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%%%% why-RSA-works/multiple-plus-one.tex
%%%% Copyright 2012 Peter Franusic.
%%%% All rights reserved.
RSA uses two integers as exponents.
One is the encryptor $e$ and the other is the decryptor $d$.
In order for RSA to work, the product $ed$ must satisfy a strict condition.
The condition is that the product $ed$ must have a \emph{multiple-plus-one} form.
The product must be able to be written in the form $k\lambda+1$.
The reason for this condition will become apparent later.
For now, however, we need to understand what the expression $k\lambda+1$ means.
%% This paragraph shall introduce Table \ref{mult-plus-one} below.
Table \ref{mult-plus-one} contains some examples of multiple-plus-one products.
Each product ends in 001.
Each product is a multiple of 1000, plus one.
In the first example, the product 174001 is equal to $174 \cdot 1000 + 1$.
%%%% multiple-plus-one table
\caption{Multiples of 1000, plus one}
%% This paragraph shall introduce $\lambda$.
The Greek letter $\lambda$ (pronounced \textsf{LAM duh})
is specified in the RSA literature.\cite{RSA-standard}
We use $\lambda$ here as an integer constant.
It typically has a huge value, almost as large as modulus $n$.
In the context of Table \ref{mult-plus-one} it has a small value, $\lambda=1000$.
The products can therefore be written like this:
911 \cdot 191 &=& 174 \lambda + 1 \\
913 \cdot 977 &=& 892 \lambda + 1 \\
917 \cdot 253 &=& 232 \lambda + 1 \\
& \vdots &
%% This paragraph shall explain that $k$ is some unspecified positive integer.
The products in the table can be written in the form $k\lambda+1$.
The symbol $k$ signifies some positive integer.
Its value is not important.
The term $k\lambda$ simply means \emph{some integer multiple of $\lambda$}.
This meaning of $k$ allows the multiple-plus-one condition to be stated succinctly.
%% This paragraph shall formally state the condition.
\paragraph{Multiple-plus-one condition:}
Given positive integers $e$, $d$, and $\lambda$,
the product $ed$ shall be an integer multiple of $\lambda$, plus one.
\begin{equation} \label{eq:inv-pair}
ed = k\lambda + 1