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\begin{solution}  If $x$=$0$ (mod $3$),then for $x$ to be prime $x$ must be $3$. If $x$=$1$(mon$3$), then $x$ has to be $4$.  The this shows that the next $x$ will be prime relative to $3$.   We can then assume $x$=$2$m$+$1$, $x$=$2$m+$1$,  and $y$=$2n$+$1$, $y$=$2n$+$1$ ,  since odd numbers can be represented as $2$m$+$1$ $2$n+$1$  and $2n$+$1$ where $m$ and $n$ are integers.   \smallskip  \noindent\emph