I'm trying to calculate d, given p=163, q=311, e=101, n=50693, φ(n)=50220
The below is my current effort - how do I calculate d from this? I'm struggling with this part - can someone point me in the right direction?
ed mod φ(n) =1
101d mod 50220 = 1
50220 = 497(101) + 23
101 = 4(23) + 9
23 = 2(9) + 5
9 = 1(5) + 4
5 = 1(4) + 1
Back substitution:
1 = 5 - 1(4)
1 = 5 - 1(9 - 1(5))
1 = 1(5) - 1(9 - 1(5))
1 = 2(5) - 1(9)
1 = 2(23-2(9)) - 1(9)
1 = 2(23-2(9)) - 1(101-4(23))
You don't really need the backsubstitution:
write $$50220 = 0 \cdot 101 + 1\cdot 50220\tag{1}\label{eq1}$$ $$101 = 1\cdot 101 + 0\cdot 50220\tag{2}\label{eq2}$$
as the first two equations, that trivially hold. Then $101$ divides $50220$ $497$ times with remainder $23$ so multiply the second equation by $497$:
$$50160 = 497\cdot 101 + 0\cdot 50220$$ and substract it from the $\eqref{eq1}$ to get (just looking at the coefficients of $101$ and $50220$:
$$23 = -497 \cdot 101 + 1\cdot 50220\tag{3}\label{eq3}$$
$60$ only divides $101$ $4$ times, so $\eqref{eq2}$ minus $4$ times equation $\eqref{eq3}$ yields:
$$9 = 1989\cdot 101 + (-4)\cdot 50220\tag{4}\label{eq4}$$
Now $9$ divides $23$ twice, so we substract $2$ times $\eqref{eq4}$ from $\eqref{eq3}$:
$$5 = -4475 \cdot 101 + 9\cdot 50220\tag{5}\label{eq5}$$
Now $5$ divides $49$ once, so compute $\eqref{eq4}-\eqref{eq5}$:
$$4 = 6464 \cdot 101 + (-13)\cdot 50220\tag{6}\label{eq6}$$
And then finally $\eqref{eq5}$ minus $\eqref{eq6}$ yields:
$$1 = -10939\cdot 101 + 22\cdot 50220\tag{7}\label{eq7}$$
(which you can just check, as all intermediate stages, using a simple calculator), and so an inverse of $101$ modulo $50220$ is $-10939$ which, reduced modulo $50220$, yields $d=39281$.
Note that the left hand numbers are just the intermediate stages of the Euclidean algorithm, that you had. You just have to track along the equations as I did to get the final Bézout identity.
