deletions | additions       diff --git a/untitled.tex b/untitled.tex  index af5521e..0566a9f 100644  --- a/untitled.tex  +++ b/untitled.tex 
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$2n$+$1$=($2$m+$1$)^$2$+$1$  $2n$+$1$=$4$m^$2$+$4$m+$1$+$2$  $2n$+$1$=$4$m^$2$+$4$m+$3$  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.  \smallskip  \noindent\emph  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.  \end{solution}