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The this shows that the next $x$ will be prime relative to $3$.   We can then assume $x$=$2$m+$1$, and $y$=$2n$+$1$ , since odd numbers can be represented as $2$n+$1$ and  $2n$+$1$ where $m$ and $n$ are integers. So the equation will be:  \smallskip  $2n$+$1$=($2$m+$1$)^$2$+$1$ $2n$+$1$=$4$m^$2$+$4$m+$1$+$2$ $2n$+$1$=$4$m^$2$+$4$m+$3$ \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