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I don't understand how to do these 2 tasks:

1) Prove that any arithmetic progression modulo $n$ has a period that divides $n$.

2) Prove that any geometric progression modulo a prime number $p$ has a period that divides $p-1$.

A progression modulo some number $n$ is when you have a progression and then you replace every $a_i$ by $a_i\mod n$.

A period is the number of elements in the smallest repeated sub-sequence, for example $...1,2,3,1,2,3...$ has period $3$.

In the first task, if we have a progression with difference $d$, and $d$ and $n$ are relatively prime, then the period will be $n$ because $1$ is the greatest common divisor and that's why all elements ($0,...,n-1$) will be repeated but I don't understand how to prove for the general case. Maybe when the gcd is some other number $k$, it means that every $k-th$ number will be present in the repeated sub-sequence and the $n/period=k$?

Pavel
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  • Note that a sequence $a_k$ has a period dividing $n$ precisely when $a_k=a_{k+n}$ – πr8 Jul 16 '16 at 12:25
  • If you can use elementary group theory then it is easy because it follows from Lagrange's theorem. – lhf Jul 16 '16 at 12:44
  • There are many ways to proceed depending on what you know. Do you know about the concept of "order" (additive and multiplicative)? Do you know any group theory? Do you know properties of gcd such as Bezout's identity or the gcd distributive law $,\gcd(ca,cb) = c \gcd(a,b)?\ $ – Bill Dubuque Jul 16 '16 at 14:47
  • @BillDubuque this task is from Gelfand, and i wanted to solve it using things given in the book. In the book he only explained that if you have n things arranged in a circle, and you start from one and count every k-th thing, then eventually you will have a collision, and that means that two numbers have a common divisor, then he explains that if 2 numbers are relatively prime, then you will have all numbers repeating in your subsequence that are 0...n-1. This is all I was given. This book is for 9th grade students. – Pavel Jul 16 '16 at 15:28
  • Thanks for giving more context. It would be very helpful if you quoted the argument by Gelfand, since that may give us a better idea of what methods you have available, and what sort of argument Gelfand has in mind for the exercise. – Bill Dubuque Jul 16 '16 at 15:41
  • @BillDubuque just found it in English online (I am reading in Russian) and unfortunately the English edition doesn't have this last chapter that talks about remainder arithmetic and p-adic numbers. As for the argument by Gelfand, there isn't any whatsoever, the book literally only concisely describes the things I told you about in the same fashion I described them. – Pavel Jul 16 '16 at 15:53
  • Could you please give more details on the part "and that means that two numbers have a common divisor" Too much is omitted to infer the exact argument Gelfand gives. – Bill Dubuque Jul 16 '16 at 16:01
  • @BillDubuque let me translate. Let n people be arranged in a circle (in places of vertices of a regular polygon) and we know that we can count every k-th person starting from someone (for example every third person starting from person one would be 1,4,7,10,...). Prove that if n and k have a common divisor, we will end up in a loop before we get to visit every person. Solution. Let k and n have a common divisor d. We add k at each step and subtract n, when we have done one cycle -- in every case we move by some number divisible by d. – Pavel Jul 16 '16 at 16:18
  • Continued... That means that we can't move by some number that isn't divisible by d. – Pavel Jul 16 '16 at 16:18
  • Then he goes on and explains why we will visit every person if the numbers are relatively prime, he simply supposes that we don't and then says that the gcd of n and k is 1, so we must visit everyone, before we get into a loop. – Pavel Jul 16 '16 at 16:20
  • I hope it makes sense. – Pavel Jul 16 '16 at 16:23
  • There is still too much missing to infer the methods used, e.g. how does he infer "so we must visit everyone"? – Bill Dubuque Jul 16 '16 at 16:40
  • @BillDubuque I know what you mean, that is unclear to me, as well, but there's no formal mathematical proof there, just this kind of explanation (once again, the book is for the 9th graders, so maybe he just wants the people reading this book to believe him, at least, which I do). Like there's no rigor, but the argument is convincing enough. – Pavel Jul 16 '16 at 16:48
  • So he doesn't say anything more than what you quoted above? That's hard to believe, because Gelfand is usually very careful pedagogically, but many essential details are missing from what is quoted above. – Bill Dubuque Jul 16 '16 at 16:50
  • @BillDubuque Gelfand is not the only author, the other one is Shen, since Gelfand is dead and this is a more recent addition to this book (as I mentioned earlier the English edition doesn't have this), it's possible that Gelfand didn't have any influence on this last chapter. What I said above is literally all. There's no rigor or anything. I have run into several problems with this book before, namely this one: http://math.stackexchange.com/questions/1852722/discriminant-of-a-polynomial-definition to me, it seemed like magic that a discriminant has this form and Gelfand doesn't explain it. – Pavel Jul 16 '16 at 16:58
  • @BillDubuque http://www.cimat.mx/ciencia_para_jovenes/bachillerato/libros/algebra_gelfand.pdf here's the English version. Go to page 104, problem 243. It says that if you express that sort of polynomial in p,q,r, it's called a discriminant. To me it seemed like they did a huge number of steps to simplify that to such a small form, but it turned out that I simply didn't know the definition of discriminant and Gelfand doesn't provide it, note that he says that the polynomial in p,q and r is the discriminant and not in x1,x2,x3, so I thought that it was some sort of magic. – Pavel Jul 16 '16 at 17:02
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    The definition is correct, but there appears to be little motivation. Note that you can do that exercise mechanically, i.e. the discriminant is a symmetric polynomial in the roots so you can use Gauss's algorithm to rewrite it as a polynomial in the elementary symmetric polynomials of the roots, which (Vieta) are the coefficients of the original polynomial. The algorithm is (conceptually) very simple, e.g. see the worked example in the prior link. – Bill Dubuque Jul 16 '16 at 17:27
  • @BillDubuque thanks for the link – Pavel Jul 16 '16 at 17:38

3 Answers3

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1. Let $a_k$ be an arithmetic progression. Then $$ a_{k+m}=a_k+mr$$

Thus $$a_{k+m} \equiv a_k \pmod{n} \Leftrightarrow \\ mr \equiv 0 \pmod{n} \Leftrightarrow \\ n|mr \\ $$

Now prove that the smallest $m$ which satisfies this relation is $$m=\frac{n}{\gcd(n,r)}$$

which is a divisor of $n$.

2. Is similar:

Let $b_k$ be an arithmetic progression. Then $$ b_{k+m}=b_k\cdot r^k$$

If some $b_k \equiv 0 \pmod{p}$ the problem is easy, otherwise Thus $$b_{k+m} \equiv b_k \pmod{p} \Leftrightarrow \\ r^m \equiv 1 \pmod{p} \\ $$

Now, by Fermat Little Theorem you have $r^{p-1} \equiv 1 \pmod{p}$. If $m$ is your period, show that $$r^{\gcd(r,p-1)}\equiv 1 \pmod{p}$$

N. S.
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Hints: For (1), if the common difference is $d$, show that the period is $\dfrac{n}{\gcd(n,d)}$. For (2), ignoring the trivial case where the common ratio is $0$ or when the sequence start with $0$, use the fact that there exists a primitive element $u$ modulo $p$ (i.e., there exists $u\not\equiv0\pmod{p}$ such that, for any $x\not\equiv0\pmod{p}$, $x\equiv u^l\pmod{p}$ for some nonnegative integer $l$).

Alternative Hints: Use the Pigeonhole Principle to establish that both sequences will eventually become periodic. Assume that $\ell$ is the smallest period. What happens if the division of $n$ in (1), or $p-1$ in (2), by $\ell$ leaves a remainder $r$ with $0<r<\ell$? (In fact, it can be shown that, for (2), if $p$ is not necessarily prime, then $\ell$ must divide $\lambda(p)$, where $\lambda$ is the Carmichael function.)

Batominovski
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Prove that any arithmetic progression modulo $n$ has a period that divides $n$.

Observe that $ |Z_n| = n $. Thus we have $ ng = 1 $, since $ \langle g \rangle \subseteq Z_n $, and from Larange's theorem, we know that $ |g| \bigm| n $, i.e. $ n = m|g| $ for some $ m $.

(to prove Larange's theorem, simply show that cosets partition the group and they have equal size)

Prove that any geometric progression modulo a prime number $p$ has a period that divides $p-1$.

Observe that $ g $ has an inverse in $ Z_n^\star $, iff. $ \gcd(g, n) = 1 $. Thus, we have $ |Z_p^\star| = p - 1 $, since primes are relatively prime to all $ n \neq p $.