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\end{solution}  \smallskip  When $2$m+$1$ is not initially $3$ then $4$m^$2$+$4$m+$3$ 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 $x$=$3$, becasue after that on side will be even   and divisible by 2 while the left side will be odd.  \noindent\emph  \end{solution}  %%%%%%%%%%%%%%%%%%%%%%%%%%%%  \begin{problem}  Alice's RSA public key is $P = (e,n) = (13,77)$.