RSA: what is $k$ in the formula to calculate $d$?

Question

In RSA, to calculate $d$, when given $\phi(n)$ and $e$, I stumbled upon this formula:

$$d = \dfrac{k \phi(n) + 1}{e}$$

But what does $k$ stand for? How to obtain the value for $k$?

Thank you,
Chris

First place to look is the same place you stumbled upon that formula. — Christoph, Dec 24 '20 at 14:02
Recall the decryption exponent $,d,$ is chosen so that $!\bmod{!\phi(n)}!:\ d\equiv 1/e,,$ i.e. $,de \equiv 1,,$ i.e. $,(de!-!1)/\phi(n),$ is an integer, denoted $,k,$ here. — Bill Dubuque, Dec 24 '20 at 14:04

score 2 · Answer 1 · answered Dec 24 '20 at 13:57

2

$d$ in an integer which satisfies

$ed\equiv 1\mod \phi(n)$,

which means that there exists an integer $k$ such that

$ed = 1 + k\cdot \phi(n).$

answered Dec 24 '20 at 13:57

Wuestenfux

score 1 · Answer 2 · answered Dec 24 '20 at 14:03

We can rearrange your equation to obtain:

$$de = k\phi(n)+1$$

or:

$$de \equiv 1 \pmod {\phi(n)}$$

Therefore $d$ is the inverse of $e$ in $\mathbb Z_{\phi(n)}$. (This only makes sense if $e$ and $\phi(n)$ are coprime, by Bezout's Lemma.)

To find this $d$ (and the less important $k$), we can use the Euclidean Algorithm.

2 Answers2