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$2$n+$1$=($4$m)($4$m)+$4$m+$3$  \smallskip  When $2$m+$1$ is not initially $3$ then $4$m^$2$+$4$m+$1$ will result in and odd number and an odd number squared will give another odd number. When solving for this the right hand side will be even since and odd plus and odd results in and even integer. The right hand side will be divisible by $2$ when m does not equal $1$ showing that $x$ and $y$ be both prime only when $x$=$3$, because after that on side will be even and divisible by $2$ while the left side will be odd.  \noindent\emph{Hint:} Consider cases, depending on the remainder of $x$ modulo $3$.  \end{problem}