Group theory – divisibility of the group of invertible functions

Suppose we have the group $$G$$ of invertible functions over a particular set $$S subseteq mathbb {R}$$ under composition. I am interested in the divisibility of such a group. Take, for example $$S =[-1, 1]$$for each $$n in mathbb {N}$$you could try to find a function $$f in S$$ so that
$$underbrace {f circ f cdots circ f} _ {n text {times}} = sin ( frac { pi x} {2})$$
In general we want to satisfy $$forall n forall x exists y y ^ n = x$$, the usual axiom of divisibility. I suspect that these groups are not divisible, but it is known that they can always be embedded in a divisible group $$overline {G}$$, The construction of $$overline {G}$$ I've read that a series of wreath products and a direct limit are required. This usually makes things very abstract and can be difficult to identify $$f$$ as an element of $$overline {G}$$,

Is there a better way to visualize this construction? My goal is to understand the elements in $$overline {G}$$ as features over some larger (possibly not even real) $$S supseteq S$$,

Number theory – divisibility properties of Pisano periods

To let $$(F_n)$$ the Fibonacci sequence and $$pi (m)$$ the pisano time of $$m$$ (ie the smallest period of $$F_n pmod {m}$$). There are many proven results over $$pi (m)$$, For example, this is known $$pi (p ^ a) mid p ^ {a-1} pi (p)$$, for any prime $$p$$ and an integer $$a geq 1$$, it is supposed that $$pi (p ^ a) = p ^ {a-1} pi (p)$$,

My question is a weak version of this assumption. For example, I want to prove that $$m ^ 2 mid pi (m)$$ for all positive integers $$m$$,

Since for $$m = p_1 ^ {a_1} cdots p_k ^ {a_k}$$it holds $$pi (m) = lcm ( pi (p_1 ^ {a_1}), lpoints, pi (p_k ^ {a_k}))$$Then it is enough to prove that
[
p^amid pi(p^{2a}),
]
For every prime $$p$$ and an integer $$a geq 1$$, There's the guess $$pi (p ^ a) = p ^ {2a-1} pi (p)$$We have some freedom, though $$a <2a-1$$ (ie. $$a> 1$$). Could someone help me?

Actually, even in the case of $$m mid pi (m ^ 3)$$ would be helpful. Thank you in advance!

Divisibility – proof strategy polynomial division

I have received the following polynomial

$$p (x) = x ^ 4 -2x ^ 3 + 6x ^ 2-10x + 5 in mathbb {Q}[x]$$

I have to prove that $$(x-1) ^ 2 | p (x)$$,

How can you prove that?

One way is to do the long polynomial division and check if the remainder is 0.

Are there other ways to observe, for example, that 1 is a root?

Divisibility of the sum of multinomials

To let $$n, m$$ and $$t$$ be positive integers. Define the multi-family sequences
$$S (n, m, t) = sum_ {k_1 + cdots + k_n = m} binom {m} {k_1, dots, k_n} ^ t$$
where the sum runs over non-negative integers $$k_1, dots, k_n$$, These numbers refer to average distances (from the origin) of random, uniform step-steps in the plane.

QUESTION. Is it always like that? $$n$$ splits $$S (n, m, t)$$?

Watch that $$S (n, m, 1) = n ^ m$$,

nt.number theory – Systems of \$ n \$ divisibility conditions on \$ n \$ Prim variables

Consider a system of $$n$$ Divisibility conditions on $$n$$ Prim variables:
$$p_i | a_ {i, 1} p_1 + dotsc + a_ {i, n} p_n, ; ; ; ; ; ; 1 leq i leq n,$$
from where $$a_ {i, j}$$ are integers. How many solutions are there at all? $$p_i$$ Prime and in one area $$[N_0,N_1]$$? (We can guess that $$N_0$$ is a bit bigger than all $$a_ {i, j}$$In other words, under what conditions are there very few solutions?

Some thoughts: If $$lbrack N_0, N_1 rbrack$$ is dyadic (ie the shape) $$lbrack N, 2N rbrack$$) then the divisibility conditions can be replaced by a system of equations $$c_i p_i = a_ {i, 1} p_1 + dotsc + a_ {i, n} p_n$$, $$c_i$$ limited. That's a system of $$n$$ Equations in $$n$$ Variables and should therefore generally have no non-trivial solutions (which means that there are no solutions, such as $$0$$ is not a prime), although of course the fact that there is $$C ^ k$$ Possibilities for $$c_i$$ can be annoying, to say the least.

Of course, something Conditions are required $$a_ {i, j}$$ For the system there are few solutions: If $$a_ {i, j} = 0$$ for all $$(i, j)$$The system has many solutions!

ag.algebraic geometry – divisibility of a divisor

To let $$X$$ to be a smooth complex projective curve and $$f colon X to Y$$ an étale Galois cover, whose Galois group $$G$$ is finite and tidy $$r$$, For all $$g in G$$, define $$Delta_g = {(x, , g cdot x) ; | ; x in X } subset X times X.$$ Then everyone $$Delta_g$$ is a smooth divisor isomorphic to the diagonal $$Delta = Delta_1$$,

In addition, the fact is that $$f$$ is étale implied $$Delta_g cap Delta_h = emptyset$$ if $$g neq h$$so that the reducible divider $$D = sum_ {g in G} Delta_g$$ is smooth (note that $$D$$ is the fiber product $$X times_Y X$$).

question, is $$Delta$$ $$r$$divisible into $$mathrm {Pic} (X times X)$$?

Number theory – divisibility check in different modules

So I am working on modular arithmetic and have encountered the following problem:

Accept that $$a cdot b mod m = c$$ and I know that $$c mod n = d$$, Can I still test for divisibility? $$c$$ With $$d$$? A quick check tells me if that works $$d mod n = 0$$ does not work – but is this review not possible or is there a general way to find out $$c | a$$? Maybe, if $$m$$ and $$n$$ are primes?

Many Thanks!

Number theory – check of divisibility with minimal bits

Suppose we get an infinite stream of integers. $$x_1, x_2, …$$,

a) Show that we can calculate whether the sum of all integers seen so far is divisible by a fixed integer $$N$$ With $$O (log N)$$ Bits of memory.

b) Leave $$N$$ an arbitrary number, and let's assume we are given $$N$$Hauptfaktorisierung: $$N = p_1 ^ {k_1} p_2 ^ {k_2} … p_r ^ {k_r}$$, How would you check if $$N$$ divides the product of all integers $$x_i$$ So far, use as little memory as possible? Make note of the number of bits used in $$k_1, …, k_r$$,

For part a) we know this for every prime number $$p ne 2, 5$$There is an integer
$$r$$ so that to see if $$p$$ divides a decimal number $$n$$we interrupt $$n$$ in $$r$$-Tuple of decimal places, add these $$r$$tuple and check if the sum is divisible by $$p$$, But $$N$$ is a fixed integer and not necessarily a prime. Is there a way to join the above sentence with any number?

For part b) for the product of $$x_i$$ (call $$y$$) is divisible by $$N$$, then $$y$$ must be divisible by everyone $$p_i ^ {k_i}$$ (call $$a_i$$, Since we get the prime factorization, we can only check if $$y$$ is divisible by $$N$$ by sharing $$y$$ of each $$a_i$$ and stop, if it fails, right? Would this lead to the use $$k_1 * … * k_r$$ Bits?

Am I at least on the right track or am I completely wrong? Any help that understands this problem would be enormously helpful. Many Thanks.

nt.number theory – integer partitions under divisibility constraint

Consider integer partitions of $$x in mathbb {N}$$ of the size $$k$$ on the condition that the partition elements are different and the ratio of an element to each smaller element is a natural number.

example $$f (x, k)$$:

• $$f (17.3) = {1, 4, 12 }$$
• $$f (14,3) = {2, 4, 8 }$$
• $$f (101,4) = {1, 2, 14, 84 }$$
• $$f (4,3) = emptyset$$

As @Henrik has pointed out, one limitation can be expressed as follows: $$x = a_1 + a_1 a_2 + ldots + a_1 cdots a_k$$ and then factor $$x$$, $$x-a_1$$, $$x-a_1 a_2$$etc. to find candidates for successive ones $$a_i$$but because there are usually several options for each successive one $$a_i$$Maybe you have to use a sophisticated search (with backtracking) or a variant of linear programming.

Were these partitions (or series) examined? Is there an efficient algorithm or method to find it? $$x$$ and $$k$$?