Create a proof of the theorem: If is an even number, then
Suppose \(n\) is even.
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Then there exists an \(m\) so that \(n = 2m\text{.}\)
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Then \(n\) is a prime number.
#paired
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Then there exists an \(m\) so that \(n = 2m + 1\text{.}\)
#paired
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Click the heels of your ruby slippers together three times.
#distractor
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So \(n = 2m + 0\text{.}\)
This is a superfluous second paragraph in this block.
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Thus \(n\equiv 0\mod 2\text{.}\)
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And a little bit of irrelevant multi-line math
\begin{align*}
c^2&a^2+b^2\\
&x^2+y^2\text{.}
\end{align*}
#distractor
