Is it a coincidence that $8^{73} \bmod{91} = 8^{29} \bmod{72}$ both equal $8$?

Question

While working on RSA I accidentally stumbled on something I find to unlikely to be a coincidence but cant see the connection:

Encrypted message $y=8$, public key $(n, e) = (91, 5)$, so I'm trying to solve the equation $x \equiv y^d \pmod{91}$ for $x$, where $e, d \in \mathbb{Z_{\phi(n)}^*}$ and $e\cdot d \mod \phi(n) = 1$.

Initially I calculated: $d = 73$ and then found $x = 8^{73} \bmod 91 = 8$ but then found another $d = 29$, so then $x = 8^{29} \bmod 72 = 8$, also (!).

Checked with Haskell:

Prelude> 8^73 `mod` 91
8
Prelude> 8^29 `mod` 72
8

The accident happened because initially I calculated $d$ with $m = 91$, instead of $72$.

$8^2\equiv-8\bmod72$, so every odd power of $8$ mod $72$ is $8$ — J. W. Tanner, Jan 28 '20 at 20:42
$8^{73} \equiv 8 \pmod{91}$ is because $\phi(91) = 72$. The easiest way I can see to explain $8^{29} \equiv 8 \pmod{72}$ is that $8^2 \equiv -8 \pmod{72}$ and therefore $8^3 \equiv 8 \pmod{72}$ — Izaak van Dongen, Jan 28 '20 at 20:42

Bill Dubuque · Answer 1 · 2020-01-29T01:22:41.140

Below let $\, i= 1,\, j= 18,\, k= 14$.

Lemma $\ \ \bbox[5px,border:1px solid #c00]{8^{\large i+4j}\bmod{91} = 8^{\large i} = 8^{\large i+2k}\bmod 72}\ $ for $\,i\in\{1,2\},\ j,k\in\Bbb N^{\phantom{|^|}}$

Proof $\ $ First $\,(8^{\large 2})^{\large 2}\!\equiv (\pm1)^{\large 2}\!\equiv 1\pmod {\!7\ \&\ 13},\,$ so $\, \color{#90f}{8^{\large 4}\equiv 1}\pmod {\!91\!=\!7\cdot 13}\,$ by CCRT, so $\!\bmod 91\!:\,\ 8^{\large i+4j^{\phantom{|^|}}}\!\!\!\equiv 8^{\large i}(\color{#90f}{8^{\large 4}})^{\large j}\equiv 8^{\large i}\color{#90f}{1}^{\large j}\equiv 8^{\large i}$.
Second, $\ 8^{\large i+2k^{\phantom{|^{|^.}}}}\!\!\!\!\bmod 72\, =\, \color{#c00}8(8^{\large i-1+\color{#0a0}{2k}}\bmod 9) = 8(8^{i-1}) = 8^{\large i}\,$ by $\bmod 9\!:\ 8^{\large\color{#0a0}2}\!\equiv (-1)^{\large 2^{\phantom{|^|}}}\!\!\!\equiv 1,\,$
by applying $\, \color{#c00}ab\bmod \color{#c00}ac\, =\, \color{#c00}a\,(b\bmod c)^{\phantom{|^|}}\!\! = $ $\!\bmod\!$ Distributive Law to factor $\,\color{#c00}{a = 8}\,$ out of the mod.

Remark $ $ So the "coincidence" arises from $\,73\bmod 4 = 1 = 29\bmod 2,\,$ i.e. in the language of the Lemma it arises from the congruence: $ \ i\!+\!4j\,\bmod\, 4\, =\, i\, =\ i\!+\!2k\,\bmod\, 2,\,$ i.e. from the fact that the exponents on $\,8\,$ both reduce to the equal value $\,i\,$ when reduced modulo the period of $\,8$.

another coincidence is that the order of $8 \pmod {91-72=19}$ is $4+2$ — , Jan 31 '20 at 00:36

Is it a coincidence that $8^{73} \bmod{91} = 8^{29} \bmod{72}$ both equal $8$?

1 Answers1