Further Number Theory G13FNT cw '11

3 Bernoulli numbers

De nition of the Bernoulli numbers Bm Motivation

We know that Pn 1 and Pn 2 1 ; but what is Pn m • k=1 k = 2 n(n+1) k=1 k = 6 n(n+1)(2n+1) k=1 k in general?

We know that P∞ −2 2 ; but what is P∞ −m in general? • n=1 n = π /6 n=1 n • Does xn + yn = zn have any solutions with x, y, z ∈ N and n ≥ 3? In all cases it turns that there is an answer which involves Bernoulli numbers. t De nition. F (t) = with F (0) = 1. et − 1 −1 t t t2 t3 , so setting makes sense. Expand as 1/F (t) = t (e − 1) = 1 + 2! + 3! + 4! + ... F (0) = 1 F (t) a Taylor series: ∞ t X tm F (t) = = B . et − 1 m m! m=0

De nition. The rational number Bm is called the mth Bernoulli Number.

F (t) is called the \exponential generating function" for the Bm.

m 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 1 1 1 0 1 0 1 0 1 0 5 0 691 0 7 0 Bm − 2 6 − 30 42 − 30 66 − 2730 6 m 16 17 18 19 20 21 22 23 24 25 26 3617 0 43867 0 174611 0 854513 0 236364091 0 8553103 Bm − 510 798 − 330 138 − 2730 6

In the problem sheet, we will show that ; hence 1 , and for F (t) − F (−t) = −t B1 = − 2 Bm = 0 all odd m > 1. m−1 X m Proposition 3.1. B = 0 for all m 2. k k > k=0

Proof. Compare the coecients on both sides of the identity t = (et − 1)F (t).

m−1 −1 X m + 1 Corollary 3.2. For all m 1, B = B . > m m + 1 k k k=0

The sum of m-th powers of consequtive integers We wish to nd a general formula for 1m + 2m + 3m + ··· + (n − 1)m.

1 ; • 1 + 2 + 3 + ··· + (n − 1) = 2 n(n − 1) 2 2 2 2 1 ; • 1 + 2 + 3 + ··· + (n − 1) = 6 n(n − 1)(2n − 1) • 13 + 23 + 33 + ··· + (n − 1)3 =?

Notation. For let m m m m. m ∈ N Sm(n) = 1 + 2 + 3 + ··· + (n − 1) Further Number Theory G13FNT cw '11

Our aim is to compute Sm(n), using Bernoulli numbers. m X m nk+1 Theorem 3.3. For all m ∈ , S (n) = B . N m k m−k k + 1 k=0

For example, 1 4 3 2 1
2. S3(n) = 4 (n − 2n + n ) = 2 n(n − 1) Proof. Evaluate Pn−1 at in two ways. Using at P∞ m tm gives A = a=0 e e = m=0 a m!

∞ X tm A = S (n). m! m m=0 Summing A as a geometric series gives

ent − 1 ent − 1 t ent − 1 A = = = F (t), et − 1 t et − 1 t which leads to ∞ !  ∞  X tk X tj A = nk+1 B . (k + 1)!  j j!  k=0 j=0 Comparing coecients gives the result.

Lemma 3.4. If p − 1 divides m, then Sm(p) ≡ −1 (mod p), otherwise Sm(p) ≡ 0 (mod p).

Proof. Let g be a primitive element modulo p. Then

p−2 m m X k m m·(p−1) (g − 1) · Sm(p) ≡ (g − 1) · (g ) = g − 1 ≡ 0 (mod p). k=0 If , then m and so . Otherwise (p − 1) - m g 6≡ 1 Sm(p) ≡ 0 (mod p) Sm(p) ≡ p − 1 ≡ −1 (mod p).

Riemann's zeta-function

De nition. The Riemann zeta-function is de ned by

∞ X 1 1 1 1 ζ(s) = = 1 + + + + ... ns 2s 3s 4s n=1

Remark. It diverges when s = 1, but we will see later that ζ(s) converges for s > 1. Here we are only interested in ζ(s) for integer values of s: we will give a formula valid for all positive even integers.

Theorem 3.5 (Euler). For all m ∈ N, (2π)2m ζ(2m) = (−1)m+1 B . 2 (2m)! 2m

Proof. See separate non-examinable handout. Further Number Theory G13FNT cw '11

Examples: Taking and using 1 gives 2 . Taking and using m = 1 B2 = 6 ζ(2) = π /6 m = 2 1 gives 4 . B4 = − 30 ζ(4) = π /90 Corollary 3.6. (i). For all , the sign of is m+1. m ∈ N B2m (−1) √ (ii). The sequence grows like 2(2m)! m
2m. |B2m| (2π)2m ∼ 4 πm · eπ (iii). For all even we have . m > 18 |Bm| > m

Proof. since it is the sum of a series of positive terms, so m+1 2 (2m)! ζ(2m) > 0 (−1) B2m = (2π)2m ζ(2m) > . Since , we have 2(2m)! . The second expression follows from Stirling's 0 ζ(2m) > 1 |B2m| > (2π)2m formula. Now for m = 9, we get B18/18 > 3 and the function B2m/2m increases quickly in m.

Congruences for Bernoulli Numbers The Theorems stated in this lecture give information about the numerators and denominat- ors of the (rational) Bernoulli numbers Bm. The rst one tells us exactly what the denomin- ator is. Notation. For set primes such that . m ∈ N ∆m = { p (p − 1) | m}

Theorem 3.7 (Clausen & von Staudt). For all even m ∈ N, X 1 C = B + ∈ . m m q Z q∈∆m

In particular, the denominator of B is precisely Q q. m q∈∆m Example: For we have , so 1 1 1 61 . In fact, m = 50 ∆50 = {2, 3, 11} B50 + 2 + 3 + 11 = B50 + 66 ∈ Z 61 . B50 + 66 = 7500866746076964366855721

Proof. Let p be any prime, the aim is to show that the denominator of Cm is coprime to p. The theorem can be proven by induction on even m. First, m = 2 is easy. Let m > 2 be even. Then Bm−1 = 0. The formula for Sm(p) can be written as

m X m pk S (p) = B · p + p B (1) m m k m−k k + 1 k=2

By induction, pBm−k has no p in the denominator. In the problem sheet, we prove that, if k k > 2, then the numerator of p /(k + 1) is divisible by p. Hence the sum is a rational number whose numerator is divisible by p. If p 6∈ ∆m, then Sm(p) ≡ 0 (mod p) by lemma 3.4. So Bm, and hence Cm, have no p in the denominator. If p ∈ ∆m, then Sm(p) + 1 is divisible by p. So 1 , and hence , have no in the denominator. Bm + p Cm p

Finally, a congruence which tells us something about the numerator of Bm. Theorem 3.8 (The Kummer Congruences). Let be even and . Then m ∈ N p∈ / ∆m B B m ≡ n (mod p − 1) =⇒ m ≡ n (mod p). m n by which we mean that the numerator of Bm Bn is divisible by . m − n p Further Number Theory G13FNT cw '11

More generally, m ≡ n (mod (p − 1)pk) implies

B B (1 − pm−1) m ≡ (1 − pn−1) n (mod pk). m n Examples: −5 is divisible by as . So is −5 B6/6−B2/2 = 63 p = 5 6 ≡ 2 (mod p−1) B10/10−B2/2 = 66 or 21335 . B18/18 − B2/2 = 7182

Regular and irregular primes and Fermat's Last Theorem

De nition. The odd prime number p is called regular if p does not divide the numerator of for all even . Otherwise is irregular. Bm m 6 p − 3 p Examples: are all regular, but is irregular: 7709321041217 and p = 3, 5, 7,..., 31 37 B32 = − 510 7709321041217 = 37 · 683 · 305065927. The signi cance of regularity comes from the following application.

Theorem 3.9 (Kummer, 1850). Let p be an odd regular prime. Then the equation

xp + yp = zp has no solution in positive integers.

Proof. Omitted (uses Algebraic Number Theory). Question: How many regular primes are there? Answer: No-one knows, but computer calculations suggest that 61% of primes are regular and 39% are irregular. What we can prove is this: Theorem 3.10. There are in nitely many irregular primes.

Proof. Take the complete list of all irregular primes . Consider Q . p1, p2, . . . , pr N = 2 i(pi − 1) By Corollary 3.6 we have because . So there exists a prime which divides |BN | > N N > 18 p the numerator of BN /N. We will show that p is irregular and not in the set {p1, p2, . . . , pr}. By Theorem 3.7, p∈ / ∆N since p divides the numerator but not the denominator of BN . So . Hence and for . (p − 1) - N p 6= 2 p 6= pi 1 6 i 6 r Take n with 0 6 n < p − 1 and n ≡ N (mod p − 1); then n is even and n > 0 since (p − 1) - N, so 2 6 n 6 p − 3. By Theorem 3.8 we have B B n ≡ N ≡ 0 (mod p), n N so p divides the numerator of Bn; hence p is irregular.

n n n Theorem 3.11 (Wiles-Taylor-. . . ). If n > 3, then x + y = z has no solution in N.