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Let $n$ be an integer $n=0,1,2,\ldots$ The divisibility of the following odd numbers (e.g. by 7) is structured as follows:

  • $7\mid[4]_7\cdot2^{3n+1}-1$
  • $7\mid[2]_7\cdot2^{3n+2}-1$
  • $7\mid[1]_7\cdot2^{3n+3}-1$

What is the general law for such divisibilities? Which algebraic structure (ideals, rings, p-adic valuations, or whatever) covers such behavior? May even I missed some residue classes - how may I show that the three above-shown cases cover all such divisibilities?

The same applies to the divisibility by five:

  • $5\mid[3]_5\cdot2^{4n+1}-1$
  • $5\mid[4]_5\cdot2^{4n+2}-1$
  • $5\mid[2]_5\cdot2^{4n+3}-1$
  • $5\mid[1]_5\cdot2^{4n+4}-1$

Of course I can show inductively that these divisibilities exists. But is there a general algebraic approach that explains this?

2 Answers2

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The set of residues mod $n$ is a ring, with the usual (modular) addition and multiplication. For example, your first divisibility relation can be written as $$2^{3n+1}\cdot 4 \equiv1\pmod{7},$$ which is easily verified for all $n$ because $2^{3n}\equiv8^n\equiv1^n\equiv1\pmod{7}$.

Servaes
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    It seems you spotted it faster than I did – Maximilian Janisch Sep 23 '20 at 09:18
  • Thank you very much. But how can I ensure that with the residue classes $[4]_7, [2]_7, [1]_7$ I covered all cases? Does the set ${[4]_7, [2]_7, [1]_7}$ may form a special closed structure as well? –  Sep 23 '20 at 09:26
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    @Eldar They are the (nonzero) squares mod $7$. They form a group; in general the units (i.e. invertible elements) of a commutative ring form a group. The squares then form a subgroup. It seems that in general you are looking at the subgroup generated by $2$. For $n=7$ this yields half of all nonzero residues, for $n=5$ this yields all residues. – Servaes Sep 23 '20 at 09:28
  • Great thanks! In the case of five the set ${[1]_5,[2]_5,[4]_5,[3]_5}$ covers all above shown divisibilities, nevertheless we do not have here the squares mod 5 (since $[3]_5$ is not a square)? –  Sep 23 '20 at 09:32
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    @EldarSultanow That is correct. In general you get all powers of $2$ modulo $n$. For $n=5$ this yields all nonzero residues. For $n=7$ it doesn't, because $2$ is a square modulo $7$, so this only yields squares. – Servaes Sep 23 '20 at 09:50
  • Thank you very much - I got it :-) –  Sep 23 '20 at 09:53
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    @EldarSultanow Correct they are not squares but the powers of $2$. I have explained this more precisely in an answer, which hopefully clarifies the matter. – Bill Dubuque Sep 23 '20 at 19:28
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    Anyone care to explain the downvote? – Servaes Sep 24 '20 at 10:14
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    I am not aware that anything has been downvoted. Since I am not an expert with the system, please let me know what I can do to fix this –  Sep 25 '20 at 17:00
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    @EldarSultanow Users with over 1000 reputation can see the number of up- and downvotes separately. This answer currently has 1 downvote. There is nothing you can do to change another user's votes. – Servaes Sep 25 '20 at 19:27
  • Thanks for your hint. So I need to move towards 1k reputation :-) –  Sep 26 '20 at 14:31
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Your divisibilities are immediate consequences of integer exponent laws and CPR = Congruence Power Rule, when viewed arithmetically in terms of congruences, namely they have the form

$\!\!\bmod 5\!:\,\ 2^{\large J} 2^{\large 4N+4-J}= 2^{\large 4N+4}= (\color{#c00}2^{\large 4})^{\large N+1}\equiv \color{#c00}1^{\large N+1}\equiv 1\ $ by $\,\color{#c00}{2^{\large 4}\equiv 1}\,$ and CPR

e.g. for $\,J=3\,$ we get $\,1\equiv \color{#0a0}{2^{\large 3}} 2^{\large 4N+1}\equiv \color{#0a0}{3}\cdot 2^{\large4 N+1}$ as in your list, by $\,\color{#0a0}{2^{\large 3}\equiv 3}$.

Essentially you have discovered the cyclic group structure of $\langle 2\rangle $ = $\{ 2^n\ :\ n\in \Bbb N\}.\,$ The same cyclicity will be true of $\langle a\rangle$ for any $a$ coprime to the modulus $m$ using $\,a^{\phi(m)} \equiv 1\pmod{\!m}\,$ by Euler (or $\,a^n\equiv 1\,$ for $\,n\,$ being the order of $\,a,\,$ which divides $\,\phi(m)\,$ by the Order Theorem).

Such divisibility properties are further algebraically reified by passing from congruence arithmetic to the corresponding quotient ring $\Bbb Z/5$. Then the above congruences become equalities when the integers are interpreted as names for their equivalence class, e.g. $\,2\,$ denotes $\,[2]_5 = 2 + 5 \Bbb Z$.

Bill Dubuque
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  • Thank you very much for this additional deep explanation. I will take a look at this cyclic group. Servaes wrote in this hint above "It seems that in general you are looking at the subgroup generated by 2." - Are we talking here about the same thing? –  Sep 24 '20 at 07:25
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    @Eldar Yes, if $\gcd(a,m)=1,$ and $,n,$ is the order of $,a,$ modulo $,m,,$ i.e. the least $,k > 0,$ with $,a^k\equiv 1,,$ then the subgroup generated by $,a,$ is $, \langle a\rangle ={1,a,a^2,a^3,\ldots a^{n-1}},,$ and $,a^{-1}\equiv a^{n-1},,$ since $,a\cdot a^{n-1}\equiv a^n\equiv 1.,$ The order must divided $,\phi(m),$ by the Order Theorem. – Bill Dubuque Sep 24 '20 at 07:35
  • Thank you again for these great insights into abstract algebra. Based on this I could investigate multiplicative groups and Euler's theorem further. And I hit a bump on the road, which I put into a related question here: https://math.stackexchange.com/questions/3862118/finding-homomorphism-between-congruences-bmod-18-and-bmod-3 Maybe you have an idea how these congruences relate to each other. –  Oct 13 '20 at 06:42