trigonometry – Divisibility of Chebyshev Polynomials

I was trying to solve a problem involving an Insect crawling on the Cartesian/Coordinate Plane.

We have an insect on the origin of the coordinate plane, who remembers a particular angle $theta.$ We can command the insect to move in terms of a given string of commands with $text{R’s}$ and $text{M’s}.$

The command $R$ is for “Rotate an angle $theta$ anticlockwise”, the command $M$ for “Move $1$ Step forward in whatever angle you are facing”

For example, if the insect remembers the angle $theta=90^{circ},$ and performs the string MRMRMRM, then it will trace a square path, and will finally get back to the origin.

I need to prove that if $theta$ is an angle such that all of the numbers;
$$cos{theta}$$
$$cos(2theta)$$
$$cos(3theta)$$
$$vdots$$

are irrational, then there is no such string of $text{M’s}$ and $text{R’s}.$

But this is not the real question I want help on, I have an argument that can reduce this to proving an easier fact.
We let $T_k$ denotes the $k^text{th}$ Chebyshev Polynomial $big($so that $T_k(cos{theta})=cos{(ktheta)}big),$ and let $n$ be a natural number.

Lemma 1: We need to prove that, for all nonnegative integers $a_0,a_1,dots,a_n,$ there exists some polynomial $p(x),$ and integers $a, b,$ and a Chebyshev Polynomial $T_m$ such that;
$$p(x)big(a_n T_n(x)+a_{n-1}T_{n-1}(x)+cdots+a_0T_0(x)big)=aT_{m}(x)+b.$$

This is what I need help on. I have tried proving this by induction, but it didn’t help.

Can you help me prove Lemma 1, since I can manage the rest. If you think it is not true, you can tell me about it. In that case you can try to answer the original question.

Thank You and Regards,

MathEagle

nt.number theory – GCD and divisibility of a polynomial

If
begin{equation}
frac {big((-n^4+27m^4-18m^2n^2)(λ-μ)+8mn^3(-2η+λ+μ)big)} {(3m^2+n^2)^2}
end{equation}

for integers $m, n(≠0, ±1)$ with $gcd(m, n) = 1$ is true, (where the numerator is the characteristic polynomial of a 3 by 3 matrix with diagonal entries are $η neq λneq μ$).

Then, is there any technique to factorize the numerators of the following equations

$frac {n^4η+8m^3n(μ-λ)-2m^2n^2big(η-2(λ+μ)big)+m^4big(η+4(λ+μ)big)} {(3m^2+n^2)^2}$, and
$frac{-2m(m^2-n^2)Big(n(μ-η)+m(η-2λ+μ)Big)} {(3m^2+n^2)^2}$ and gives the same numerator as in the first equation ?

Note that in all the above equations, we have $m, n(≠0, ±1)$ with $gcd(m, n) = 1.$

nt.number theory – divisibility of polynomials over partitions

This is a continuation of my earlier MO question.

Given an integer partition $ lambda = ( lambda_1, dots, lambda _ { ell ( lambda)}) $ of $ n $ Where $ ell ( lambda) $ is the length of $ lambda $, associate its conjugate partition $ lambda & # 39; $. Designate with $ lambda & # 39; & # 39; = lambda & # 39 ;, $ 0 found by adding an extra zero to the right end of $ lambda & # 39; $. Also define the following two numbers $ a ( lambda & # 39; & # 39;) _ j = lambda_j & # 39; & # 39; – lambda_ {j + 1} & # 39; & # 39; $ to the $ j = 1,2, dots, ell ( lambda & # 39;) $ and that too
$ b ( lambda & # 39; & # 39;) = # {j: a ( lambda & # 39; & # 39;) _ j> 0 } $.

For example when $ lambda = (4,2,1) $ then $ lambda & # 39; = (3,2,1,1) $ and $ lambda & # 39; & # 39; = (3,2,1,1,0) $ and $ a ( lambda & # 39; & # 39;) = (1,1,0,1) $ and $ b ( lambda & # 39; & # 39;) = 3 $.

Consider the polynomials
$$ f_n (q): = sum _ { lambda vdash n} (q-1) ^ {b ( lambda & # 39; & # 39;) – 1} , q ^ { ell ( lambda) -b ( lambda & # 39; & # 39;)}. tag1 $$

Designate with $ t_n $ the biggest $ t $ so that $ q ^ t $ Splits $ f_n (q) $.

QUESTION 1. Is it true that $ t_n in {0,1,2 } $?

QUESTION 2. (stronger) Is it true that the infinite product $ t_1t_2t_3 cdots = 0 prod_ {k = 1} ^ { infty} 01 ^ {2k} 02 ^ k $?

python text is printed based on the divisibility of the number

The following code outputs a text based on the divisibility of the number. If the number is not divisible by 3.5 and 7 here, it will be printed out. Does using f string implicitly make the type box from integers to strings?

def convert(number):
result = ""
if (number % 3 == 0):
    result += "Pling"
if (number % 5 == 0):
    result += "Plang"
if (number % 7 == 0):
    result += "Plong"
if(number % 7 != 0 and number % 5 != 0 and number%3 != 0):
    result = f{'number'}
return result

Prove the divisibility test by $ 7,11,13 for numbers with more than six digits

Prove the divisibility test $ 7,11,13 for numbers with more than six digits


Attempt:

We know that $ 7 cdot 11 cdot 13 = $ 1001, The for a six-digit number, for example, $ 120,544 $Let's write it as
$$ 120544 = 120120 + 424 = 120 cdot1001 + 424 $$
So we only check the divisibility of $ 424 $ by $ 7,11,13,

Know a number with more than six digits, for example: $ 270060340 $.

$$ 270060340 = 270270270 – 209930 $$
$$ = 270 cdot (1001001) – 209930 $$
$$ = 270 cdot (1001000) + (270 – 209930) = 270 cdot (1001000) – 209660 $$

So we check the divisibility of $ 209660 = 209209 + $ 451, or only $ 451 $,

The test says, however, that: z $ 270060340 $we group three digits from the right:
$ 270, 60, 340 $
then check the divisibility of $ 340 + 270 – (60) $,

How can you prove that?

Divisibility of certain polynomials

Consider the finite sums
$$ F_n (q) = sum_ {k = 1} ^ nq ^ { binom {k} 2} $$
with exponents the triangular numbers $ binom {k} $ 2, When $ n $ strange, it seems that $ F_n (q) $ does not factor over $ mathbb {Z} (q) $, On the other hand, if $ n = $ 2 million is just

QUESTION. is it true that $ F_ {2m} (q) $ is divisible by the product
$$ prod_ {j geq0} (1 + q ^ {m / 2 ^ j}) $$
where the product lasts as long as $ m / 2 ^ j $ is an integer.

Examples. Here is an example:
begin {align}
(1 + q ^ 2) (1 + q) , vert & , F_4 (q); qquad (1 + q ^ 3) , , vert , F_6 (q); \
(1 + q ^ 4) (1 + q ^ 2) (1 + q) , , vert & , F_8 (q); qquad (1 + q ^ 6) (1 + q ^ 3) , , vert , F_ {12} (q).
end

Divisibility when a is odd.

Suppose that $ a | (4b + 5c) $ and $ a | (2 b + 2 c). $ Prove that if $ a $ is so strange $ a | b $ and $ a | c $

So, since $ a | (2 b + 2 c) $ that implies $ a | (4b + 4c) $ in order to $ a | (4b + 5c) $ After deducting two dividends, we receive directly $ a | c $ but I can not understand why it should share $ b $ if $ a $ is odd.

Probability – Divisibility of the square from the sum of two random variables

To let $ I_1, I_2 $ be two independent random variables.
$ I $ is her sum, $ I = I_1 + I_2 $, Now consider the square of $ I $.

$ I ^ 2 = (I_1 + I_2) ^ 2 $

We are interested in when $ I ^ 2 $ can be represented as the sum of two independent random variables. In other words, we can find two independent random variables $ X_1, X_2 $ so that
$ I ^ 2 = (I_1 + I_2) ^ 2 = X_1 + X_2 $

Suppose none of $ X_1, X_2 $ are constants.

In general, it's up NOT true.

Question 1: If $ I_1, I_2 $ are infinitely divisible, will be true above?

Question 2: If $ I_1, I_2 $ follows, what kind of special distribution then becomes true above?