deletions | additions       diff --git a/Primality.tex b/Primality.tex  index 4b30885..e356f4b 100644  --- a/Primality.tex  +++ b/Primality.tex 
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We note that $n\in\mathbb{P}$ that are not $2$ appear to have two, larger primes introduced as divisible factors. (sometimes more? $13$. [semiprime?].  \\\\  \begin{theorem}[Theorem 1]  $\forall \; p\in\mathbb{P}, n\in\mathbb{Z}\;:\;1;pn$ is divisible by $1;p$.  \end{theorem}  \begin{proof}[Proof]  Automatically true for $n=1$. \\  No proof for other $n$, only based on the evidence above. \\  \end{proof}  \\\\  \begin{theorem}[Theorem 2]  $\forall \; p_i\in\mathbb{P}, n\in\mathbb{Z}\;:\;1;p_1p_2\cdotp_Nn$ is divisible by $1;p_1,1;p_2,...,1;p_N$.  \end{theorem}  \begin{proof}[Proof]  We may single out any of the $p_i$ and make $1;p_in$, if $Theorem \; 1$ holds, then it is divisible by $1;p_i$, \forall i\in1,\codots,N.  \end{proof}  \\\\  $10$,$12$,$14$, may or may not have some curious relationship as thier numbers are $9091,9901,909091$...$99990001$, Think [False concatenation of $90$ -> $(90;;3)+1$], [think added permutations], does this apply to $333667$ as $(3;3|6;3)+1$. This sounds ridiculous which is why it's exciting.  Addition to this idea: Try "Any number $n$ whose $1;n$ spare representation is of the form (9,0,+1), multiplied to another $m$ whose $1;m$ spare representation is of the form $1;a$, gives that $1;nm$ has spare represntation the form (9,0,+1).