residues on the previous slide now shows (1). 26 / 35

66 Congruences for -harmonic numbers Theorem (Shi-Pan, 2007) p 1 i=1 p 1 i=1 1 p 1 [i] 2 ( 1) + p ( 1)2 [p] mod [p] 2 1 [i] 2 (p 1)(p 5) ( 1) 2 mod [p] / 35

67 Congruences for -harmonic numbers Theorem (Shi-Pan, 2007) p 1 i=1 p 1 i=1 1 p 1 [i] 2 ( 1) + p ( 1)2 [p] mod [p] 2 1 [i] 2 (p 1)(p 5) ( 1) 2 mod [p] 12 Example (p = 5) ( ) ( ) [i] 2 = i=1 ( + 1) 2 ( 2 + 1) 2 ( ) 2 27 / 35

68 Congruences for -harmonic numbers Theorem (Shi-Pan, 2007) p 1 i=1 p 1 i=1 1 p 1 [i] 2 ( 1) + p ( 1)2 [p] mod [p] 2 1 [i] 2 (p 1)(p 5) ( 1) 2 mod [p] 12 Example (p = 5) ( ) ( ) [i] 2 = i=1 ( + 1) 2 ( 2 + 1) 2 ( ) 2 Euivalent congruences can be given for This choice actually appears a bit more natural p 1 i=1 i [i] n 27 / 35

69 An exemplatory proof We wish to prove p 1 i=1 i [i] 2 p (1 )2 mod [p] Ling-Ling Shi and Hao Pan A -Analogue of Wolstenholme s Harmonic Series Congruence The American Mathematical Monthly, 144(6), / 35

70 An exemplatory proof We wish to prove Write: p 1 i=1 i [i] 2 p (1 )2 mod [p] p 1 i=1 i [i] 2 p 1 = (1 ) 2 i=1 i (1 i ) 2 } {{ } =:G() Ling-Ling Shi and Hao Pan A -Analogue of Wolstenholme s Harmonic Series Congruence The American Mathematical Monthly, 144(6), / 35

71 An exemplatory proof [p] = We wish to prove Write: p 1 p 1 i=1 ( ζ m ) m=1 i [i] 2 p (1 )2 mod [p] p 1 i=1 i [i] 2 p 1 = (1 ) 2 i=1 Hence we need to prove: G(ζ m ) = p i (1 i ) 2 } {{ } =:G() for m = 1, 2,..., p 1 Ling-Ling Shi and Hao Pan A -Analogue of Wolstenholme s Harmonic Series Congruence The American Mathematical Monthly, 144(6), / 35

72 An exemplatory proof [p] = We wish to prove Write: p 1 p 1 i=1 ( ζ m ) m=1 i=1 i [i] 2 p (1 )2 mod [p] p 1 i=1 i [i] 2 p 1 = (1 ) 2 i=1 i (1 i ) 2 } {{ } =:G() Hence we need to prove: G(ζ m ) = p2 1 for m = 1, 2,..., p 1 12 p 1 G(ζ m ζ mj p 1 ) = (1 ζ mj ) 2 = ζ j (1 ζ j ) 2 = G(ζ) i=1 Ling-Ling Shi and Hao Pan A -Analogue of Wolstenholme s Harmonic Series Congruence The American Mathematical Monthly, 144(6), / 35

73 An exemplatory proof II Define G(, z) = p 1 i=1 i (1 i z) 2 We need G(ζ, 1) = p / 35

74 An exemplatory proof II Define G(, z) = p 1 i=1 i (1 i z) 2 We need G(ζ, 1) = p p 1 G(ζ, z) = ζ i ζ ki (k + 1)z k i=1 k=0 29 / 35

75 An exemplatory proof II Define G(, z) = p 1 i=1 i (1 i z) 2 We need G(ζ, 1) = p p 1 G(ζ, z) = ζ i ζ ki (k + 1)z k = i=1 k=0 p 1 kz k 1 ζ ki k=1 i=1 29 / 35

76 An exemplatory proof II Define G(, z) = p 1 i=1 i (1 i z) 2 We need G(ζ, 1) = p p 1 G(ζ, z) = ζ i ζ ki (k + 1)z k = i=1 k=0 p 1 kz k 1 ζ ki k=1 k=1 i=1 = p pkz k 1 k=1 kz k 1 p 1 ζ ki = i=1 { p 1 1 if p k otherwise 29 / 35

77 An exemplatory proof II Define G(, z) = p 1 i=1 i (1 i z) 2 We need G(ζ, 1) = p p 1 G(ζ, z) = ζ i ζ ki (k + 1)z k = i=1 k=0 p 1 kz k 1 ζ ki k=1 k=1 i=1 = p pkz k 1 k=1 = p2 z p 1 (1 z p ) 2 1 (1 z) 2 kz k 1 p 1 ζ ki = i=1 { p 1 1 if p k otherwise 29 / 35

78 An exemplatory proof II Define G(, z) = p 1 i=1 i (1 i z) 2 We need G(ζ, 1) = p p 1 G(ζ, z) = ζ i ζ ki (k + 1)z k = i=1 k=0 p 1 kz k 1 ζ ki k=1 k=1 i=1 = p pkz k 1 k=1 kz k 1 p 1 ζ ki = i=1 { p 1 1 if p k otherwise = p2 z p 1 (1 z p ) 2 1 as z 1 (1 z) 2 p / 35

79 An exemplatory proof II Define G(, z) = p 1 i=1 i (1 i z) 2 We need G(ζ, 1) = p p 1 G(ζ, z) = ζ i ζ ki (k + 1)z k i=1 k=0 This is beautifully generalized in: p 1 = kz k 1 ζ ki Karl Dilcher Determinant expressions for -harmonic k=1 k=1 i=1 = p pkz k 1 k=1 kz k 1 p 1 ζ ki = i=1 { p 1 1 congruences and degenerate Bernoulli numbers Electronic Journal of Combinatorics 15, 2008 if p k otherwise = p2 z p 1 (1 z p ) 2 1 as z 1 (1 z) 2 p / 35

80 A -analog of Ljunggren s congruence Theorem (S, 2010) For primes p 5: ( ) ( ) ap a bp b p2 ( a b + 1 )( b ) p (p 1) 2 mod [p] 3 30 / 35

81 A -analog of Ljunggren s congruence Theorem (S, 2010) For primes p 5: ( ) ( ) ap a bp b p2 ( a b + 1 )( b ) p (p 1) 2 mod [p] 3 Example Choosing p = 13, a = 2, and b = 1, we have ( ) 26 = ( 13 1) 2 + ( ) 3 f() 13 where f() = is an irreducible polynomial with integer coefficients. 30 / 35

82 A -analog of Ljunggren s congruence Theorem (S, 2010) For primes p 5: ( ) ( ) ap a bp b p ( )( ) a b + 1 p 2 1 b (p 1) 2 mod [p] Example Choosing p = 13, a = 2, and b = 1, we have ( ) 26 = ( 13 1) 2 + ( ) 3 f() where f() = is an irreducible polynomial with integer coefficients / 35

83 Just coincidence? ( ) ap bp ( ) ( a a b b + 1 p2 )( b ) p (p 1) 2 mod [p] 3 Ernst Jacobsthal (1952) proved that Ljunggren s classical congruence holds modulo p 3+r where r is the p-adic valuation of ( ) ( )( ) a a b + 1 ab(a b) = 2a. b b It would be interesting to see if this generalization has a nice analog in the -world. 31 / 35

84 What happens for composite numbers? ( ) ap bp ( ) ( a a b b + 1 p2 )( b ) p (p 1) 2 mod [p] 3 Example (n = 12, a = 2, b = 1) ( ) 24 = (12 1) ( ) 3 f() }{{} Φ 12 () where f() = is an irreducible polynomial with integer coefficients. 32 / 35

85 What happens for composite numbers? ( ) ap bp ( ) ( a a b b + 1 p2 )( b ) p (p 1) 2 mod [p] 3 Example (n = 12, a = 2, b = 1) ( ) 24 = (12 1) ( ) 3 f() }{{} Φ 12 () where f() = is an irreducible polynomial with integer coefficients. Ljunggren s -congruence holds modulo Φ n () 3 if (n, 6) = 1 over integer coefficient polynomials! otherwise we get rational coefficients 32 / 35

86 Can we do better than modulo p 3? Are there primes p such that ( ) 2p 1 1 mod p 4? p 1 Such primes are called Wolstenholme primes. The only two known are and McIntosh, 1995: up to 10 9 C. Helou and G. Terjanian On Wolstenholme s theorem and its converse Journal of Number Theory 128, / 35

87 Can we do better than modulo p 3? Are there primes p such that ( ) 2p 1 1 mod p 4? p 1 Such primes are called Wolstenholme primes. The only two known are and McIntosh, 1995: up to 10 9 Infinitely many Wolstenholme primes are conjectured to exist. However, no primes are conjectured to exist for modulo p 5. C. Helou and G. Terjanian On Wolstenholme s theorem and its converse Journal of Number Theory 128, / 35

88 Can we do better than modulo p 3? Are there primes p such that ( ) 2p 1 1 mod p 4? p 1 Such primes are called Wolstenholme primes. The only two known are and McIntosh, 1995: up to 10 9 Infinitely many Wolstenholme primes are conjectured to exist. However, no primes are conjectured to exist for modulo p 5. Conjecturally, Wolstenholme s congruence characterizes primes: ( ) 2n 1 1 mod n 3 n is prime n 1 C. Helou and G. Terjanian On Wolstenholme s theorem and its converse Journal of Number Theory 128, / 35

89 Some open problems Extension to Jacobsthal s result? Extension to ( ) ap bp ( ) ] a [1 ab(a b) p3 b 3 B p 3 mod p 4, and insight into Wolstenholme primes? Is there a nice -analog for Gauss congruence? ( ) (p 1)/2 2a mod p (p 1)/4 where p = a 2 + b 2 and a 1 mod 4. Generalized to p 2 and p 3 by Chowla-Dwork-Evans (1986) and by Cosgrave-Dilcher (2010) 34 / 35

90 THANK YOU! Slides for this talk will be available from my website: Victor Kac and Pokman Cheung Quantum Calculus Springer, 2002 A -analog of Ljunggren s binomial congruence Proceedings of FPSAC, / 35