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The multiplicative group $Z^*_{\varphi(87)}$ of integers módulo $\varphi(87)$ has 24 elements, which are $$\{1, 3, 5, 9, 11, 13, 15, 17, 19, 23, 25, 27, 29, 31, 33, 37, 39, 41, 43, 45, 47, 51, 53, 55\}.$$ Some are their own multiplicative inverses and these are $\{1, 13, 15, 27, 29, 41, 43, 55\}$. (For example, $13\cdot13 = 169 \equiv 1 \bmod{\varphi(87)}$, where $\varphi(87) = 56$. They seem like twin primes. I mean --- we can match them up as in $(13, 15)$, $(27, 29)$, $(41, 43)$ and $(55, 57 \equiv 1 \bmod{\varphi(87)})$.

Is there some well-known structure in this group that explains the fact?

Shaun
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user1145880
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    Which ring or group has $56$ elements? And if we count the elements in the displayed set do we get - as claimed - $56$ elements? In which sense is $13$ "its own multiplicative inverse"? (Hard to understand what "their own ... inverses" mean.) Please edit to make clear which is the setting, the framework for the question. – dan_fulea Nov 15 '23 at 00:46
  • Thank you for the corrections. – user1145880 Nov 15 '23 at 01:32

1 Answers1

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$x^{-1}\equiv x\iff x^2\equiv 1$, which has a unique twin root pair $\,x\equiv -1,1\,$ modulo an odd prime $p.\,$ But $\!\bmod 8\!:\ x^2\equiv 1\!\iff\! x\,$ is odd so we have four twin pairs $\,-1,1;\,$ $\,1,3;\,$ $\,3,5;\,$ $\,5,7,\,$ which by CRT combine to four twin pairs $\!\bmod 8p,\,$ e.g. in OP where $p = 7$ combining the root $-1\bmod 7\,$ with all of the roots $\,7,5,3,1\bmod 8\,$ we get all of the twins root pairs that you listed, namely $\!\bmod 56\!:\,$ $ (-1,7) \equiv -1 \equiv 55,\,$ $\,(-1,5) \equiv13,\,$ $(-1,3) \equiv27,\,$ $\,(\color{#0a0}{-1},\color{#c00}1)\ \equiv\color{#0af}{41}\,$ [the notation means $\,x\equiv \color{#0a0}{-1}\pmod{\!7},\,x\equiv \color{#c00}1\pmod{\!8}\iff x\equiv \color{#0af}{41}\pmod{\!56}\,$ by CRT].

Or directly: $\bmod 8p\!:\ (x\!-\!1)^2\equiv (x\!+\!1)^2\!$ $\iff 4x\equiv 0\!$ $\iff\! 8p\mid 4x $ $\iff\! 2p\mid x\!$ $\iff\! x\equiv 0,2p,4p,6p\pmod{\!8p},\,$ which is $\,0,14,28,42\pmod{\!56}\,$ when $\,p=7$ as in OP.

Bill Dubuque
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    See here for how to use CRT to find all roots of $,x^2\equiv 1\pmod{n}\ \ $ – Bill Dubuque Nov 15 '23 at 01:28
  • Wonderful. Thanks so much for the reference regarding the CRT. – user1145880 Nov 15 '23 at 01:36
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    Just a quibble: perhaps it's better to say "Sun-Ze's theorem" (even though that is still not entirely accurate), rather than "Chinese remainder theorem"... Just like we don't say that quadratic reciprocity is "the European theorem"... :) – paul garrett Nov 15 '23 at 02:14