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Does symmetry for (M2-M1) hold for all $P$, where $P$ is prime? Please read the info-graphic for clarification. Right click on the image and "open in a new tab". I cannot find a proof showing that for all $P$, their sorted multiplicative inverse pairs will have symmetrical plots. ----- I'm not asking for a proof, I'm asking if there is a proof... so an answer of yes or no is fine.

$P$ = 47; (multiplicative inverse pairs sorted and in correct column)

$P$ = 47; (multiplicative inverse pairs sorted and NOT in correct column)

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    Does what hold for all $P$? You're not making a statement, so we can't tell you what should hold. Also, I feel completely lost here: If you have conjecture or theorem, a) explicitly state it, b) explain what you're considerations so far are, and c) what exactly your current question is! – Marcus Müller Dec 19 '21 at 23:40
  • I was hoping you'd read the information in the picture. Does the "symmetry" hold for all sorted multiplicative inverse pairs? I think it does. –  Dec 19 '21 at 23:45
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    OK, I read your steps, and I read your observation. But it's an observation, not a conjecture! It wasn't clear to me that it's what you want to show. OK, "I think it does": so tell us why! Honestly, also, maybe try to formalize things beyond plots, which are mostly a visual thing to make things easier to spot; you want to say something about $M_{2}-M_1$, so maybe make that statement explicit by writing down in formulas what should hold; I bet that's what you did anyway already, before posting this! – Marcus Müller Dec 19 '21 at 23:49
  • Also, I might not really be too deep into number theory/algebra, so I would (and maybe people would) enjoy if you'd actually define what a "multiplicative pair for a given prime $P$" is. My guess is that it's something that its a pair of elements of $\mathrm{GF}_P$, that if multiplied gives some specific numbers, but I can't seem to make out what they actually do. For example, I'd expected them to be defined as $${(M_1, M_2): M_1,M_2\in \mathrm{GF}_P, M_2 > M_1, M_2\cdot M_1 \equiv 1 \mod P},$$ but that doesn't agree with all the table entries, I think? – Marcus Müller Dec 19 '21 at 23:56
  • The first table in Step 1 is the "correct" pairing of multiplicative inverse pairs where (M1 X M2) has a remainder of 1 when divided by 47 and the plot of (M2 - M1) is non-symmetrical. But in Step 2 column M2 is sorted and the multiplicative inverse pairs are out of order. The plot of M2-M1 is "now" symmetrical. I believe this holds for any Prime Number. –  Dec 19 '21 at 23:59
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    please don't add it as comments, but edit your question to make it as clear as possible! Thanks. Generally, Text in pictures is already not great, it's hard to read. – Marcus Müller Dec 20 '21 at 00:00
  • "Text in pictures is already not great, it's hard to read." Also, it may be invisible to search engines. We want content here to be findable. – Gerry Myerson Dec 20 '21 at 04:59
  • Right click on the image and "open in new tab" to read it. –  Dec 20 '21 at 05:26
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    downvote: after being asked to properly formalize, more hard-to-read images were added without any further statements or clarification. – Marcus Müller Dec 20 '21 at 23:11
  • Let me be clear. If I place random numbers in M1 column, and random numbers in M2 column such that all of M1 and M2 are "mutually exclusive"... and I sort both columns and then perform (M2-M1), the plot will not be "symmetrical". But it appears from my empirical study, I deduce that all sorted multiplicative inverse pairs when M2 - M1 is applied... will produce symmetrical plots for any prime number $P$. I have not found a proof for this. If there is a proof, someone can simply answer YES or NO. I'm not even asking anyone to solve it. Just a simple YES or NO. How hard can that be? –  Dec 21 '21 at 03:54

1 Answers1

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Notice that if $ab\equiv1$ then $(-a)(-b)\equiv1$. This means that for each pair of inverses $(a,b)$ with $a\le b$, there is a corresponding pair of inverses $(P-a,P-b)$ with $P-a\ge P-b$. So the set of numbers in your second column just consists of the negatives (mod P) of the numbers in your first column.

As a result, after you sort the columns separately, the $n$th element from the top in the second column is just $P-a$ where $a$ is the $n$th element from the bottom in the first column. If the difference on the $n$th row from the top is $(P-a)-b$, then the difference on the $n$th row from the bottom is $(P-b)-a$. These expressions are equal to each other, so your plots are symmetrical.

Karl
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  • Thank you. I had to go through my table and verify your explanation or (proof). I have been looking for some proof of this and never could find anything. I re-read it a few times to ensure I see the logic. I sincerely appreciate the help. I'm an EE, but did a minor in math with numerical analysis and proofs a while back. --- Your logic also applies not only to remainders of 1, but also remainders of 2,3,4,5 ... , ($P$ -1). And that is a good thing. –  Dec 21 '21 at 05:12
  • If this answers your question, pEYEp, then you have the option of "accepting" it, by clicking in the check mark next to it. – Gerry Myerson Dec 22 '21 at 23:19