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\theoremstyle{plain}
\newtheorem{theorem}{Theorem}
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\begin{center}
\vskip 1cm{\LARGE\bf On Lower Order Extremal Integral Sets 
\vskip .1in
Avoiding Prime Pairwise Sums \\
}
\vskip 1cm
\large
Ram Krishna Pandey \\
Department of Mathematics\\
Indian Institute of Technology, Patna \\
Patliputra Colony, Patna - 800013\\
India \\
\href{mailto:ram@iitp.ac.in}{\tt ram@iitp.ac.in}\\
\end{center}

\vskip .2in

\begin{abstract}
Let $A$ be a subset of $\{1,2, \ldots, n\}$ such that the sum of no
two distinct elements of $A$ is a prime number. Such a subset is
called a prime-sumset-free subset of $\{1,2, \ldots, n\}$. A
prime-sumset-free subset is called an extremal prime-sumset-free
subset of $\{1,2, \ldots, n\}$ if $A \cup \{a\}$ is not a
prime-sumset-free subset for any $a \in \{1,2, \ldots, n\} \setminus
A$. We prove that if $n \geq 10$ then there is no any extremal
prime-sumset-free subset of $\{1,2, \ldots, n\}$ of order $2$ and if
$n \geq 13$ then there is no any extremal prime-sumset-free subset
of $\{1,2, \ldots, n\}$ of order $3$. Moreover, we prove that
for each integer $k \geq 2$, there exists a $n_k$ such that if $n
\geq n_k$ then there does not exist any extremal prime-sumset-free
subset of $\{1,2, \ldots, n\}$ of length $k$. Furthermore, for some
small values of $n$, we give the orders of all extremal prime-sumset-free subset
of $\{1,2, \ldots, n\}$ along with an example of
each order, and we give all extremal prime-sumset-free subsets of $\{1,2,
\ldots, n\}$ of orders $2$ and $3$ for $n \leq 13$.
\end{abstract}


\section{Introduction}
Let $A$ be a subset of $\{1,2, \ldots, n\}.$ Then a combinatorial
problem posed by Chen \cite{chen} is to study the subsets
$A$ such that
\begin{equation}
(A \hat{+} A) \cap P = \emptyset, \label{eq1}
\end{equation}
where $A \hat{+} A = \{a+b : a, b \in A, a \neq b\}$ and $P$ is the
set of all prime numbers. Such a set is called a \emph{prime-sumset-free
subset} of the set $\{1,2, \ldots, n\}.$ If we replace the set $P$
above by a given set $T$ of positive integers then $A$ is called a
\emph{$T$-sumset-free set}. Chen \cite{chen} determined all
prime-sumset-free subsets of $\{1,2, \ldots, n\}$ with the largest
cardinality. Let the largest cardinality be $U_n.$ Chen \cite{chen}
proved the following theorem.
\begin{theorem}
For all $n \geq 1$ we have $U_n = \lfloor \frac{1}{2}(n+1)\rfloor.$
Furthermore, if $A \subset \{1,2, \ldots, n\}$ is a
prime-sumset-free set with $|A|= U_n$, then all elements of $A$ have
the same parity.
\end{theorem}

With this, the natural and more challenging question coming in mind
is that what is the largest cardinality of $A \subset \{1,2, \ldots,
n\}$ satisfying (\ref{eq1}) when $A$ contains elements of both parities?

A prime-sumset-free subset $A$ of $\{1,2, \ldots, n\}$ is called an
extremal prime-sumset-free subset of $\{1,2, \ldots, n\}$ if $A \cup
\{a\}$ is not a prime-sumset-free subset for any $a \in \{1,2,
\ldots, n\}\setminus A.$ Let $(PF_k(n))_{k \geq 1}$ be the sequence
of cardinalities of all the extremal prime-sumset-free subsets of
$\{1,2, \ldots, n\}$ with $PF_1(n) &gt; PF_2(n) &gt; \cdots$ . Then by the
theorem of Chen we have $PF_1(n) = U_n = \lfloor
\frac{1}{2}(n+1)\rfloor.$

In this paper we provide the sequence $(PF_k(n))$ for a few small
values of $n$ and we study the finite monotonic strictly decreasing
sequence $PF_k(n)$ with the largest term $\lfloor
\frac{1}{2}(n+1)\rfloor$ and the smallest term $2$ (if it exists)
from the lower end of the sequence for all sufficiently large values
of $n$. In particular, we show that if $n \geq 10$ then there is no
extremal prime-sumset-free subset of $\{1,2, \ldots, n\}$ of order
$2$ and if $n \geq 13$ then there is no extremal prime-sumset-free
subset of $\{1,2, \ldots, n\}$ of order $3$. That is in the case $n
\geq 10$ the sequence $PF_k(n)$ will not take the value $2$ and
hence will terminate at $3$ (if it exists) and in the case $n \geq
13$ the sequence $PF_k(n)$ will not take the value $3$ and hence
will terminate at $4$ (if it exists) as $PF_k(n) \neq 1$ for any
$n.$ We also prove that for each integer $k \geq 2$, there exists a
$n_k$ such that if $n \geq n_k$ then there does not exist any
extremal prime-sumset-free subset of $\{1,2, \ldots, n\}$ of length
$k$.


\section{Main Results}
Sometimes we use $[n]$ for $\{1,2, \ldots, n\}$ and PSFS for
prime-sumset-free subset in this section. Observe that any proper
subset of an extremal prime-sumset-free subset of $\{1,2, \ldots,
n\}$ of order $l$ can not be an extremal prime-sumset-free subset of
$\{1,2, \ldots, n\}$. Thus if we have extremal prime-sumset-free
subsets of $\{1,2, \ldots, n\}$ of both orders $k$ and $l$ where $k
&lt; l$ then the extremal prime-sumset-free subset of $\{1,2, \ldots,
n\}$ of order $k$ will not be a subset of the extremal
prime-sumset-free subset of $\{1,2, \ldots, n\}$ of order $l.$ In
the table below we provide the sequence $(PF_k(n))$ and an example
of extremal prime-sumset-free subset of $[n]$ for each term of the
sequence for all $1 \leq n \leq 14.$

\begin{table}[H]
\begin{center}
\begin{tabular}{|c|c|}
  \hline
  % after \\: \hline or \cline{col1-col2} \cline{col3-col4} ...
  $n$ &amp; Sequence $PF_k(n)$ with an extremal prime-sumset-free subset of $[n]$ for each term\\
  \hline
  1 &amp; not defined \\
  \hline
  2 &amp; not defined \\
  \hline
  3 &amp; $(PF_k(3))\equiv (2)$; (\{1,3\}) \\
  \hline
  4 &amp; $(PF_k(4))\equiv (2)$; (\{1,3\}) \\
  \hline
  5 &amp; $(PF_k(5))\equiv (3,2)$; (\{1,3,5\}, \{2,4\})\\
  \hline
  6 &amp; $(PF_k(6))\equiv (3,2)$; (\{1,3,5\}, \{4,5\})\\
  \hline
  7 &amp; $(PF_k(7))\equiv (4,3,2)$; (\{1,3,5,7\}, \{2,4,6\}, \{4,5\}) \\
  \hline
  8 &amp; $(PF_k(8))\equiv (4,3,2)$; (\{1,3,5,7\}, \{2,7,8\}, \{4,5\}) \\
  \hline
  9 &amp; $(PF_k(9))\equiv (5,4,3,2)$; (\{1,3,5,7,9\}, \{2,4,6,8\},\{2,7,8\}, \{4,5\}) \\
  \hline
  10 &amp; $(PF_k(10))\equiv (5,3)$; (\{1,3,5,7,9\},\{2,7,8\}) \\
  \hline
  11 &amp; $(PF_k(11))\equiv (6,5,4,3)$; (\{1,3,5,7,9,11\}, \{2,4,6,8,10\}, \{4,5,10,11\}, \{1,7,8\}) \\
  \hline
  12 &amp; $(PF_k(12))\equiv (6,4,3)$; (\{1,3,5,7,9,11\}, \{4,5,10,11\}, \{1,7,8\}) \\
  \hline
  13 &amp; $(PF_k(13))\equiv (7,6,4)$; (\{1,3,5,7,9,11,13\}, \{2,4,6,8,10,12\}, \{1,7,8,13\}) \\
  \hline
  14 &amp; $(PF_k(14))\equiv (7,5,4)$; (\{1,3,5,7,9,11,13\}, \{1,7,11,13,14\}, \{3,9,12,13\}) \\
  \hline
\end{tabular}
\end{center}
\caption{$PF_k(n)$ and extremal-prime-sumset-free subset of $[n]$
for $1 \leq n \leq 14$}
\end{table}

\medskip

After having a look at the above table a natural question coming in
mind is that for a given positive integer $n$ does there exist an
extremal prime-sumset-free subset of $[n]$ of each order $l$, where
$2 \leq l &lt; \lfloor \frac{1}{2}(n+1)\rfloor$? We shall see that if
$n \geq 10$ then there is no extremal prime-sumset-free subset of
$[n]$ of order $2$ and if $n \geq 13$ then there is no extremal
prime-sumset-free subset of $[n]$ of order $3$. Before we prove
these results we give all extremal prime-sumset-free subset of $[n]$
of order $2$ and $3$ for each $n$ where $1 \leq n \leq 13$ in the
following table.

\medskip

\begin{table}[H]
\begin{center}
\begin{tabular}{|c|c|c|}
  \hline
  % after \\: \hline or \cline{col1-col2} \cline{col3-col4} ...
  $n$ &amp; All extremal PSFSs of $[n]$ of order $2$ &amp; All extremal PSFSs of $[n]$ of order $3$\\
  \hline
  3 &amp; \{1,3\} &amp; not defined\\
  \hline
  4 &amp; \{1,3,\}, \{2,4\} &amp; not defined\\
  \hline
  5 &amp; \{2,4\}, \{4,5\} &amp; \{1,3,5\}\\
  \hline
  6 &amp; \{6,3\}, \{4,5\} &amp; \{1,3,5\}, \{2,4,6\}\\
  \hline
  7 &amp; \{2,7\}, \{6,3\}, \{4,5\} &amp; \{2,4,6\}\\
  \hline
  8 &amp; \{6,3\}, \{4,5\} &amp; \{1,7,8\}, \{2,7,8\}\\
  \hline
  9 &amp;  \{4,5\} &amp; \{1,7,8\}, \{2,7,8\}, \{3,6,9\}\\
  \hline
  10 &amp; not exists &amp; \{1,7,8\}, \{2,7,8\}, \{3,6,9\}, \{4,5,10\}\\
  \hline
  11 &amp; not exists &amp; \{1,7,8\}, \{2,7,8\}, \{3,6,9\}\\
  \hline
  12 &amp; not exists &amp; \{1,7,8\}, \{2,7,8\}\\
  \hline
  13 &amp; not exists &amp; not exists\\
  \hline
\end{tabular}
\end{center}
\caption{Extremal prime-sumset-free subset of $[n]$ of
orders 2 and 3 for $1 \leq n \leq 13$}
\end{table}

\begin{theorem}
If $n \geq 10$ then there does not exist any extremal
prime-sumset-free subset of $[n]$ of order $2$.
\end{theorem}

\begin{proof}
The proof is by induction on $n$. If $n = 10$ then there does not exist
any extremal prime-sumset-free subset of $[10]$ of order $2$.
Indeed, we know that if $n \geq 6$ then an extremal
prime-sumset-free subset of order $2$ (if exists) will contain
integers of opposite parity. Therefore, the only possibilities of
two elements subsets of $[10]$ to be an extremal prime-sumset-free
subset are the following sets:
\[\{1,8\}, \{3,6\}, \{5,4\}, \{5,10\}, \{7,2\}, \{7,8\}, \{9,6\}\]
But none of the above sets is an extremal prime-sumset-free subset
as each one is a subset of some prime-sumset-free subsets of order
$3$ given below.
\[\{1,8,7\}, \{3,6,9\}, \{5,4,10\}, \{7,2,8\}.\]
Hence the theorem is true for $n=10.$ Now let the theorem is true
for $k \geq 10.$ Without loss of generality we can assume that $k$
is odd. Take all subsets $\{k+1, a\}$ where $a$ is odd and $1\leq a
\leq k$ such that $k+1+a$ is a composite integer. We shall show that
there exists an integer $t$ such that $t \in
[k+1]\setminus\{k+1,a\}$ and both $k+1+t$ and $a+t$ are composite.

\medskip

{\bf CASE I:} $1\leq a \leq k-1.$ By induction the subset $\{k,
a+1\}$ is not an extremal prime-sumset-free subset of $[k]$. Hence
there exists an integer $l \in [k]\setminus\{k,a+1\}$ such that all
$l+k, k+a+1$ and $a+1+l$ are composite integers. Now if $l$ is odd
then taking $t=l+1$ and if $l$ is even then taking $t=l-1,$ we see
that $\{k+1,a,t\}$ is a prime-sumset-free subset of $[k+1].$ Hence
the theorem is true in this case.

\medskip

{ \bf CASE II:} $a = k.$ The subset $\{k+1, a\}$ becomes $\{k+1,
k\}$. In this case the integer $t$ is given in the following table
depending on $k$.
\begin{table}[H]
\begin{center}
\begin{tabular}{|c|c|}
  \hline
  % after \\: \hline or \cline{col1-col2} \cline{col3-col4} ...
  Units digit of $k$ &amp; $t$ \\
  \hline
  1 &amp; 4 \\
  \hline
  3 &amp; 2 \\
  \hline
  5 &amp; 10 \\
  \hline
  7 &amp; 8 \\
  \hline
  9 &amp; 6 \\
  \hline
\end{tabular}
\end{center}
\caption{$t$ as a function of $k$}
\end{table}

\medskip

This completes the proof of the theorem.
\end{proof}


\begin{theorem}
If $n \geq 13$ then there does not exist any extremal
prime-sumset-free subset of $[n]$ of order $3$.
\end{theorem}

\begin{proof}
Let $A = \{a_1, a_2, a_3\}$ be a prime-sumset-free subset of $[n]$
of order $3$. We are required to show that $A$ is not an extremal
prime-sumset-free subset of $[n]$. If all elements of $A$ are of
same parity then clearly, $A$ is not an extremal prime-sumset-free
subset of $[n]$ as $n \geq 13.$ So, let us assume that $A$ contains
elements of both parities.

\medskip

{\bf CASE I:} $A$ contains two odd integers and an even integer.
Without loss of generality let $a_1$ be even.

\medskip

{\bf Subcase (i):} $2 \leq a_1 \leq 10.$ First let $a_2, a_3 \in
\{1,3,5, \ldots, 13\}.$ In this case the only possibilities of the
set $A$ for each $a_1$ is $\{2,7,13\},$ $\{4,5,11\},$ $\{6,3,9\},$
$\{8,1,7\},$ $\{8,1,13\},$ $\{8,7,13\},$ $\{10,5,11\}$. None of the
above subsets is an extremal prime-sumset-free subset because of the
following prime-sumset-free subsets.
\[\{2,7,13,8\}, \{4,5,11,10\},  \{6,3,9,12\}, \{8,1,7,13\}.\]
Secondly, let $a_2, a_3$ be any odd numbers in $A$ not necessarily
from the set $\{1,3,5, \ldots, 13\}$ and the set $A$ be distinct
from the sets given in the above list: $\{2,7,13\}$, $\{4,5,11\},$
$\{6,3,9\},$ $\{8,1,7\},$ $\{8,1,13\},$ $\{8,7,13\},$ $\{10,5,11\}$.
We have the following:\\
$2 \in A \Rightarrow A \cup \{a\}$ is a prime-sumset-free subset for
some $a \in \{7,13\}$,\\
$4 \in A \Rightarrow A \cup \{a\}$ is a prime-sumset-free subset for
some $a \in \{5,11\}$,\\
$6 \in A \Rightarrow A \cup \{a\}$ is a prime-sumset-free subset for
some $a \in \{3,9\}$,\\
$8 \in A \Rightarrow A \cup \{a\}$ is a prime-sumset-free subset for
some $a \in \{1,7,13\}$,\\
$10 \in A \Rightarrow A \cup \{a\}$ is a prime-sumset-free subset
for some $a \in \{5,11\}$.\\
Consequently, $A$ is not an extremal prime-sumset-free subset. Thus
we have seen that if $2 \leq a_1 \leq 10$ then $A$ is not an
extremal prime-sumset-free subset.


\medskip

{\bf Subcase (ii):} $12 \leq a_1 \leq 28.$ In the list below, for
each $a_1$ we provide at least three distinct odd integers from
$[n]$ such that their sum with $a_1$ give composite integers. Hence
if $12 \leq a_1 \leq 28$, we do not have extremal prime-sumset-free
subsets of $[n]$ of length $3$.

\medskip

List: $\{12,3,9,13\},$ $\{14,1,7,11\},$ $\{16,5,9,11\},$
$\{18,3,7,9\},$ $\{20,1,5,15\},$ $\{22,3,5,13\},$ $\{24,1,3,11\},$
$\{26,1,9,19\},$ $\{28,5,7,17\}$.

\medskip

{\bf Subcase (iii):} $a_1 \geq 30.$ Depending on $a_1$ we have at
least three distinct odd integers from $[n]$ such that their sum
with $a_1$ give composite integers. Hence if $a_1 \geq 30$, we do
not have extremal prime-sumset-free subsets of $[n]$ of length $3$.
\begin{table}[H]
\begin{center}
\begin{tabular}{|c|c|}
  \hline
  % after \\: \hline or \cline{col1-col2} \cline{col3-col4} ...
  Units digit of $a_1$ &amp; Three distinct odd integers \\
  \hline
  0 &amp; 5,15,25 \\
  \hline
  2 &amp; 3,13,23 \\
  \hline
  4 &amp; 1,11,21 \\
  \hline
  6 &amp; 9,19,29 \\
  \hline
  8 &amp; 7,17,27 \\
  \hline
\end{tabular}
\end{center}
\caption{Examples for Subcase (iii)}
\end{table}

\medskip

{\bf CASE II:} $A$ contains two even integers and an odd integer.
Without loss of generality let $a_1$ be odd.

\medskip

{\bf Subcase (i):} $1 \leq a_1 \leq 11.$ First let $a_2, a_3 \in
\{2,4,6, \ldots, 12\}.$ In this case the only possibility of the set
$A$ for each $a_1$ is $\{3,6,12\}, \{5,4,10\}, \{7,2,8\},
\{9,6,12\}, \{11,4,10\}$. None of the above subsets is an extremal
prime-sumset-free subset because of the following prime-sumset-free
subsets.
\[\{3,6,12,9\}, \{5,4,10,11\}, \{7,2,8,13\}.\]
Secondly, let $a_2, a_3$ be any even numbers of $A$ not necessarily
from the set $\{2,4, \ldots, 12\}$ and the set $A$ be distinct from
the sets given in the above list: $\{3,6,12\}$, $\{5,4,10\},$
$\{7,2,8\},$ $\{9,6,12\},$ $\{11,4,10\}$. We have the
following:\\
$1 \in A$ and $A = \{1,8,14\}$ then $A$ is not an extremal
prime-sumset-free subset because $\{1,8,14,7\}$ is a prime-sumset-free subset.\\
$1 \in A$ and  $A \neq \{1,8,14\} \Rightarrow A \cup \{a\}$ is a
prime-sumset-free subset for some $a \in \{8,14\}$,\\
$3 \in A \Rightarrow A \cup \{a\}$ is a prime-sumset-free subset for
some $a \in \{6,12\}$,\\
$5 \in A \Rightarrow A \cup \{a\}$ is a prime-sumset-free subset for
some $a \in \{4,10\}$,\\
$7 \in A \Rightarrow A \cup \{a\}$ is a prime-sumset-free subset for
some $a \in \{2,8\}$,\\
$9 \in A \Rightarrow A \cup \{a\}$ is a prime-sumset-free subset for
some $a \in \{6,12\}$,\\
$11 \in A \Rightarrow A \cup \{a\}$ is a prime-sumset-free subset
for some $a \in \{4,10\}$.\\
Consequently, $A$ is not an extremal prime-sumset-free subset.


\medskip

{\bf Subcase (ii):} $13 \leq a_1 \leq 29.$ In the list below, for
each $a_1$ we provide at least three distinct even integers from
$[n]$ such that their sum with $a_1$ give composite integers. Hence
if $13 \leq a_1 \leq 29$, we do not have extremal prime-sumset-free
subsets of $[n]$ of length $3$.

\medskip

List: $\{13,2,8,12\},$ $\{15,6,10,12\},$ $\{17,4,8,10\},$
$\{19,6,8,16\},$ $\{21,4,6,14\},$ $\{23,2,4,12\},$ $\{25,2,10,20\},$
$\{27,8,18,6\},$ $\{29,4,6,16\}$.

\medskip

{\bf Subcase (iii):} $a_1 \geq 31.$ Depending on $a_1$ we have at
least three distinct even integers from $[n]$ such that their sum
with $a_1$ give composite integers. Hence if $a_1 \geq 31$, we do
not have extremal prime-sumset-free subsets of $[n]$ of length $3$.
\begin{table}[H]
\begin{center}
\begin{tabular}{|c|c|}
  \hline
  % after \\: \hline or \cline{col1-col2} \cline{col3-col4} ...
  Units digit of $a_1$ &amp; Three distinct even integers \\
  \hline
  1 &amp; 4,14,24 \\
  \hline
  3 &amp; 2,12,22 \\
  \hline
  5 &amp; 10,20,30 \\
  \hline
  7 &amp; 8,18,28 \\
  \hline
  9 &amp; 6,16,26 \\
  \hline
\end{tabular}
\end{center}
\caption{Examples for Subcase (iii)}
\end{table}

This completes the proof of the theorem.
\end{proof}

Inspired by the above two theorems, we have the following:
\begin{theorem}
For each integer $k \geq 2$, there exists a $n_k$ such that if $n
\geq n_k$ then there does not exist any extremal prime-sumset-free
subsets of $[n]$ of length $k$.
\end{theorem}
One can consult \cite{nivenetal} for definitions, notation and
results used in the proof below.
\begin{proof}
Suppose that $A = \{a_1, a_2, \ldots, a_k\}$ be a prime-sumset-free
subset of $[n]$. Clearly, by the definition of psfs we have that
$a_i + a_j = m_{ij}$ is a composite integer for each $1 \leq i,j
\leq k.$ Let $1 &lt; m^{i}_j &lt; m_{ij}$ and $m^{i}_j | m_{ij}$ for each
fixed $1 \leq i \leq k$ and for each $1 \leq j \leq k.$ Let us
consider the following $k$ simultaneous congruences
\begin{eqnarray*}\label{1}
    x &amp;\equiv&amp; -a_1 \pmod {m^{i}_1}\\
    x &amp;\equiv&amp; -a_2 \pmod {m^{i}_2}\\
    &amp;\vdots&amp;\\
    x &amp;\equiv&amp; -a_k \pmod {m^{i}_k}\\
\end{eqnarray*}
for each $1 \leq i \leq k.$ Each one of the $k$ simultaneous
congruences is consistent being $a_i$ is the common solution for the
$i$th simultaneous congruence for each $1 \leq i \leq k.$ Set
\[a^{i}_{k+1} = a_i + t [m^{i}_1, m^{i}_2, \ldots, m^{i}_k],\]
for each $1 \leq i \leq k$ and $t \geq 0$ . We see that
$a^{i}_{k+1}$ is a solution of the $i$th simultaneous congruence and
hence all $a_1 + a^{i}_{k+1}, a_2 + a^{i}_{k+1}, \ldots, a_k +
a^{i}_{k+1}$ are composite integers for each $1 \leq i \leq k$.
Choosing $t$ minimum such that $a^{i}_{k+1} \not \in \{a_1, a_2,
\ldots, a_k\}$, we see that for $n \geq n_k = \max_{i}a^{i}_{k+1} :=
a_{k+1}$ we have all $a_1 + a_{k+1}, a_2 + a_{k+1}, \ldots, a_k +
a_{k+1}$ are composite integers. Consequently, the set $\{a_1, a_2,
\ldots, a_{k+1}\}$ is a prime-sumset-free subset of $[n]$ and hence
$A$ is not an extremal prime-sumset-free subset of $[n]$ of length
$k$. This proves the theorem.
\end{proof}

\section{Acknowledgments}
The author is very much thankful to the
anonymous referee for his/her useful comments/suggestions
(especially for the proof of the last theorem) to make the paper in
a better format.



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Y. G. Chen,
 Integer sequences avoiding prime pairwise sums, {\it J.
Integer Sequences}, {\bf 11} (2008),  
\href{http://www.cs.uwaterloo.ca/journals/JIS/VOL11/Chen/chen67.html}{Article 08.5.6}.

\bibitem {nivenetal}
I. Niven, H. S. Zuckerman, and H. L. Montgomery,
 {\it An Introduction to the Theory of Numbers}, John Wiley \&amp;
 Sons, 1991.
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\noindent 2000 {\it Mathematics Subject Classification}: Primary
11A41, Secondary 11B75, 05D05.

\noindent \emph{Keywords:} primes, sumsets, simultaneous congruences.

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\noindent
Received March 16 2012;
revised version received  June 1 2012.
Published in {\it Journal of Integer Sequences}, June 12 2012.

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