Lehmer’s Conjecture on the Non-vanishing of Ramanujan’s Tau Function

Will Y. Lee

Abstract In this paper we prove Lehmer’s conjecture on Ramanujan’s tau function, namely τ(n) ≠ 0 for each n ≥ 1 by investigating the additive group structure attached to τ(n) with the aid of unique factorization theorem.

Note added Sept. 11, 2014 by the uploader, Doron Zeilberger: I was told that the present proof is incomplete. Will Lee informed me that he is currently working on fixing it.

1 Let E_{k} (k = 2, 4, …) be the normalized Eisenstein series ([4 : 108 − 122]) given by

E_{k} = 1 − \frac{2k}{B_{k}} ∑_{n=1}^{∞} σ_{k−1}(n)q^{n}  (1)

where q := e^{i2πz} (ℑ(z) > 0), B_{k} the Bernoulli number defined by

\frac{x}{e^{x} − 1} = ∑_{k=0}^{∞} B_{k} \frac{x^{k}}{k!}

and σ_{k−1}(n) the divisor function:

σ_{k−1}(n) := ∑_{d|n} d^{k−1}.

For an elliptic curve given by

y^{2} = 4x^{3} − g_{2}(z)x − g_{3}(z)  (2)

12000 Mathematics Subject Classification. Primary 11L40; Secondary 11Y XX

where g_{2}(z) = 120ζ(4)E_{4}(z), g_{3}(z) = 280ζ(6)E_{6}(z) and E_{k}(z) given by equation (1) and ζ(k) is Riemann zeta function:

ζ(k) := ∑_{n=1}^{∞} \frac{1}{n^{k}} .

A simple calculation ([1 : 14], [4 : 112]) shows the discriminant Δ(z) := 4^{4}(x_{1} − x_{2})^{2}(x_{2} − x_{3})^{2}(x_{3} − x_{1})^{2}, where x_{1}, x_{2} and x_{3} are the roots the right side of equation (2), is given by

Δ(z) = g_{2}(z)^{3} − 27g_{3}(z)^{2} = \frac{(2π)^{12}}{1728} (E_{4}(z)^{3} − E_{6}(z)^{2}).  (3)

On the other hand Jacobi’s theorem ([4 : 122]) asserts that

(2π)^{−12}Δ(z) = q ∏_{n=1}^{∞} (1 − q^{n})^{24}.  (4)

From equation (4), Ramanujan has defined his tau function τ(n) ([1], [2], [3], [4 : 122], [5]− [7]) by

q ∏_{n=1}^{∞} (1 − q^{n})^{24} := ∑_{n=1}^{∞} τ(n)q^{n}.  (5)

Notice that each τ(n) (n ≥ 1) has an integer value. In a series of papers ([5] − [7]), D.H. Lehmer investigated the properties of τ(n) for n ≤ 300, proved that τ(n) ≠ 0 for n < 3316799, later for n < 214928639999 ([1 : 22]). He also showed that if τ(n) = 0 then n must be a prime. He then conjectured, what is nowadays known as Lehmer’s conjecture ([6]) that

τ(n) ≠ 0 for each n ≥ 1.  (6)

A simple calculation ([3 : 21 − 22], [4 : 122 − 123]) shows

τ(n) = \frac{65}{756}σ_{11}(n) + \frac{691}{756}σ_{5}(n) − \frac{691}{3} ∑_{j=1}^{n−1} σ_{5}(j)σ_{5}(n − j).  (7)

Since Lehmer’s conjecture is equivalent to 3τ(n) ≠ 0 for each n ≥ 1, we write

A(n) := \frac{65}{252}σ_{11}(n) + \frac{691}{252}σ_{5}(n); B(n) := 691 ∑_{j=1}^{n−1} σ_{5}(j)σ_{5}(n − j).  (8)

Then 3τ(n) = A(n)−B(n). Observe that A(n) takes on integer value for each n ≥ 1 since both τ(n) and B(n) do. Now Lehmer’s conjecture is, in view of equations (7), (8) and the unique factorization theorem, equivalent to:

A(n) ≠ B(n) for each n ≥ 1.  (9)

Recent calculation by Bosman confirms Lehmer’s conjecture for n ≤ 22798241520242687999. In this paper we prove equation (9) by showing that {∑_{k=0}^{q−1} ka_{i,k} mod q}_{k=0}^{q−1} forms an additive group of order q modulo q for q | A(p), q > p, p ≡ −1 mod 691, [a_{i,k}]_{0≤i,k≤q−1} q × q-matrix, with the aid of the unique factorization theorem, the pigeonhole principle and the remainder theorem. We prove equation (9) first for prime p then for p^{α}, α ≥ 2 and finally for any composite number n. Since 11 ∤ 690 and since (p + 1) | (p^{11} + 1), the following Lemma 1 evidently holds.

Lemma 1 Let A(p) be given by equation (8). Then the following two conditions (10) and (11) are equivalent:

A(p) ≡ 0 mod 691.  (10)
p ≡ −1 mod 691.  (11)

If 691 ∤ A(p) or equivalently p does not satisfy equation (11) then we trivially have A(p) ≠ B(p) by equation (8). It suffices therefore to prove Lehmer’s conjecture for prime p satisfying equation (10) or (11). In what follows, prime p satisfies either equation (10) or (11). We first prove:

Lemma 2 Let p satisfy equation (10) or (11). Then A(p) has at least one prime factor q greater than p.

Proof. Write A(p) from equation (8) as

\begin{aligned}A(p) & = \frac{65}{252}(1 + p^{11}) + \frac{691}{252}(1 + p^{5}) \\ & = 3 + K_{5}p^{5} \end{aligned}  (12)

where K_{5} := \frac{691}{252} + \frac{65}{252}p^{6} is an integer with p^{5} < K_{5} < p^{6} since A(p) is an integer with p^{10} < A(p) < p^{11}. Suppose A(p) has no prime factor greater than p. A(p) then is written via the unique factorization theorem as

A(p) = 2^{e_{0}}q_{1}^{e_{1}} … q_{m}^{e_{m}}, q_{i} < p, e_{i} ≥ 1 (1 ≤ i ≤ m).  (13)

Notice that A(p) has an even factor 2^{e_{0}} by substituting equation (11) into equation (8). Write x = [x] + {x} where [x] represents the integral part of x and {x} the nonintegral part of x. Since p^{10} < A(p) < p^{11}, the representation for A(p) in the base p is uniquely given from equation (13) by

A(p) = ∑_{i=0}^{10} b_{i}p^{i}, b_{i} := [p\{\frac{A(p)}{p^{i+1}}\}] (0 ≤ i ≤ 10).  (14)

We show that each b_{i} ≠ 0 (0 ≤ i ≤ 10). Indeed we prove

b_{i} ≥ 5 (1 ≤ i ≤ 10), b_{0} ≠ 0.  (15)

The fact that b_{0} ≠ 0 follows from p ∤ A(p). Likewise b_{10} ≥ 5 follows from p ∤ A(p) and b_{10} ≥ [\frac{65}{252}p] ≥ [\frac{p}{252}] ≥ 5 in view of equations (14) and (17), where p ≥ 1381. ¿From equation (8) or (12), we readily have

\frac{65}{252}p^{11} < A(p) < (\frac{65}{252} + 3p^{−6})p^{11}.  (16)

Equation (16) is equivalent to

\frac{65}{252}p^{10−i} < \frac{A(p)}{p^{i+1}} < (\frac{65}{252} + 3p^{−6})p^{10−i} (1 ≤ i ≤ 9).  (17)

Write

65p^{10−i} = 252Q_{i} + R_{i}, Q_{i} ≥ 1, 1 ≤ R_{i} ≤ 251 (1 ≤ i ≤ 9).  (18)

Substitution of equation (18) into equation (17) reveals

Q_{i} + \frac{R_{i}}{252} < \frac{A(p)}{p^{i+1}} < (Q_{i} + \frac{R_{i}}{252} + 3p^{4−i}) (1 ≤ i ≤ 9).  (19)

Inequality (19) implies \{\frac{A(p)}{p^{i+1}}\} ≥ \frac{R_{i}}{252} ≥ \frac{1}{252} (1 ≤ i ≤ 9) and hence we have since p ≥ 1381

b_{i} = [p\{\frac{A(p)}{p^{i+1}}\}] ≥ [\frac{p}{252}] ≥ 5 (1 ≤ i ≤ 9).  (20)

This establishes inequality (15). Rewrite equation (14) as

A(p) = L_{5}p^{5} + ∑_{i=0}^{4} b_{i}p^{i}  (21)

where L_{5} := ∑_{i=0}^{5} b_{5+i}p^{i}. Subtraction of equation (21) from equation (12) with rearrangement of terms leads us to

\begin{aligned}p^{5} & ≤ (K_{5} − L_{5})p^{5} \\ & = (b_{0} − 3) + ∑_{i=1}^{4} b_{i}p^{i} \\ & < (p − 1) ∑_{i=0}^{4} p^{i} \\ & = p^{5} − 1. \end{aligned}  (22)

Since each b_{i} ≥ 1 (0 ≤ i ≤ 10) from equation (15), K_{5} − L_{5} ≥ 1 follows from the the second line equality of equation (22) regardless of the value of b_{0} ≠ 0. Since 1 ≤ b_{i} ≤ p − 1 (0 ≤ i ≤ 10) from equation (15) and since b_{0} − 3 < p − 1, the third line inequality follows. Inequality (22) is absurd. Consequently the assumption that A(p) has no prime factor > p is false. This establishes Lemma 2.

It is easy to check our proof for Lemma 2 works for all primes p > 252 with inequality (15) replaced by 1 ≤ b_{i} (0 ≤ i ≤ 10). A simple computation reveals Lemma 2 also holds for primes p ≤ 252. Consequently Lemma 2 holds for all primes p. Thus the assumption that the prime p generated by equation (11) in Lemma 2 is redundant. Let q be an odd prime prime factor of A(p) greater than p. Existence of such a prime q is guaranteed by Lemma 2. Construct matrix [a_{i,k}]_{0≤i,k≤q−1} as follows:

a_{i,k} := p−1 ; ∑ ; \begin{matrix}j = 1 \\ i691σ_{5}(j)σ_{5}(p − j) ≡ k mod q \end{matrix} 1.  (23)

Since σ_{5}(j)σ_{5}(p − j) = σ_{5}(p − j)σ_{5}(p − (p − j)), we have from equation (23)

a_{i,k} = 2 (p−1)/2 ; ∑ ; \begin{matrix}j = 1 \\ i691σ_{5}(j)σ_{5}(p − j) ≡ k mod q \end{matrix} 1.  (24)

Then the matrix [a_{i,k}]_{0≤i, k≤q−1} has the following properties:

a_{i,k} ≡ 0 mod 2  (0 ≤ i, k ≤ q − 1).  (25)
a_{0,0} = p − 1, a_{0,k} = 0  (1 ≤ k ≤ q − 1).  (26)
a_{i,0} = a_{j,0}  (1 ≤ i ≠ j ≤ q − 1).  (27)
a_{i,k} = a_{q−i,q−k}  (1 ≤ i, k ≤ q − 1).  (28)
i691 ∑_{j=1}^{p−1} σ_{5}(j)σ_{5}(p − j) ≡ ∑_{k=1}^{q−1} ka_{i,k} mod q (1 ≤ i ≤ q − 1).  (29)
∑_{k=1}^{q−1} ka_{i,k} ≡ i ∑_{k=1}^{q−1} ka_{1,k} mod q (1 ≤ i ≤ q − 1).  (30)

Notice that given a_{1,k} (1 ≤ k ≤ q − 1), a_{i,k} (2 ≤ i ≤ q − 1, 1 ≤ k ≤ q − 1) are reshuffles of a_{1,k}(1 ≤ k ≤ q − 1) and vice versa determined by

a_{i,k} = a_{1,i^{−1}k mod q} ⟺ a_{1,k} = a_{i,ik mod q} (1 ≤ i, k ≤ q − 1).  (31)

For each i = 1, 2, …, q − 1, write f_{j_{i}} := i691σ_{5}(j)σ_{5}(p − j) mod q. Then f_{j_{i}} = f_{p−j_{i}} (1 ≤ i ≤ q − 1) from equation (23). Define for each i = 1, 2, …, q − 1:

\begin{aligned}S_{i,l} & := & \{k : a_{i,k} = 2l\} (1 ≤ k ≤ q, 1 ≤ l ≤ q_{0}) \\ ⟺ & = & \{(j_{i_{1}}, j_{i_{2}}, …, j_{i_{l}}) : f_{j_{i_{1}}} = f_{j_{i_{2}}} = ⋯ = f_{j_{i_{l}}} = k\} (1 ≤ j_{i_{1}} < j_{i_{2}} < ⋯ < j_{i_{l}} ≤ \frac{(p−1)}{2}) \\ S_{i,l} & = & ∅ (1 ≤ l ≤ q_{0}) for l > q_{0}. \end{aligned}

(32)

Since q > p and since a_{i,k} (0 ≤ i, k ≤ q − 1) cannot be too large even number from equations (23) and (24), a positive integer q_{0} < q − 1 exists, depending on p and q, satisfying the last line of equation (32). It is clear from equation (32) with the aid of equation (31) that for each l = 1, 2, …, q_{0} :

S_{i,l} = S_{j,l} (1 ≤ i < j ≤ q − 1).  (33)

For each q | A(p) with q > p, we then have from equations (31) − (33) that

∑_{l=0}^{q_{0}} | S_{i,l} | = q (1 ≤ i ≤ q − 1).  (34)
∑_{k=1}^{q−1} a_{i,k} = ∑_{l=1}^{q_{0}} 2l | S_{i,l} | = p − 1 (1 ≤ i ≤ q − 1).  (35)

Equation (35) reads when q | A(p) with q < p that:

∑_{k=0}^{q−1} a_{i,k} = p − 1 (1 ≤ i ≤ q − 1).  (36)

Equations (25) − (31) readily follow from equations (23) and (24). Equation (29) is a restatement of the remainder theorem in view of equations (23), (32) and (35). Equations (34), (35) and (36) follow from the pigeonhole principle. Since we exclusively use S_{1,l} (1 ≤ l ≤ q_{0}) in what follows, we show the following inequality:

| S_{1,l−1} | > l | S_{1,l} | (2 ≤ l ≤ q_{0}).  (37)

To prove inequality (37) we use the second line of equation (32) for the definition of S_{1,l}. Consider the map β : S_{1,l} ↦ S_{1,l−1} × S_{1,l−1} × ⋯ × S_{1,l−1} given by

\begin{aligned}β(j_{1}, j_{2}, …, j_{l}) := & ((β_{1}(j_{1}), β_{2}(j_{1}), …, β_{l−1}(j_{1})), (β_{1}(j_{2}), β_{2}(j_{2}), …, β_{l−1}(j_{2})), …, \\ & (β_{1}(j_{l}), β_{2}(j_{l}), …, β_{l−1}(j_{l}))) \end{aligned}

(38)

such that for each i = 1, 2, …, l:

\begin{aligned}β_{1}(j_{i}) & := min_{j_{i_{k}}}\{| j_{i} − j_{i_{k}} |: a_{1,j_{i_{k}}} = 2(l − 1) \\ f_{β_{1}(j_{i})} & = f_{β_{2(j_{i})}} = ⋯ = f_{β_{l−1}(j_{i})}. \end{aligned}  (39)

In the second line of equation (39), β_{k}(j_{i}) (2 ≤ k ≤ l − 1, 1 ≤ i ≤ l) are uniquely determined once β_{1}(j_{i}) (1 ≤ i ≤ l) is determined by the first line of equation (39). Observe that (β_{1}(j_{i}), β_{2}(j_{i}), …, β_{l−1}(j_{i})) ∈ S_{1,l−1} (1 ≤ i ≤ l) are distinct from equations (32) and (39). To show that the map β given by equation (38) maps S_{1,l} into a proper subset of S_{1,l−1}, write

\begin{aligned}β_{1}(j′_{1}) & := min_{j_{1_{k}}}\{| j_{1} − j_{1_{k}} |: a_{1,j_{1_{k}}} = 2(l − 1), β_{1}(j_{1}) ≠ β_{1}(j′_{1}) \\ f_{β_{1}(j′_{1})} & = f_{β_{2}(j′_{1})} = ⋯ = f_{β_{l−1}(j′_{1})}. \end{aligned}  (40)

Observe that (β_{1}(j′_{1}), β_{2}(j′_{1}), …, β_{l−1}(j′_{1})) ∈ S_{1,l−1} and distinct from (β_{1}(j_{i}), β_{2}(j_{i}), …, β_{l−1}(j_{i})) (1 ≤ i ≤ l) from equations (32), (39) and (40). Equations (39) and (40) imply that the map β : S_{1,l} ↦ S_{1,l−1} × S_{1,l−1} × ⋯ × S_{1,l−1} given by equation (38) maps S_{1,l} into a proper subset of S_{1,l−1} in a fashion of 1 to l. This establishes inequality (37). See Table 1 for examples of primes p with q | A(p), q > p, satisfying inequality (37), where q_{0} ≤ 4. Lehmer’s conjecture therefore is equivalent via equation (29) for i = 1 to:

∑_{k=0}^{q−1} ka_{1,k} ≢ 0 mod q.  (41)

Since both A(p) and B(p) are even and divisible by 691, we have (A(p), B(p)) ≥ 1382. Suppose q divides both A(p) and B(p). Then by equation (29), we have:

∑_{k=0}^{q−1} ka_{i,k} ≡ 0 mod q (0 ≤ i ≤ q − 1).  (42)

Clearly equation (42) is equivalent by equation (30) to:

∑_{k=0}^{q−1} ka_{1,k} ≡ 0 mod q.  (43)

Since ∑_{k=0}^{q−1} ka_{0,k} = 0 ≡ 0 mod q by equation (26), it follows that {∑_{k=0}^{q−1} ka_{i,k} mod q}_{i=0}^{q−1} = {0}, the trivial additive group modulo q. Conversely, equation (42) or (43) implies both q | A(p) and q | B(p) by equation (29). On the other hand, since nonzero a_{i,k} (0 ≤ i ≤ q − 1) is even and ≥ 2 from equation (25), with the aid of the unique factorization theorem, equation (42) or (43) is equivalent to:

min_{1≤i<j≤q−1}(∑_{k=0}^{q−1} ka_{i,k}, ∑_{k=0}^{q−1} ka_{j,k}) = 2q.  (44)

Consequently equation (42), (43) or (44) completely characterizes common odd prime factors of both A(p) and B(p). We thus have:

Lemma 3 The following conditions are equivalent:

(i) q divides both A(p) and B(p).

(ii) ∑_{k=0}^{q−1} ka_{i,k} ≡ 0 mod q (0 ≤ i ≤ q − 1).

(iii) ∑_{k=0}^{q−1} ka_{1,k} ≡ 0 mod q.

(iv) {∑_{k=0}^{q−1} ka_{i,k} mod q}_{i=0}^{q−1} = {0}, the trivial additive group modulo q.

(v) min_{1≤i<j≤q−1}(∑_{k=0}^{q−1} ka_{i,k}, ∑_{k=0}^{q−1} ka_{j,k}) = 2q.

Lemma 4 (Main Lemma) Let p satisfy equation (10) or (11) and let q | A(p) with q > p. Then {∑_{k=0}^{q−1} ka_{i,k} mod q}_{i=0}^{q−1} forms an additive group of order q modulo q.

Proof. Let a_{i,k} (0 ≤ i, k ≤ q − 1) be defined by equation (23). We have for each i = 1, 2, …, q − 1 :

\begin{aligned}& ∑_{k=0}^{q−1} ka_{i,k} & + & ∑_{k=0}^{q−1} ka_{q−i,k} \\ = & ∑_{k=1}^{q−1} ka_{i,k} & + & ∑_{k=1}^{q−1} ka_{i,q−k}  by (28) \\ = & ∑_{k=1}^{q−1} ka_{i,k} & + & ∑_{k=1}^{q−1}(q − k)a_{i,k} \\ = & q ∑_{k=1}^{q−1} a_{i,k} \\ = & q ∑_{l=1}^{q_{0}} 2l | S_{1,l} | & by & (35) \\ = & q(p − 1) & by & (35) \end{aligned}  (45)

Notice that equation (45) holds regardless of {∑_{k=0}^{q−1} ka_{i,k} mod q}_{i=0}^{q−1} being trivial or not. We claim that {∑_{k=0}^{q−1} ka_{i,k}}_{i=0}^{q−1} are all distinct. To show the claim observe that {S_{1,l}}_{l=0}^{q_{0}} are disjoint from equation (32). Since a_{i,k} = a_{1,i^{−1}k mod q} from equation (31), we have for each l = 1, 2, …, q_{0} :

∑_{k∈S_{1,l}}ka_{i,k}  = ∑_{k∈S_{1,l}}ka_{1,i^{−1}k mod q} =
∑_{k∈S_{1,l}}ik (mod q) a_{1,k} = 2l ∑_{k∈S_{1,l}}ik (mod q). (46)

It is evident for each 1 ≤ i ≠ j ≤ q − 1 and each l (1 ≤ l ≤ q_{0}) that:

∑_{k∈S_{1,l}}ik (mod q) ≠ ∑_{k∈S_{1,l}}jk (mod q).  (47)

For each 1 ≤ i ≠ j ≤ q − 1, conjunction of equations (46) and (47) leads us to

∑_{k=0}^{q−1} ka_{i,k}
= ∑_{l=1}^{q_{0}} 2l ∑_{k∈S_{1,l}}ik (mod q) by (46)
≠ ∑_{l=1}^{q_{0}} 2l ∑_{k inS_{1,l}}jk (mod q) by (37) & (47)
= ∑_{k=0}^{q−1} ka_{j,k} by (46).

(48)

Equation (48) establishes the claim. Since ∑_{k=0}^{q−1} ka_{1,k} mod q is a generator for the additive group {∑_{k=0}^{q−1} ka_{i,k} mod q}_{i=0}^{q−1} from equation (30) if it is nontrivial, it suffices therefore to show that

∑_{k=0}^{q−1} ka_{1,k} ≢ 0 mod q.  (49)

Write

C_{i} := ∑_{k=0}^{q−1} ka_{i,k} (1 ≤ i ≤ q − 1).  (50)

Notice that {C_{i}}_{i=1}^{q−1} are distinct from equation (48). Rename C_{i} (1 ≤ i ≤ q − 1) again as C_{i} (1 ≤ i ≤ q − 1) in ascending order as follows:

C_{1} < C_{2} < ⋯ < C_{q−1}.  (51)

We claim that there is at least one pair {C_{j}, C_{j+1}} (1 ≤ j ≤ q − 2) from equation (51) such that

C_{j+1} − C_{j} < q − 1 for some j (1 ≤ j ≤ q − 2).  (52)

Assume equation (52) is false. We then have:

C_{q−1}
:= max_{1≤i≤q−1} ∑_{k=1}^{q−1} ka_{i,k} by (51)
:= ∑_{k=1}^{q−1} ka_{i_{0},k} for some i_{0} (1 ≤ i_{0} ≤ q − 1)
= C_{1} + ∑_{k=1}^{q−2}(C_{k+1} − C_{k})
≥ C_{1} + ∑_{k=1}^{q−2}(q − 1) by assumption
= C_{1} + (q − 2)(q − 1)
> (q − 2)(q − 1).

(53)

On the other hand, we estimate C_{q−1} from equations (23) and (46). Since each nonzero a_{i_{0},k} (0 ≤ i_{0} ≤ q −1) is even ≥ 2 from equation (25), there are at most (p−1)/2 -numbers of nonzero a_{i_{0},k} ≥ 2 (0 ≤ k ≤ q −1). Notice that each nonzero a_{i_{0},k} is a small even number due to equations (32) and (35) with 2 ≤ a_{i_{0},k} ≤ 2q_{0} (0 ≤ k ≤ q − 1). It follows that there are at least (q − (p − 1)/2) -numbers of a_{i_{0},k} = 0 (0 ≤ k ≤ q − 1). We then have:

C_{q−1}
= ∑_{k=0}^{q−1} ka_{i_{0},k}
= ∑_{k=0}^{q−1} i_{0}k (mod q) a_{1,k} by (31)
= ∑_{l=1}^{q_{0}} 2l( ∑_{k∈S_{1,l}}i_{0}k (mod q)) by (46)
< (∑_{l=1}^{q_{0}} 2l | S_{1,l} |)(q − 1)
= (p − 1)(q − 1) by (35)
< (q − 2)(q − 1).

(54)

In the last line of inequality (54), we use the assumption p+1 < q and hence p−1 < q −2. The last line of inequality (54) contradicts inequality (53). This establishes inequality (52). For j chosen from inequality (52), since each nonzero a_{i,k} ≥ 2 (1 ≤ i ≤ q − 1, 0 ≤ k ≤ q − 1), we then have:

2 ≤ (C_{j}, C_{j+1}) = (C_{j}, C_{j+1} − C_{j}) < q − 1.  (55)

Equation (55) implies C_{j} := ∑_{k=0}^{q−1} ka_{u,k} and C_{j+1} := ∑_{k=0}^{q−1} ka_{v,k} for some u, v (1 ≤ u, v ≤ q − 1), have no common factor q, which leads to q ∤ ∑_{k=0}^{q−1} ka_{1,k} in view of equation (30), thereby proving equation (49). Consequently, each ∑_{k=0}^{q−1} ka_{i,k} (1 ≤ i ≤ q − 1) has no factor q from equations (30) and (49). We thus have:

∑_{k=0}^{q−1} ka_{i,k} ≢ 0 mod q, 1 ≤ i ≤ q − 1.  (56)

Equation (56) is equivalent that the map:

{∑_{k=0}^{q−1} ka_{i,k} mod q}_{i=0}^{q−1} ⟼ ℤ/qℤ

is an isomorphism. Furthermore equations (45) and (56) reveal the structure of the additive group {∑_{k=0}^{q−1} ka_{i,k} mod q}_{i=0}^{q−1} which is nontrivial, namely

∑_{k=0}^{q−1} ka_{i,k} + ∑_{k=0}^{q−1} ka_{q−i,k} ≡ 0 mod q, 1 ≤ i ≤ q − 1.  (57)

Equations (56) and (57) show ∑_{k=0}^{q−1} ka_{i,k} mod q and ∑_{k=0}^{q−1} ka_{q−i,k} mod q are additive inverse to each other modulo q for each i = 1, 2, … q − 1. Clearly ∑_{k=0}^{q−1} ka_{0,k} = 0 ≡ 0 (mod q) is the additive identity modulo q from equation (26). This completes the proof of Lemma 4. Since p ∤ A(p) from equation (12), conjunction of Lemma 3 and Lemma 4 leads us to:

Corollary 5 Let 691 | A(p). An odd prime q divides both A(p) and B(p) only if q < p.

¿From Lemma 4, we have in particular for i = 1 :

B(p) = 691 ∑_{j=1}^{p−1} σ_{5}(j)σ_{5}(p − j) ≡ ∑_{k=0}^{q−1} ka_{1,k} ≢ 0 mod q by (29) & (56).  (58)

Equation (58) implies q ∤ B(p) and hence A(p) ≠ B(p) and τ(p) = (A(p) − B(p))/3 ≠ 0 via the unique factorization theorem if 691 | A(p). If 691 ∤ A(p), then since 691 | B(p) from equation (8), we trivially have A(p) ≠ B(p) and τ(p) = (A(p) − B(p))/3 ≠ 0 via the unique factorization theorem in this case too. We thus have:

Theorem 6 τ(p) ≠ 0 for each prime p.

For 691 | A(p) and q | A(p) with q > p, since {∑_{k=0}^{q−1} ka_{i,k}}_{i=0}^{q−1} are distinct from equation (48) and since q ∤ ∑_{k=0}^{q−1} ka_{i,k} (1 ≤ i ≤ q − 1) from Lemma 4, we have ∑_{k=0}^{q−1} ka_{i,k} = 2^{s}t (s ≥ 1, t = odd, 1 ≤ i ≤ q − 1), where q ∤ t from Lemma 4. Since q > p, and since each nonzero a_{i,k} ≥ 2 from equation (25), there is at least one i (1 ≤ i ≤ q − 1) such that ∑_{k=0}^{q−1} ka_{i,k} = 2t, q ∤ t. We thus have from Lemma 1 (statement (v))with the aid of unique factorization theorem:

Corollary 7 Suppose p satisfies equation (10) or (11). Let q | A(p) with q > p. Then

min_{1≤i<j≤q−1}(∑_{k=0}^{q−1} ka_{i,k}, ∑_{k=0}^{q−1} ka_{j,k}) = 2.

Now let α ≥ 2. Then equations (10) and (11) are no longer equivalent. As in the case of α = 1, since A(p^{α}) ≡ 3 mod p^{5} and p^{11α−1} < A(p^{α}) < p^{11α} from equation (8), an almost identical proof of Lemma 2 works for α ≥ 2, where in equation (14), the upper limit for the sum is replaced by 11α − 1. We thus have:

Lemma 8 Let 691 | A(p^{α}) for α ≥ 2. There is at least one prime q | A(p^{α}) with q > p^{α}.

For q | A(p^{α}), construct matrix [a_{i,k}]_{0≤i, k≤q−1} exactly the same way as in equation (23). Then properties (25) − (31), (33) − (37) hold with p replaced by p^{α}. Likewise almost identical proof of Lemma 4 works for α ≥ 2. We thus have:

Lemma 9 Let 691 | A(p^{α}) for α ≥ 2. Let q | A(p^{α}) with q > p^{α}. Then {∑_{k=0}^{q−1} ka_{i,k} mod q}_{i=0}^{q−1} forms an additive group of order q modulo q.

In particular for i = 1 from Lemma 9 and equation (29), we have for α ≥ 2

B(p^{α}) = 691 ∑_{j=1}^{p^{α}−1} σ_{5}(j)σ_{5}(p^{α} − j) ≡ ∑_{k=0}^{q−1} ka_{1,k} ≢ 0 mod q.  (59)

Equation (59) implies q ∤ B(p^{α}) and hence A(p^{α}) ≠ B(p^{α}) and τ(p^{α}) = (A(p^{α}) − B(p^{α}))/3 ≠ 0 by the unique factorization theorem. If 691 ∤ A(p^{α}), since 691 | B(p^{α}) from equation (8), we then trivially have A(p^{α}) ≠ B(p^{α}) and τ(p^{α}) = (A(p^{α}) − B(p^{α}))/3 ≠ 0 via the unique factorization theorem in this case too. We thus have:

Theorem 10 τ(p^{α}) ≠ 0 for each α ≥ 2. Finally we show that τ(n) ≠ 0 for any positive integer n. Theorem 11 (Lehmer’s Conjecture) τ(n) ≠ 0 for each n ≥ 1. Proof. Since τ(1) = 1, it suffices to prove the theorem when n is composite from Theorem 6 and Theorem 10. Write

n = p_{0}^{s_{0}}p_{1}^{s_{1}} … p_{u}^{s_{u}}, p_{0} := 2, s_{0} ≥ 0, s_{j} ≥ 1, 1 ≤ j ≤ u.

Since τ(n) is multiplicative ([ 1 : 92 − 93 ], [ 2 : 52 − 53 ], [ 4 : 122 ], [ 5 ], [ 6 ]), Theorem 11 readily follows from Theorem 6 or Theorem 10 , namely

\begin{aligned}τ(n) & = & ∏_{j=0}^{u} τ(p_{j}^{s_{j}}) \\ & ≠ & 0. \end{aligned}  (60)

This completes the proof. Suppose for each α ≥ 1,

A(p^{α}) ≡ 0 mod 691.  (61)

Equation (61) is equivalent to:

p^{(α+1)} ≡ 1 mod 691 and (p − 1, 691) = 1.  (62)

Equation (62) implies the following periodicity theorem modulo 691:

Theorem 12 (periodicity modulo 691) Suppose 691 | A(p^{α}) for α ≥ 1. Then we have:

A(p^{α+k(α+1)}) ≡ 0 mod 691, k = 0, 1, 2, …

The values of α satisfying the periodicity of A(p^{α}) ≡ 0 mod 691 for each α ≥ 1 have gaps in view of equation (62) and Fermat’s little theorem, namely A(p^{α}) ≢ 0 mod 691 if and only if the factors of α + 1 do not divide 690 = 2.3.5.23. Thus A(p^{α}) ≢ 0 mod 691 for α in the following set S of numbers:

S := {6, 10, 12, 16, 18, 28, 30, 36, 40, 42, 46, 48, 52, 58, … }

Needless to say A(p^{α}) ≠ B(p^{α}) and hence τ(p^{α}) ≠ 0 for each α ∈ S by equation (8) with the aid of the unique factorization theorem.

Remark 13 If an odd prime q | A(p^{α}), α ≥ 1 with q < p^{α}, as long as {∑_{k=0}^{q−1} ka_{i,k} mod q}_{i=0}^{q−1} forms an additive group of order q modulo q, then q ∤ B(p^{α}) by Lemma 4 or Lemma 9. It follows that A(p^{α}) ≠ B(p^{α}) and hence τ(p^{α}) = (A(p^{α}) − B(p^{α}))/3 ≠ 0 in this case too. For 691 | A(p), computer simulation reveals A(p) has at least one odd prime factor q ≠ 691, q | A(p) with q < p for which q ∤ B(p) for each prime p ≤ 1100000 except p = 186569, 290219, 464351, 671651. Let 691 | A(p) and let A_{1}(p) be the product of prime divisors q | A(p) for which q < p with their respective powers and A_{2}(p) the product of prime divisors q | A(p) for which q > p with their respective powers. Computer simulation shows C_{1}p^{2} < A_{1}(p) < C_{2}p^{5} and C_{3}p^{6} < A_{2}(p) < C_{4}p^{10} with absolute constants C_{1}, C_{2}, C_{3}, C_{4} < 1 for primes p ≤ 1100000.

In Table 1, we list primes p such that both 691 and q divide A(p) with q > p and the cardinality | S_{1,l} | (1 ≤ l ≤ 5), thereby confirming inequality (37) with q_{0} ≤ 4. Notice that in Table 1, each prime p with the associated prime q | A(p) with q > p, satisfies equations (34) and (35). Computer simulation reveals that the majority of respective relatively large odd prime factors less than p of both A(p) and B(p) are distinct. Likewise an overwhelming majority of common odd prime factors of both A(p) and B(p) for which 691 | A(p) are relatively small apart from 691, thereby confirming Corollary 5. In Table 2, we list primes p ≤ 3000000 such that 691 | A(p) and the odd prime factors of (A(p), B(p)) are ≥ 11. Acknowledgment. We are deeply grateful to the referee who pointed out the obscurity of the additive group structure of {∑_{k=0}^{q−1} ka_{i,k} mod q}_{i=0}^{q−1} in our original manuscript. We are thankful to P. Deligne, S. Durbha, J. Gerver, H. Li and M. Nerurkar for many lively discussions. We are graeful to D. Zeiberger for his encouragement. The proof of Lemma 2 is due to P. Deligne ([ 9 ]).

Table 1

\begin{matrix}p & q & | S_{1,0} | & | S_{1,1} | & | S_{1,2} | & | S_{1,3} | & | S_{1,4} | & | S_{1,5} | \\ 8291 & 216113 & 212008 & 4065 & 40 & 0 & 0 & 0 \\ 29021 & 1357091 & 1342657 & 14358 & 76 & 0 & 0 & 0 \\ 30403 & 1283839 & 1268731 & 15015 & 93 & 0 & 0 & 0 \\ 34549 & 789673 & 772578 & 16918 & 175 & 2 & 0 & 0 \\ 51133 & 112919 & 89995 & 20474 & 2267 & 174 & 9 & 0 \\ 53897 & 371549 & 345582 & 25014 & 925 & 28 & 0 & 0 \\ 96739 & 392957 & 347376 & 42917 & 2543 & 118 & 3 & 0 \end{matrix}

Table 2

\begin{matrix}p & (A(p), B(p)) \\ 547271 & 2.3.11.691 \\ 610843 & 2.3.17.691 \\ 988129 & 2.3.5.13.691 \\ 1112509 & 2.3.5.23.691 \\ 1336393 & 2.3.101.691 \\ 1405493 & 2.3.113.691 \\ 1716463 & 2.3^{2}.23.691 \\ 1875373 & 2.23.691 \\ 1940327 & 2^{2}.3^{2}.13.691 \\ 2126897 & 2.3^{3}.19.691 \\ 2128279 & 2^{2}.5.11.691 \end{matrix}

18

\begin{matrix}p & (A(p), B(p)) \\ 2161447 & 2^{2}.23.691 \\ 2198761 & 2.43.691 \\ 2447521 & 2.23.691 \\ 2479307 & 2.23.691 \\ 2538733 & 2.11.691 \\ 2542879 & 2^{4}.3.5.23.691 \\ 2956097 & 2.23.691 \end{matrix}

References

1. Apostol, Tom, Modular Functions And Dirichlet Series, Springer (1997).

2. Berndt, B., Number Theory in the Spirit of Ramanujan, AMS (2006).

3. Iwaniec, H., Topics in Classical Automorphic Forms, AMS (1997), 13 − 22.

4. Koblitz, N., Int. to Elliptic Curves And Modular Functions, Springer (1993), 108 − 123.

5. Lehmer, D.H., Ramanujan’s Function τ(n), Duke Math. J. 10 (1943), 483 − 492.

6. Lehmer, D.H., The Vanishing of Ramanujan’s Function τ(n), Duke Math. J. 14 (1947), 429 − 433.

7. Lehmer, D.H., Note on the Distribution of Ramanujan’s τ Function, Math. Comp.24 (1970), 741 − 743.

8. Hua, L.K., Int. to Number Theory, Springer (1982) 204 − 216.

9. Deligne, P., Personal Correspondence.

Rutgers University-Camden Camden, NJ 08102 USA