Using Euclidean Algorithm in RSA?

Question

it leverages multiplication and subtraction, which humans are fairly good at, to make fractions like 15996751/3870378 reducible. Also useful in scaling equations down to their simplest integer representation in one step, given with extra integers, GCD(C,GCD(A,B)) is equivalent to GCD(A,B,C). It's been around for over two thousand years and mathematicians have bent it to many purposes , but just the first bit should be justification. it leverages multiplication and subtraction, which humans are fairly good at, to make fractions like 15996751/3870378 reducible. Also useful in scaling equations down to their simplest integer representation in one step, given with extra integers, GCD(C,GCD(A,B)) is equivalent to GCD(A,B,C). It's been around for over two thousand years and mathematicians have bent it to many purposes , but just the first bit should be justification. ind a and b?

Answer 1

You would have to work backwards from applying the Euclidean algorithm. To show what I mean I'll use your question as an example. If we apply the Euclidean algorithm we get (as you've already shown): $$1680=71(23)+47$$ $$71=47(1)+24$$ $$47=24(1)+23$$ $$24=23(1)+1$$ Now what I mean by working backwards is we will start with our final remainder and express it in terms of the previous one. We will then try to eliminate it by replacing it with the remainder previous to our previous one until we are back to our original equation (i.e.,): $$1=24-23$$ $$=24-(47-24)=24(2)-47$$ $$=71(2)-(3)47$$ $$=71(71)+(-3)1680$$ This gives us a particular solution to the linear Diophantine equation you are working with, but there are infinitely many solutions in this case.

Answer 2

Here is the continued fraction way of writing this. It is, of course, equivalent to the "back-substitution" way one often sees. The convergents can bbe written in a single row of fractions, as below.

$$ \gcd( 1680, 71 ) = ??? $$

$$ \frac{ 1680 }{ 71 } = 23 + \frac{ 47 }{ 71 } $$ $$ \frac{ 71 }{ 47 } = 1 + \frac{ 24 }{ 47 } $$ $$ \frac{ 47 }{ 24 } = 1 + \frac{ 23 }{ 24 } $$ $$ \frac{ 24 }{ 23 } = 1 + \frac{ 1 }{ 23 } $$ $$ \frac{ 23 }{ 1 } = 23 + \frac{ 0 }{ 1 } $$ Simple continued fraction tableau:
$$ \begin{array}{cccccccccccc} & & 23 & & 1 & & 1 & & 1 & & 23 & \\ \frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 23 }{ 1 } & & \frac{ 24 }{ 1 } & & \frac{ 47 }{ 2 } & & \frac{ 71 }{ 3 } & & \frac{ 1680 }{ 71 } \end{array} $$ $$ $$ $$ \begin{array}{ccc} \frac{ 1 }{ 0 } & \mbox{digit} & 23 \\ \frac{ 23 }{ 1 } & \mbox{digit} & 1 \\ \frac{ 24 }{ 1 } & \mbox{digit} & 1 \\ \frac{ 47 }{ 2 } & \mbox{digit} & 1 \\ \frac{ 71 }{ 3 } & \mbox{digit} & 23 \\ \frac{ 1680 }{ 71 } & \mbox{digit} & 0 \\ \end{array} $$

$$ 1680 \cdot 3 - 71 \cdot 71 = -1 $$

Answer 3

This answer outlines one implementation of the Euclidean Algorithm that can be used to find a solution: $$ \begin{array}{r} &&23&1&1&1&23\\\hline 1&0&1&-1&2&\color{#C00}{-3}&71\\ 0&1&-23&24&-47&\color{#C00}{71}&-1680\\ \color{#090}{1680}&\color{#090}{71}&47&24&23&\color{#C00}{1}&0 \end{array} $$ which says that $$ \color{#C00}{-3}\cdot\color{#090}{1680}+\color{#C00}{71}\cdot\color{#090}{71}=\color{#C00}{1} $$