congruences Shanghai, July 2013 Simple proofs of Ramanujan s partition congruences Michael D. Hirschhorn

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1 s s m.hirschhorn@unsw.edu.au

2 Let p(n) be the number of s of n. For example, p(4) = 5, since we can write 4 = 4 = 3+1 = 2+2 = = and there are 5 such representations. It was known to Euler that p(n)q n = 1 E(q) s

3 where E(q) = (1 q)(1 q 2 )(1 q 3 ), and by convention, p(0) = 1. Euler also knew that E(q) = 1 q q 2 +q 5 +q 7 q 12 q 15 +q 22 +q = ( 1) n q (3n2 +n)/2. It follows that (1 q q 2 +q 5 +q 7 q 12 q 15 +q 22 +q ) p(n)q n = 1, s from which it follows that for n > 0, p(n) = p(n 1)+p(n 2) p(n 5) p(n 7) +p(n 12)+p(n 15)++.

4 Thus, p(1) = p(0) = 1, p(2) = p(1)+p(0) = 2, p(3) = p(2)+p(1) = 3, p(4) = p(3)+p(2) = 5, and so on. P. MacMahon, in 1917, calculated p(n) for n up to 200, and presented his results thus: Look at the bottom row. Ramanujan did, and made the exciting discovery that p(5n +4) 0 (mod 5). s

5 He also discovered that p(7n+5) 0 (mod 7) and p(11n+6) 0 (mod 11). And these were just the simplest of his conjectures. (There is nothing quite so simple for other primes.) Ramanujan proved these three s, but his proof of the mod 11 is much deeper than his proofs of the mod 5 and mod 7 s. The purpose of this talk is to give simple proofs of all three s. The proofs of the mod 5 and mod 7 s are simpler versions of proofs, and the proof of the mod 11 is new. s

6 2 Two proofs of the mod 5 We start with two well known identities, proofs of which can be found in Hardy & Wright, to the Theory of Numbers, for example. Euler s identity s E = (1 q)(1 q 2 )(1 q 3 ) = 1 q q 2 +q 5 +q 7 q 12 q 15 +q 22 +q = ( 1) n q (3n2 +n)/2 and Jacobi s identity J = E 3 = 1 3q +5q 3 7q 6 +9q 10 11q q 21 15q q 36 19q 45 + = ( 1) n (2n +1)q (n2 +n)/2.

7 First we notice that the exponents in the series for E, namely 0, 1, 2, 5, 7, 12, 15 and so on are all congruent to 0, 1 or 2 modulo 5. So we can write E E 0 +E 1 +E 2, where for i = 0, 1, 2, E i comprises those terms in E in which the exponent is congruent to i (mod 5). Similarly, we notice that, modulo 5, J J 0 +J 1, where J i comprises terms in which the exponent is congruent to i (mod 5), since (n 2 +n)/2 0,1 or 3 (mod 5), but 2n+1 0 (mod 5) precisely when (n 2 +n)/2 3 (mod 5). s

8 Also, it is easy to check that so (1 q) 5 1 q 5 (mod 5) E 5 = (1 q) 5 (1 q 2 ) 5 (1 q 3 ) 5 (1 q 5 )(1 q 10 )(1 q 15 ) = E(q 5 ). So now we have, modulo 5, p(n)q n = 1 E = E4 E 5 E4 E(q 5 ) EJ E(q 5 ) (E 0 +E 1 +E 2 )(J 0 +J 1 ) E(q 5 ) E 0J 0 +E 0 J 1 +E 1 J 0 +E 1 J 1 +E 2 J 0 +E 2 J 1 E(q 5, ) s

9 and if we look for terms in which the exponent is congruent to 4 (mod 5), we find p(5n+4)q 5n+4 0 and the mod 5 is proved! (This is a simplified version of proof.) s

10 An alternative proof using only J goes as follows. Modulo 5, so p(n)q n = 1 E = E9 E 10 = (E3 ) 3 (E 5 ) 2 J3 E(q 5 ) 2 (J 0 +J 1 ) 3 E(q 5 ) 2 J3 0 +3J2 0 J 1 +3J 0 J 2 1 +J3 1 E(q 5 ) 2, p(5n+4)q 5n+4 0. s

11 3 Proof of the mod 7 We have E = E 0 +E 1 +E 2 +E 5, where E i comprises those terms of E in which the exponent is congruent to i (mod 7), and, modulo 7, J J 0 +J 1 +J 3, where J i comprises those terms of J in which the exponent is congruent to i (mod 7). Note that 2n+1 0 (mod 7) whenever (n 2 +n)/2 6 (mod 7). s

12 It follows that, modulo 7, p(n)q n = 1 E = E6 E 7 = (E3 ) 2 E 7 = J2 E 7 (J 0 +J 1 +J 3 ) 2 E(q 7 ) = J2 0 +2J 0J 1 +J J 0J 3 +2J 1 J 3 +J 2 3 E(q 7 ) So if we look for those terms in which the exponent is 5 (mod 7), we find p(7n+5)q 7n+5 0., s

13 4 Proof of the mod 11 We have and, modulo 11, E = E 0 +E 1 +E 2 +E 5 +E 7 +E 15 J J 0 +J 1 +J 3 +J 6 +J 10, since 2n+1 0 (mod 11) whenever (n 2 +n)/2 4 (mod 11). So we have p(n)q n = 1 E = E21 E 22 = (E3 ) 7 (E 11 ) 2 = J7 (E 11 ) 2 J 7 E(q 11 ) 2 (J 0 +J 1 +J 3 +J 6 +J 10 ) 7 E(q 11 ) 2. s

14 If we now extract those terms in which the exponent is congruent to 6 (mod 11), we find p(11n +6)q 11n+6 P E(q 11 ) 2, where P is a polynomial in J 0, J 1, J 3, J 6 and J 10 of degree 7. s

15 Indeed, modulo 11, P 7J0 6 J 6 +10J0 5 J2 3 +J4 0 J 1J 6 J 10 +8J0 3 J3 1 J 3 +2J0 3 J 1J3 2 J 10 +8J0J 3 6J J0J 2 1J 2 3 J6 2 +3J0J 2 1J 2 6 J J0J 2 3J 2 6J J0J 2 3 J 6 J J)0 2 J J 0 J1 6 +J 0 J1J 4 3 J 10 +2J 0 J1J 2 3J J 0 J1J 2 3J J 0 J 1 J 3 J6 4 +2J 0 J 1 J6J J 0 J3J 4 6 J 10 +8J 0 J3 3 J J5 1 J2 6 +2J3 1 J 3J6 2 J 10 +8J1 3 J 6J J2 1 J5 3 +8J 1 J3 3 J3 6 +3J 1J3 2 J2 6 J2 10 +J 1J 3 J 6 J J 1J J6 3 J 10 +7J 3 J J6J We want to show that P 0. s

16 The crucial idea We have J 4 = E 12 = E 11 E E(q 11 )(E 0 +E 1 +E 2 +E 5 +E 7 +E 15 ). That is, (J 0 +J 1 +J 3 +J 6 +J 10 ) 4 E(q 11 )(E 0 +E 1 +E 2 +E 5 +E 7 +E 15 ). If we extract those terms in which the exponents are 3, 6, 8, 9 and 10 (mod 11), we find J 3 0 J 3 +J 0 J J 0J 1 J 3 J 6 +7J 2 1 J2 6 +3J 3J 2 6 J 10 +J 6 J , J 3 0 J 6 +7J 2 0 J2 3 +6J 0J 1 J 6 J 10 +J 3 1 J 3 +3J 1 J 2 3 J 10 +J 3 6 J 10 0, 3J 0 J 2 1J 6 +6J 0 J 3 J 6 J 10 +J 0 J J 2 1J 2 3 +J 1 J 3 6 +J 3 3J 10 0, 3J 2 0J 3 J 6 +7J 2 0J J 0 J J 1 J 3 J 6 J 10 +J 3 1J 6 +J 1 J , J 3 0J 10 +6J 0 J 1 J 3 J 6 +3J 0 J 1 J J 1 J 3 3 +J 3 J J 2 6J s

17 Now we make the observation that P (7J 1 J 3 J 10 +5J0 2 J 3 +7J1 3 ) (J0J 3 3 +J 0 J1 3 +6J 0 J 1 J 3 J 6 +7J1J J 3 J6J J 6 J10) 3 +(7J0 3 +7J 0 J 1 J 10 +5J6J 2 10 ) (J0J J0J J 0 J 1 J 6 J 10 +J1J J 1 J3J J6J 3 10 ) +(7J 0 J 3 J 6 +5J 0 J J3 3 ) (3J 0 J1 2 J 6 +6J 0 J 3 J 6 J 10 +J 0 J J2 1 J2 3 +J 1J6 3 +J3 3 J 10) +(5J1J J 3 J 6 J 10 +7J10) 3 (3J0J 2 3 J 6 +7J0J J 0 J3 3 +6J 1 J 3 J 6 J 10 +J1J 3 6 +J 1 J10) 3 +(7J 0 J 1 J 6 +5J 1 J3 2 +7J6) 3 (J0 3 J 10 +6J 0 J 1 J 3 J 6 +3J 0 J 1 J10 2 +J 1J3 3 +J 3J6 3 +7J2 6 J2 10 ) s and so P 0.