~fd/ersei.net

c8136fdd60d5fedd4d624fb3a9f5fe789bf9d8d7 — Ersei Saggi a month ago 225e583
Fix RSA post
1 files changed, 1 insertions(+), 1 deletions(-)

M pages/03.blog/27.rsa-basics/item.en.md
M pages/03.blog/27.rsa-basics/item.en.md => pages/03.blog/27.rsa-basics/item.en.md +1 -1
@@ 151,7 151,7 @@ where h and k are two positive integers. This statement is true for any e and d 
ed \equiv 1 \bmod ((p-1)(q-1))
[/tex]

because [texi](p-1)(q-1)[/texi] is divisible by \lambda (pq), and therefore also by [texi]p-1[/texi] and [texi]q-1[/texi].
because [texi](p-1)(q-1)[/texi] is divisible by [texi]\lambda (pq)[/texi], and therefore also by [texi]p-1[/texi] and [texi]q-1[/texi].

By the properties of the modulus operator, checking if [texi]a \equiv b \bmod pq[/texi] is equivalent to checking if [texi]a \equiv b \bmod p[/texi] and [texi]a \equiv b \bmod q[/texi] separately.