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\begin{center}{\Large \bf Lehmer's Conjecture on the Non-vanishing of Ramanujan's Tau Function}

\vspace{0.5cm}
 Will Y. Lee\\
%Department of Mathematical Sciences, Rutgers University,
%Camden, NJ 08102 USA
\end{center}

\begin{abstract}
 In this paper we prove Lehmer's conjecture on Ramanujan's
tau function, namely $\tau(n) \ne 0$ for each $ n \ge 1$ by
investigating the additive group structure attached to $\tau(n)$
with the aid of unique factorization theorem.
\\
\end{abstract}

{\bf 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.

%\section{Introduction}
\footnote{$2000$ Mathematics Subject Classification. Primary
$11L40$; Secondary $11YXX$}
 Let $E_k\;(k=2,4,\dots)$ be the normalized Eisenstein series
 ($[4:108-122]$) given by\\
 \begin{equation}
 E_k = 1 -
 \frac{2k}{B_k}\sum_{n=1}^{\infty}\sigma_{k-1}(n)q^n
 \end{equation}
 where $q:= e^{i2\pi z}\;(\Im(z) > 0), B_k$ the Bernoulli number defined
 by \\
 \[\frac{x}{e^{x}-1} = \sum_{k=0}^{\infty}
 B_k\frac{x^k}{k!}\]\\
and $\sigma_{k-1}(n)$ the divisor function:\\
\[\sigma_{k-1}(n):= \underset{d|n}\sum \;d^{k-1}.\]
 For an elliptic curve given by\\
\begin{equation}
y^2 = 4x^3 - g_2(z)x - g_3(z)
\end{equation}
where $g_2(z) = 120\zeta(4)E_4(z), g_3(z) = 280\zeta(6)E_6(z)$ and
$E_k(z)$ given by equation $(1)$ and $\zeta(k)$ is Riemann zeta function: \\
\[\zeta(k):= \sum_{n=1}^{\infty}\;\frac{1}{n^k}.\]
 A simple calculation ($[1:14],[4:112]$) shows the discriminant
 $\varDelta(z):= 4^{4}(x_1-x_2)^2(x_2-x_3)^2(x_3-x_1)^2,$  where $x_1, x_2$ and $x_3$ are the roots
 the right side of equation $(2)$, is given by\\
\begin{equation}
\varDelta(z) = g_2(z)^3 -27g_3(z)^2 =
\frac{(2\pi)^{12}}{1728}(E_4(z)^3 - E_6(z)^2).
\end{equation}
 On the other hand Jacobi's theorem$\;([4:122]$) asserts that
\begin{equation}
(2\pi)^{-12}\varDelta(z) = q\prod_{n=1}^{\infty}\;(1-q^n)^{24}.
\end{equation}
 From equation $(4),$ Ramanujan has defined his tau function
 $\tau(n)\;([1],[2],[3],[4:122],[5]-[7])$ by\\
\begin{equation}
q\prod_{n=1}^{\infty}\;(1-q^n)^{24} :=
\sum_{n=1}^{\infty}\;\tau(n)q^n.
\end{equation}
 Notice that each $\tau(n)\;(n \ge 1)$ has an integer value. In a
 series of papers ($[5]-[7]$), D.H. Lehmer investigated the
 properties of $\tau(n)$ for $n \le 300$, proved that $\tau(n) \ne
 0$ for $n < 3316799$, later for $n < 214928639999\;([1:22])$. He also
 showed that if $\tau(n) = 0$ then $n$ must be a prime. He then
 conjectured, what is nowadays known as Lehmer's conjecture ($[6]$) that\\
 \begin{equation}
 \tau(n) \ne 0 \text{\;\;for\;\; each\;\;} n \ge 1.\\
 \end{equation}
 A simple calculation$\;([3:21-22],[4:122-123]$) shows\\
\begin{equation}
\tau(n) = \frac{65}{756}\sigma_{11}(n) +
\frac{691}{756}\sigma_{5}(n) -
\frac{691}{3}\sum_{j=1}^{n-1}\;\sigma_{5}(j)\sigma_{5}(n-j).
\end{equation}
 Since Lehmer's conjecture is equivalent to $3\tau(n) \ne 0$ for
 each $n \ge 1$, we write\\
\begin{equation}
A(n):= \frac{65}{252}\sigma_{11}(n) +
\frac{691}{252}\sigma_{5}(n);\; B(n):=
691\sum_{j=1}^{n-1}\;\sigma_{5}(j)\sigma_{5}(n-j).
\end{equation}
Then $3\tau(n) = A(n) - B(n).$ Observe that $A(n)$ takes on integer
value for each $n \ge 1$ since both $\tau(n)$ and $B(n)$ do. Now
Lehmer's conjecture is, in view of equations $(7), (8)$ and the
unique factorization theorem, equivalent to:\\
\begin{equation}
A(n) \ne B(n) \text{\;\;for\;\;each\;\;} n \ge 1.\\
\end{equation}
 Recent calculation by Bosman confirms Lehmer's conjecture for $n
 \le 22798241520242687999.$ In this paper we prove equation $(9)$ by showing
that $\{\sum_{k=0}^{q-1}ka_{i,k}\;\mod q\}_{k=0}^{q-1}$ forms an
additive group of order $q$ modulo $q$ for $q\mid A(p),\;q>p,\; p\;
\equiv\; -1\; \mod 691,\;[a_{i,k}]_{0\le i,k\le q-1}$\; $q\times
q$-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 $p^{\alpha}, \alpha \ge 2$ and
finally for any composite number $n.$ Since $11 \nmid 690$ and since
$(p+1)\mid (p^{11}+1)$, the following Lemma $1$ evidently holds. \\
\bl{lemma}Let $A(p)$ be given by equation $(8)$. Then the following
two conditions $(10)$
and $(11)$ are equivalent:\\
\begin{eqnarray}
A(p) &\equiv& 0 \;\;\;\;\;\;\;\mod\; 691. \\
p &\equiv& -1 \;\;\;\;\mod\; 691.
\end{eqnarray}
\el If $691 \nmid A(p)$ or equivalently $p$ does not satisfy
equation $(11)$ then we trivially have $A(p) \ne B(p)$ 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:\\
\bl{lemma}Let $p$ satisfy equation $(10)$ or $(11)$. Then $A(p)$ has
at least one prime factor $q$ greater than $p$. \el \pf Write
$A(p)$ from equation $(8)$ as\\
\begin{equation}\begin{array}{ll}
A(p) &= \frac{65}{252}(1 + p^{11}) + \frac{691}{252}(1 + p^5)\\
&= 3 + K_5p^5 \end{array}
\end{equation}
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
\begin{equation}
A(p) = 2^{e_0}q_1^{e_1}\dots q_m^{e_m},\;q_i < p,\;e_i \ge 1\;(1\le
i\le m).
\end{equation}
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
\begin{equation}
A(p) = \sum_{i=0}^{10}b_ip^i,\;b_i:=
[p\{\frac{A(p)}{p^{i+1}}\}]\;(0\le i\le 10).
\end{equation}
We show that each $b_i \ne 0\;(0\le i\le 10)$. Indeed we prove
\begin{equation}
b_i \ge 5\;(1\le i\le 10),\;b_0 \ne 0.
\end{equation}
The fact that $b_0 \ne 0$ follows from $p \nmid A(p).$ Likewise
$b_{10} \ge 5$ follows from $p \nmid A(p)$ and $b_{10} \ge
[\frac{65}{252}p] \ge [\frac{p}{252}] \ge 5$ in view of equations
$(14)$ and $(17)$, where $p \ge 1381$.
>From equation $(8)$ or $(12)$, we readily have \\
\begin{equation}
\frac{65}{252}p^{11} < A(p) < (\frac{65}{252} + 3p^{-6})p^{11}.
\end{equation}
Equation $(16)$ is equivalent to \\
\begin{equation}
\frac{65}{252}p^{10-i} < \frac{A(p)}{p^{i+1}} < (\frac{65}{252} +
3p^{-6})p^{10-i}\;(1\le i\le 9).
\end{equation}
Write\\
\begin{equation}
65p^{10-i} = 252Q_i + R_i,\; Q_i \ge 1,\;1\le R_i\le 251\;(1\le i\le
9).
\end{equation}
Substitution of equation $(18)$ into equation $(17)$ reveals\\
\begin{equation}
Q_i + \frac{R_i}{252} < \frac{A(p)}{p^{i+1}} < (Q_i +
\frac{R_i}{252}+3p^{4-i})\;(1\le i\le 9).
\end{equation}
Inequality $(19)$ implies $\{\frac{A(p)}{p^{i+1}}\} \ge
\frac{R_i}{252} \ge \frac{1}{252}\;(1\le i \le 9)$ and hence we have
since $p \ge 1381$\\
\begin{equation}
b_i = [p\{\frac{A(p)}{p^{i+1}}\}] \ge [\frac{p}{252}] \ge 5\;(1\le
i\le 9).
\end{equation}
This establishes inequality $(15)$. Rewrite equation $(14)$ as\\
\begin{equation}
A(p) = L_5p^5 + \sum_{i=0}^4b_ip^i
\end{equation}
where $L_5:= \sum_{i=0}^5b_{5+i}p^i$. Subtraction of equation $(21)$
from equation $(12)$ with rearrangement of terms leads us to\\
\begin{equation}\begin{array}{ll} p^5 &\le
(K_5 - L_5)p^5\\ &= (b_0-3) + \sum_{i=1}^{4}b_ip^i\\
&< (p-1)\sum_{i=0}^{4}p^i\\
&= p^5 -1.
\end{array}
\end{equation}
Since each $b_i \ge 1\;(0\le i\le 10)$ from equation $(15)$, $K_5 -
L_5 \ge 1$ follows from the the second line equality of equation
$(22)$ regardless of the value of $b_0 \ne 0$. Since $1 \le b_i \le
p-1\;(0\le i\le 10)$ 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 $1 \le b_i\;(0\le i\le
10)$. A simple computation reveals Lemma $2$ also holds for primes
$p \le 252$. 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 $[a_{i,k}]_{0\le i,k\le q-1}$ as follows:\\
\begin{equation}
a_{i,k}:= \sum_{\begin{array}{c}{j =1}\\
i691\sigma_{5}(j)\sigma_{5}(p-j)\equiv k\;\mod\;q
\end{array}}^{p-1}1.
\end{equation}\\
Since $\sigma_{5}(j)\sigma_{5}(p-j) = \sigma_{5}(p-j)\sigma_{5}(p-(p-j))$, we have from equation $(23)$\\
\begin{equation}
a_{i,k} = 2\sum_{\begin{array}{c}{j =1}\\
i691\sigma_{5}(j)\sigma_{5}(p-j)\equiv k\; \mod\; q
\end{array}}^{(p-1)/2}1.
\end{equation}\\
 Then the matrix $[a_{i,k}]_{0\le i,\;k\le
q-1}$ has the following properties:\\
\begin{eqnarray}
a_{i,k} &\equiv &0 \;\;\; \mod\;2\qquad\qquad\;\;(0\le i,k\le q-1).\\
a_{0,0} &=&p - 1,\;a_{0,k} = 0\;\;\;\;\;\;\;\;\;(1\le k \le q-1).\\
a_{i,0} &=&a_{j,0}   \qquad\qquad\qquad\;\;\;\;\;(1 \le i \ne j \le q-1).\\
a_{i,k} &=&a_{q-i,q-k} \qquad\qquad\quad\;\;(1 \le i,k
\le q-1). \\
i691\sum_{j=1}^{p-1}\sigma_{5}(j)\sigma_{5}(p-j) &\equiv&
\sum_{k=1}^{q-1}ka_{i,k} \;\;\mod \;q\quad\;\;(1 \le i \le
q-1).\\
\sum_{k=1}^{q-1}ka_{i,k} &\equiv
&i\sum_{k=1}^{q-1}ka_{1,k}\;\;\mod\;q\quad(1 \le i \le q-1).
\end{eqnarray}
Notice that given $a_{1,k}\;(1\le k\le q-1),\;a_{i,k}\;(2\le i\le
q-1,1\le k\le q-1)$ are reshuffles of $a_{1,k}(1\le k\le q-1)$ and
vice versa determined by\\
\begin{equation}
a_{i,k} = a_{1,i^{-1}k\;\mod q} \Longleftrightarrow a_{1,k} =
a_{i,ik\;\mod q}\;\;(1 \le i, k \le q-1).
\end{equation}
For each $i = 1,2,\dots,q-1,$ write $f_{j_i}:=
i691\sigma_5(j)\sigma_5(p-j)\;\mod q$. Then $f_{j_i}=
f_{p-j_i}\;(1\le i\le q-1)$ from equation $(23)$. Define for each $i
= 1,2,\dots,q-1$:
\begin{equation}\begin{array}{lll}
S_{i,l}&:= &\{k:a_{i,k} = 2l\}\;(1\le k\le q,1\le l\le q_0)\\
\Longleftrightarrow &=&
\{(j_{i_1},j_{i_2},\dots,j_{i_l}):\;f_{j_{i_1}}=f_{j_{i_2}}=\dots
=f_{j_{i_l}}=k\}\;(1\le j_{i_1}<j_{i_2}<\dots
<j_{i_l}\le\frac{(p-1)}{2})\\
S_{i,l} &= &\varnothing\;(1\le l\le q_0)\;\text{for}\; l >
q_0.\end{array}\\
\end{equation}
Since $q > p$ and since $a_{i,k}\;(0\le i,k\le q-1)$ 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 $l = 1,2,\dots,q_0:$\\
\begin{equation}
S_{i,l} = S_{j,l}\;(1\le i < j \le q-1).
\end{equation}
For each $q \mid A(p)$ with $q > p$, we then have from equations $(31)-(33)$ that\\
\begin{eqnarray}
&\sum_{l=0}^{q_0}\mid S_{i,l}\mid &= q\;(1\le i\le q-1).\\
&\sum_{k=1}^{q-1}a_{i,k} &= \sum_{l=1}^{q_0}2l\mid S_{i,l}\mid\;=
p-1\;(1 \le i \le q-1).
\end{eqnarray}
Equation $(35)$ reads when $ q\mid A(p)$ with $q < p$ that:\\
\begin{equation}
\sum_{k=0}^{q-1}a_{i,k} = p-1\;(1\le i\le q-1).
\end{equation}
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 $S_{1,l}\;(1\le l\le q_0)$ in what follows, we show the
following inequality:\\
\begin{equation}
\mid S_{1,l-1}\mid\; >\; l\mid S_{1,l}\mid\;(2 \le l\le q_0).
\end{equation}
To prove inequality $(37)$ we use the second line of equation $(32)$
for the definition of $S_{1,l}$. Consider the map $\beta: S_{1,l}
\mapsto S_{1,l-1} \times S_{1,l-1} \times\dots \times S_{1,l-1}$ given by\\
\begin{equation}\begin{array}{ll}
\beta(j_1,j_2,\dots,j_l):=
&((\beta_1(j_1),\beta_2(j_1),\dots,\beta_{l-1}(j_1)),
(\beta_1(j_2),\beta_2(j_2),\dots,\beta_{l-1}(j_2)),\dots,\\
&(\beta_1(j_l),\beta_2(j_l),\dots,\beta_{l-1}(j_l)))
\end{array}
\end{equation}
such that for each $i = 1,2,\dots,l$:\\
\begin{equation}\begin{array}{lll}
&\beta_1(j_i)&:= \underset{j_{i_k}}{\min}\{\mid j_i-j_{i_k}\mid: a_{1,j_{i_k}} = 2(l-1)\\
&f_{\beta_1(j_i)} &= f_{\beta_{2(j_i)}} = \dots =
f_{\beta_{l-1}(j_i)}.
\end{array}\end{equation}
In the second line of equation $(39),\; \beta_k(j_i)\;(2\le k\le
l-1,1\le i\le l)$ are uniquely determined once $\beta_1(j_i)\;(1\le
i\le l)$ is determined by the first line of equation $(39)$. Observe
that $(\beta_1(j_i),\beta_2(j_i),\dots,\beta_{l-1}(j_i))\; \in
S_{1,l-1}\;(1\le i\le l)$ are distinct from equations $(32)$ and
$(39)$. To show that the map $\beta$ given by equation $(38)$ maps
$S_{1,l}$ into a proper subset of $S_{1,l-1}$, write\\
\begin{equation}\begin{array}{lll}
&\beta_1(j_1')&:= \underset{j_{1_k}}{\min}\{\mid j_1-j_{1_k}\mid:
a_{1,j_{1_k}} = 2(l-1),\;\beta_1(j_1) \ne \beta_1(j_1')\\
&f_{\beta_1(j_1')} &= f_{\beta_2(j_1')} = \dots =
f_{\beta_{l-1}(j_1')}.
\end{array}\end{equation}
Observe that
$(\beta_1(j_1'),\beta_2(j_1'),\dots,\beta_{l-1}(j_1'))\;\in
S_{1,l-1}$ and distinct from
$(\beta_{1}(j_i),\beta_2(j_i),\dots,\beta_{l-1}(j_i))\\
(1\le i\le l)$ from equations $(32),(39)$ and $(40)$. Equations
$(39)$ and $(40)$ imply that the map $\beta: S_{1,l} \mapsto
S_{1,l-1} \times S_{1,l-1} \times\dots \times S_{1,l-1}$ 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 $q \mid A(p),\;q > p$,
satisfying inequality $(37)$, where $q_0 \le 4$.\\
 Lehmer's conjecture therefore is equivalent via equation $(29)$ for $i = 1$ to:\\
\begin{equation}
\sum_{k=0}^{q-1}ka_{1,k} \mathrel{\not\equiv} 0\;\;\mod q.
\end{equation}
Since both $A(p)$ and $B(p)$ are even and divisible by $691$, we
have $(A(p),B(p)) \ge 1382$. Suppose $q$ divides both $A(p)$ and
$B(p)$. Then by equation $(29)$, we have:\\
\begin{equation}
\sum_{k=0}^{q-1}ka_{i,k} \equiv 0 \;\;\mod q\;(0 \le i \le q-1).
\end{equation}
Clearly equation $(42)$ is equivalent by equation $(30)$ to:\\
\begin{equation}
\sum_{k=0}^{q-1}ka_{1,k} \equiv 0 \;\;\mod q.
\end{equation}
 Since $\sum_{k=0}^{q-1}ka_{0,k} = 0 \equiv 0 \;\mod q$ by
equation $(26)$, it follows that $\{\sum_{k=0}^{q-1}ka_{i,k}\;\mod
q\}_{i=0}^{q-1} = \{0\}$, the trivial additive group modulo $q$.
Conversely, equation $(42)$ or $(43)$ implies both $q \mid A(p)$ and
$q \mid B(p)$ by equation $(29)$. On the other hand, since nonzero
$a_{i,k}\;(0\le i\le q-1)$ is even and $\ge 2$ from equation $(25)$,
with the aid of the unique factorization theorem, equation $(42)$ or
$(43)$ is equivalent to:\\
\begin{equation}
\underset{1\le i<j\le
q-1}{\min}(\sum_{k=0}^{q-1}ka_{i,k},\;\sum_{k=0}^{q-1}ka_{j,k}) =
2q.
\end{equation}
 Consequently equation $(42),(43)$ or $(44)$
completely characterizes common odd prime factors of
both $A(p)$ and $B(p)$. We thus have:\\
\bl{lemma}The following conditions are equivalent:\\
\begin{itemize}\item[(i)] $q$ divides both $A(p)$ and $B(p)$.\\
\item[(ii)] $\sum_{k=0}^{q-1}ka_{i,k} \equiv 0 \;\;\mod q\;(0 \le i \le
q-1)$.\\
\item[(iii)] $\sum_{k=0}^{q-1}ka_{1,k} \equiv 0 \;\;\mod q.$\\
\item [(iv)] $\{\sum_{k=0}^{q-1}ka_{i,k}\;\mod
q\}_{i=0}^{q-1} = \{0\}$, the trivial additive group modulo $q$.
\item [(v)] $\underset{1\le i<j\le
q-1}{\min}(\sum_{k=0}^{q-1}ka_{i,k},\;\sum_{k=0}^{q-1}ka_{j,k}) =
2q$.
\end{itemize}
\el \bl{lemma}(Main Lemma) Let $p$ satisfy equation $(10)$ or $(11)$
and let $q \mid A(p)$ with $q > p$. Then
$\{\sum_{k=0}^{q-1}ka_{i,k}\;\mod q\}_{i=0}^{q-1}$ forms an additive
group of order $q$ modulo $q$.\el \pf Let $a_{i,k}\;(0\le i,k\le
q-1)$ be defined by equation $(23)$. We
have for each $i = 1,2,\dots,q-1:$\\
\begin{equation}\begin{array}{lllll}&&\sum_{k=0}^{q-1}ka_{i,k}
&+ &\sum_{k=0}^{q-1}ka_{q-i,k}\\
&= &\sum_{k=1}^{q-1}ka_{i,k} &+
&\sum_{k=1}^{q-1}ka_{i,q-k}\quad\quad\text {\;\;by\;\;} (28)\\
&= &\sum_{k=1}^{q-1}ka_{i,k} &+
&\sum_{k=1}^{q-1}(q-k)a_{i,k}\\
&= &q\sum_{k=1}^{q-1}a_{i,k}\\
&= &q\sum_{l=1}^{q_0}2l\mid S_{1,l}\mid &\text{by} &(35)\\
&= &q(p-1) &\text{by} &(35)
\end{array}
\end{equation}
Notice that equation $(45)$ holds regardless of
$\{\sum_{k=0}^{q-1}ka_{i,k}\;\mod q\}_{i=0}^{q-1}$ being trivial or
not. We claim that $\{\sum_{k=0}^{q-1}ka_{i,k}\}_{i=0}^{q-1}$ 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 $l = 1,2,\dots,q_0:$\\
\begin{equation}\begin{array}{lllll}
&\underset{k\in S_{1,l}}{\sum}ka_{i,k} &= &\underset{k\in
S_{1,l}}{\sum}ka_{1,i^{-1}k\;\mod q}  &= \\
&\underset{k\in S_{1,l}}{\sum}ik\;(\mod q)\;a_{1,k}  &=
&2l\underset{k\in S_{1,l}}{\sum}ik\;(\mod q).
\end{array}
\end{equation}
It is evident for each $1\le i\ne j\le q-1$ and each $l\;(1\le l\le
q_0)$
that:\\
\begin{equation}
\underset{k\in S_{1,l}}{\sum}ik\;(\mod q)  \ne \underset{k\in
S_{1,l}}{\sum}jk\;(\mod q).
\end{equation}
For each  $1
\le i \ne j \le q-1$, conjunction of equations $(46)$ and $(47)$ leads us to\\
\begin{equation}\begin{array}{ll}
&\sum_{k=0}^{q-1}ka_{i,k}\\
= &\sum_{l=1}^{q_0}2l\underset{k\in S_{1,l}}{\sum}ik\;(\mod q)\;\;\text{by}\;(46)\\
\ne &\sum_{l=1}^{q_0}2l\underset{k\ in S_{1,l}}{\sum}jk\;(\mod q)\;\;\text{by}\;(37)\;\&\;(47) \\
= &\sum_{k=0}^{q-1}ka_{j,k}\;\;\text{by}\;(46).
\end{array}
\end{equation}
Equation $(48)$ establishes the claim. Since
$\sum_{k=0}^{q-1}ka_{1,k}\;\mod q$ is a generator for the additive
group $\;\{\sum_{k=0}^{q-1}ka_{i,k}\;\mod q\}_{i=0}^{q-1}$ from
equation $(30)$ if it is nontrivial, it suffices therefore to show that\\
\begin{equation}
\sum_{k=0}^{q-1}ka_{1,k} \mathrel{\not \equiv} 0\;\mod q.
\end{equation}
Write\\
\begin{equation}
C_i:= \sum_{k=0}^{q-1}ka_{i,k}\;(1 \le i \le q-1).
\end{equation}
Notice that $\{C_i\}_{i=1}^{q-1}$ are distinct from equation $(48)$.
Rename $C_i\;(1\le i\le q-1)$ again as $C_i\;(1\le i\le q-1)$ in
ascending order as follows:
\begin{equation}
C_1 < C_2 < \dots < C_{q-1}.
\end{equation}
We claim that there is at least one pair $\{C_j, C_{j+1}\}\;(1\le
j\le q-2)$ from equation $(51)$ such that
\begin{equation}
C_{j+1} - C_j < q - 1\;\text{for\;some\;}j\;(1\le j\le q-2).
\end{equation}
Assume equation $(52)$ is false. We then have:\\
\begin{equation}\begin{array}{lll}
&&C_{q-1}\\
 &:= &\underset{1\le i\le
q-1}{\max}\sum_{k=1}^{q-1}ka_{i,k}\;\;\text{by}\;(51)\\
&:=
&\sum_{k=1}^{q-1}ka_{i_{0},k}\;\;\text{for\;some}\;\;i_{0}\;(1\le
i_{0}\le q-1)\\
&= &C_1 + \sum_{k=1}^{q-2}(C_{k+1} - C_k)\\
&\ge &C_1 + \sum_{k=1}^{q-2}(q-1)\;\;\text{by\;assumption}\\
&= &C_1 + (q-2)(q-1)\\&> &(q-2)(q-1).
\end{array}
\end{equation}
On the other hand, we estimate $C_{q-1}$ from equations $(23)$ and
$(46)$. Since each nonzero $a_{i_0,k}\;(0\le i_{0}\le q-1)$ is even
$\ge 2$ from equation $(25)$, there are at most $(p-1)/2\;$-numbers
of nonzero $a_{i_{0},k} \ge 2\;(0\le k\le q-1)$. Notice that each
nonzero $a_{i_{0},k}$ is a small even number due to equations $(32)$
and $(35)$ with $2 \le a_{i_{0},k} \le 2q_0\;(0\le k\le q-1)$. It
follows that there are at least $(q-(p-1)/2)\;$-numbers of
$a_{i_{0},k} = 0\;(0\le k\le q-1)$. We then have:\\
\begin{equation}\begin{array}{ll}
&C_{q-1}\\
= &\sum_{k=0}^{q-1}ka_{i_{0},k}\\
= &\sum_{k=0}^{q-1}i{_0}k\;(\mod q)\;a_{1,k}\;\;\text{\;\;by}\;(31)\\
= &\sum_{l=1}^{q_0}2l(\underset{k \in S_{1,l}}{\sum}i_{0}k\;(\mod
q))\;\text{by}\;(46)\\
< &(\sum_{l=1}^{q_0}2l\mid S_{1,l}\mid)(q-1)\\
= &(p-1)(q-1)\;\;\text{\;\;by}\;(35)\\
< &(q-2)(q-1).
\end{array}
\end{equation}
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 $a_{i,k}
\ge 2\;(1\le i\le q-1,\;0\le k \le q-1)$,
we then have:\\
\begin{equation}
2 \le (C_j,\;C_{j+1}) = (C_j, \;C_{j+1}-C_j) < q - 1.
\end{equation}
Equation $(55)$ implies $C_{j}: = \sum_{k=0}^{q-1}ka_{u,k}$ and
$C_{j+1}: = \sum_{k=0}^{q-1}ka_{v,k}$ for some $u,v\;(1\le u,\;v\le
q-1),$ have no common factor $q$, which leads to $q \nmid
\sum_{k=0}^{q-1}ka_{1,k}$ in view of equation $(30)$, thereby
proving equation $(49)$. Consequently, each
$\sum_{k=0}^{q-1}ka_{i,k}\;(1\le i\le q-1)$
has no factor $q$ from equations $(30)$ and $(49)$. We thus have:\\
\begin{equation}
\sum_{k=0}^{q-1}ka_{i,k}\;\mathrel{\not \equiv}\;0\;\mod q,\;1\le i
\le q-1.
\end{equation}
Equation $(56)$ is equivalent that the map:\\
\[\{\sum_{k=0}^{q-1}ka_{i,k}\;\mod q\}_{i=0}^{q-1} \longmapsto \mathbb{Z}/q\mathbb{Z}\]
is an isomorphism. Furthermore equations $(45)$ and $(56)$ reveal
the structure of the additive group
$\{\sum_{k=0}^{q-1}ka_{i,k}\;\mod
q\}_{i=0}^{q-1}$ which is nontrivial, namely\\
\begin{equation}
\sum_{k=0}^{q-1}ka_{i,k} + \sum_{k=0}^{q-1}ka_{q-i,k}
\equiv\;0\;\mod q, \;1\le i\le q-1.
\end{equation}
Equations $(56)$ and $(57)$ show $\sum_{k=0}^{q-1}ka_{i,k}\;\mod q$
and $\sum_{k=0}^{q-1}ka_{q-i,k}\;\mod q$ are additive inverse to
each other modulo $q$ for each $i = 1,2,\dots q-1$. Clearly
$\sum_{k=0}^{q-1}ka_{0,k} = 0 \equiv 0 \;(\mod q)$ is the additive
identity modulo $q$ from equation $(26)$.
This completes the proof of Lemma $4$.\\
Since $p \nmid A(p)$ from equation $(12)$, conjunction of Lemma $3$
and Lemma $4$ leads us to:\\
\bc{Corollary}Let $691 \mid A(p)$. An odd prime $q$ divides both
$A(p)$ and $B(p)$ only if $q < p$.\ec
>From Lemma $4$, we have in particular for $i = 1:$\\
\begin{equation}
B(p) = 691\sum_{j=1}^{p-1}\sigma_{5}(j)\sigma_{5}(p-j) \equiv
\sum_{k=0}^{q-1}ka_{1,k} \mathrel{\;\not}\equiv 0\;\mod q
\text{\;\;by}\;(29)\;\&\;(56).
\end{equation}
Equation $(58)$ implies $q \nmid B(p)$ and hence $A(p) \ne B(p)$ and
$\tau(p) = (A(p) - B(p))/3 \ne 0$ via the unique factorization
theorem if $691 \mid A(p)$. If $691 \nmid A(p)$, then since $691
\mid B(p)$ from equation $(8)$, we trivially have $A(p) \ne B(p)$
and $\tau(p) = (A(p) - B(p))/3 \ne 0$ via the
unique factorization theorem in this case too. We thus have: \\
\bt{theorem}$\tau(p) \ne 0$ for each prime $p$.\et For $691\mid
A(p)$ and $q\mid A(p)$ with $q > p$, since
$\{\sum_{k=0}^{q-1}ka_{i,k}\}_{i=0}^{q-1}$ are distinct from
equation $(48)$ and since $q\nmid \sum_{k=0}^{q-1}ka_{i,k}\; (1\le
i\le q-1)$ from Lemma $4$, we have $\sum_{k=0}^{q-1}ka_{i,k} =
2^{s}t\;(s \ge 1,\;t =$ odd,$\;1\le i\le q-1)$, where $q \nmid t$
from Lemma $4$. Since $q > p$, and since each nonzero $a_{i,k} \ge
2$ from equation $(25)$, there is at least one $i\;(1\le i\le q-1)$
such that $\sum_{k=0}^{q-1}ka_{i,k} = 2t,\;q\nmid t$. We thus have
from Lemma $1\;$(statement $(v)$)with the
aid of unique factorization theorem:\\
\bc{corollary} Suppose $p$ satisfies equation $(10)$ or $(11)$. Let
$q\mid A(p)$ with $q > p$. Then
 \[\underset{1\le i<j\le
q-1}{\min}(\sum_{k=0}^{q-1}ka_{i,k},\;\sum_{k=0}^{q-1}ka_{j,k}) =
2.\] \ec Now let $\alpha \ge 2$. Then equations $(10)$ and $(11)$
are no longer equivalent. As in the case of $\alpha = 1$, since
$A(p^{\alpha}) \equiv 3\; \mod p^5$ and $p^{11\alpha-1} <
A(p^{\alpha}) < p^{11\alpha}$ from equation $(8)$, an almost
identical proof of Lemma $2$ works for $\alpha \ge 2$, where in
equation $(14)$, the upper limit for the sum is replaced by
$11\alpha -1$. We thus have:\\
\bl{lemma}Let $691 \mid A(p^{\alpha})$ for $\alpha \ge 2$. There is
at least one prime $q \mid A(p^{\alpha})$ with $q > p^{\alpha}$.\el
For $q \mid A(p^{\alpha})$, construct matrix $[a_{i,k}]_{0\le
i,\;k\le q-1}$ exactly the same way as in equation $(23)$. Then
properties $(25)-(31)$, $(33)- (37)$ hold with $p$ replaced by
$p^{\alpha}$. Likewise almost identical proof of Lemma $4$ works for
$\alpha \ge 2$. We thus have:\\
\bl{lemma}Let$\;691 \mid A(p^{\alpha})$ for $\alpha \ge 2$. Let $q
\mid A(p^{\alpha})$ with $q > p^{\alpha}$. Then
$\{\sum_{k=0}^{q-1}ka_{i,k}\;\mod q\}_{i=0}^{q-1}$ forms an additive
group of order $q$ modulo $q$.\el In particular for $i = 1$ from
Lemma $9$ and equation $(29)$, we have for $\alpha \ge 2$\\
\begin{equation}
B(p^{\alpha}) =
691\sum_{j=1}^{p^{\alpha}-1}\sigma_5(j)\sigma_5(p^{\alpha}-j) \equiv
\sum_{k=0}^{q-1}ka_{1,k} \mathrel{\;\not}\equiv 0  \;\;\mod q.
\end{equation}
Equation $(59)$ implies $q \nmid B(p^{\alpha})$ and hence
$A(p^{\alpha}) \ne B(p^{\alpha})$ and $\tau(p^{\alpha}) =
(A(p^{\alpha}) - B(p^{\alpha}))/3 \ne 0$ by the unique factorization
theorem. If $691 \nmid A(p^{\alpha})$, since $691 \mid
B(p^{\alpha})$ from equation $(8)$, we then trivially have
$A(p^{\alpha}) \ne B(p^{\alpha})$ and $\tau(p^{\alpha}) =
(A(p^{\alpha}) - B(p^{\alpha}))/3 \ne 0$ via the
unique factorization theorem in this case too. We thus have:\\
\bt{theorem}$\tau(p^{\alpha}) \ne 0$ for each $\alpha \ge 2$.\et
Finally we show that $\tau(n) \ne 0$ for any positive integer $n$.
\bt{theorem}(Lehmer's Conjecture) $\tau(n) \ne 0$ for each $n \ge
1$. \et \pf Since $\tau(1) = 1$, it suffices to prove the theorem
when $n$ is composite from Theorem $6$ and Theorem $10$.
Write\\
\[
n = p_0^{s_0}p_1^{s_1}\dots p_u^{s_u},\;p_0:= 2,\;s_0 \ge 0,\;s_j
\ge 1,
\;1\le j \le u.\] \\
Since $\tau(n)$ is
multiplicative$\;([\;1:92-93\;],[\;2:52-53\;],[\;4:122\;],[\;5\;],[\;6\;])$,
Theorem $11$ readily follows from Theorem $6\;$or\;Theorem $10$ ,
namely
\begin{equation}
\begin{array}{lll}
\tau(n) &= &\prod_{j=0}^u\tau(p_{j}^{s_{j}}) \\
&\ne &0.
\end{array}
\end{equation}
This completes the proof.\\
Suppose for each $\alpha \ge 1,$\\
\begin{equation}
A(p^{\alpha}) \equiv 0 \;\;\;\;\mod 691.\\
\end{equation}
Equation $(61)$ is equivalent to:\\
\begin{equation}
p^{(\alpha+1)} \equiv 1 \;\;\mod 691\;\;\text{and}\;(p-1,\;691) = 1.\\
\end{equation}
Equation $(62)$ implies the following periodicity theorem
modulo $691$:\\
\bt{theorem}(periodicity modulo $691$) Suppose $691 \mid
A(p^{\alpha})$
for $\alpha \ge 1$. Then we have:\\
\[A(p^{\alpha+k(\alpha+1)}) \equiv 0 \;\;\mod 691,\;k=0,1,2,\dots\]\\
\et The values of $\alpha$ satisfying the periodicity of
$A(p^{\alpha}) \equiv 0 \;\mod 691$ for each $\alpha \ge 1$ have
gaps in view of equation $(62)$ and Fermat's little theorem, namely
$A(p^{\alpha}) \mathrel{\not \equiv} 0 \;\mod 691$ if and only if
the factors of $\alpha+1$ do not divide $690 = 2.3.5.23$. Thus
$A(p^{\alpha}) \mathrel{\not \equiv} 0 \;\mod
691$ for $\alpha$ in the following set $S$ of numbers:\\
\[S:= \{6,10,12,16,18,28,30,36,40,42,46,48,52,58,\dots\}\]\\
Needless to say $A(p^{\alpha}) \ne B(p^{\alpha})$ and hence
$\tau(p^{\alpha}) \ne 0$ for each $\alpha \in S$
by equation $(8)$ with the aid of the unique factorization theorem.\\
\br{remark}If an odd prime $q \mid A(p^{\alpha}),\;\alpha \ge 1$
with $q < p^{\alpha}$, as long as $\{\sum_{k=0}^{q-1}ka_{i,k}\;\mod
q\}_{i=0}^{q-1}$ forms an additive group of order $q$ modulo $q$,
then $q \nmid B(p^{\alpha})$ by Lemma $4$ or Lemma $9$. It follows
that $A(p^{\alpha}) \ne B(p^{\alpha})$ and hence $\tau(p^{\alpha}) =
(A(p^{\alpha})- B(p^{\alpha}))/3 \ne 0$ in this case too. For $691
\mid A(p)$, computer simulation reveals $A(p)$ has at least one odd
prime factor $q \ne 691,\;q \mid A(p)$ with $q < p$ for which $q
\nmid B(p)$ for each prime $p \le 1100000$ except $p = 186569,
290219, 464351, 671651$. Let $691 \mid A(p)$ and let $A_1(p)$ be the
product of prime divisors $q \mid A(p)$ for which $q < p$ with their
respective powers and $A_2(p)$ the product of prime divisors $q \mid
A(p)$ for which $q > p$ with their respective powers. Computer
simulation shows $C_1p^2 < A_1(p) < C_2p^{5}$ and $C_3p^{6} < A_2(p)
< C_4p^{10}$ with absolute constants $C_1, C_2, C_3,C_4 < 1$ for
primes $p \le 1100000$.\er In Table $1$, we list primes $p$ such
that both $691$ and $q$ divide $A(p)$ with $q > p$ and the
cardinality $\mid S_{1,l}\mid\;(1\le l\le 5)$, thereby confirming
inequality $(37)$ with $q_0 \le 4$. Notice that in Table $1$, each
prime $p$ with the associated prime $q\mid A(p)$ 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 $691 \mid A(p)$ are relatively small apart from
$691,$ thereby confirming Corollary $5$. In Table $2$, we list
primes $p \le 3000000$ such that $691 \mid A(p)$ and the odd prime
factors
of $(A(p),\;B(p))$ are $\ge 11$.\\
Acknowledgment. We are deeply grateful to the referee who pointed
out the obscurity of the additive group structure of
$\{\sum_{k=0}^{q-1}ka_{i,k}\;\mod q\}_{i=0}^{q-1}$ 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$\;([\;9\;])$.\\
\begin{center}{\bf Table $1$}
\vspace{5 pt}
\end{center}
\begin{center}
\begin{tabular}{||c|c|c|c|c|c|c|c||}
\hline p &q &$\mid S_{1,0}\mid$ &$\mid S_{1,1}\mid$ &$\mid
S_{1,2}\mid$ &$\mid S_{1,3}\mid$
&$\mid S_{1,4}\mid$ &$\mid S_{1,5}\mid$\\
\hline 8291 &216113 &212008 &4065 &40 &0 &0 &0 \\
\hline 29021 &1357091 &1342657 &14358 &76 &0 &0 &0\\\hline 30403
&1283839 &1268731 &15015 &93 &0 &0 &0\\\hline 34549 &789673 &772578
&16918
&175 &2 &0 &0\\
\hline 51133 &112919 &89995 &20474 &2267 & 174 &9 &0\\\hline 53897
&371549 &345582 &25014 &925 &28 &0 &0\\\hline 96739 &392957 &347376
&42917 &2543 &118 &3 &0\\\hline
\end{tabular}
\end{center}
\vspace{8 pt}
\begin{center}{\bf Table $2$} \vspace{5 pt}
\end{center}
\begin{center}
\begin{tabular}{||c|c||}
\hline p  &$(A(p),\;B(p))$\\
\hline 547271 &2.3.11.691\\ \hline 610843 &2.3.17.691\\
\hline 988129 &2.3.5.13.691\\
\hline 1112509 &2.3.5.23.691\\ \hline 1336393
&2.3.101.691\\ \hline 1405493 &2.3.113.691\\
\hline 1716463 &$2.3^{2}.23.691$\\ \hline 1875373 &2.23.691\\ \hline
1940327
&$2^2.3^{2}.13.691$\\ \hline 2126897 &$2.3^{3}.19.691$\\
\hline 2128279 &$2^{2}.5.11.691$\\\hline\end{tabular}
\end{center}\newpage
\begin{center}
\begin{tabular}{||c|c||}
\hline p  &$(A(p),\;B(p))$\\
\hline 2161447 &$2^{2}.23.691$\\
\hline 2198761 &2.43.691\\ \hline 2447521 &2.23.691\\
\hline
\hline 2479307 &2.23.691\\ \hline 2538733 &2.11.691\\
\hline
\hline 2542879 &$2^{4}.3.5.23.691$\\ \hline 2956097 &2.23.691\\
\hline
\end{tabular}
\end{center}
\vspace{5 pt}
\begin{center}
{\large \bf References}
\end{center}
\begin{itemize}
\item[1.] Apostol, Tom, Modular Functions And Dirichlet Series, Springer $\;(1997).$\\
\item[2.] Berndt, B., Number Theory in the Spirit of Ramanujan, AMS $\;(2006)$.\\
\item[3.] Iwaniec, H., Topics in Classical Automorphic Forms, AMS
$\;(1997),\;13-22$.\\
\item[4.] Koblitz, N., Int. to Elliptic Curves And Modular Functions, Springer$\;(1993),\\ 108-123.$\\
\item[5.] Lehmer, D.H., Ramanujan's Function $\tau(n)$, Duke Math. J. $10\;(1943),\\483-492.$\\
\item[6.] Lehmer, D.H., The Vanishing of Ramanujan's Function $\tau(n)$, Duke Math. J.
$14\\(1947),\;429-433.$\\
\item[7.] Lehmer, D.H., Note on the Distribution of Ramanujan's $\tau$ Function, Math. Comp.$24\\
(1970),\;741-743.$
\item[8.] Hua, L.K., Int. to Number Theory, Springer$\;(1982)\;204-216$.\\
\item[9.] Deligne, P., Personal Correspondence.
\end{itemize}
\begin{center}
Rutgers University-Camden\\
Camden, NJ 08102 USA
\end{center}
\end{document}



