Fleshed out stub. Added fortify and stub for attack.

whitepaper/Dissertation.tex

@@ -297,7 +297,7 @@ As the prime generation routine generates primes of equal length, this property
 	We see that $(1 + n)^n \equiv 1 \mod n^2$ from binomial expansion. So $1 + n$ is invertible as required.
 \end{proof}
 
-The selection of such $g$ is ideal, as the binomial expansion property helps to optimise exponentiation. Clearly, from the same result, $g^m = 1 + mn$. This operation is far easier to perform, as it can be performed without having to take the modulus to keep the computed value within range (for $m$ of sufficiently small size).
+The selection of such $g$ is ideal, as the binomial expansion property helps to optimise exponentiation. Clearly, from the same result, $g^m = 1 + mn$. This operation is far easier to perform, as it can be performed without having to take the modulus to keep the computed value within range.
 
 \subsection{Encryption}