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?

Infinite divisibility of log normals – MathOverflow

TL; DR: What is the low point of a piece of logarithmic normal distribution?

We know that logarithmic normals are infinitely divisible. What would be the low point of a root of lognormal?

More precisely, let us assume that $ X $ is a log normal. Given an integer $ k> = 2 $, we now, since there is $ X_1, …, X_k $ so that:

$$ mathcal {L} (X) = mathcal {L} ( sum limits_ {i = 1} ^ {k} X_i) $$

From where $ mathcal {L} $ denotes the distribution of a random variable.

$$ Question: $$ Is there a way to simulate directly from $ X_i $Law?

Divisibility – What makes it so difficult to prove that $ 3 $ shares no odd perfect number?

To let $ sigma (x) $ denote the Sum of divisors the positive integer $ x $, The frequency Index from $ x $ is then given by the formula $ I (x) = sigma (x) / x $,

A number $ N in mathbb {N} $ it is said that Perfect if $ sigma (N) = 2N $, Equivalent, if $ N $ is perfect $ I (N) = 2 $,

Euler showed that one odd perfect number $ M $, if available, must necessarily have the form $ M = p ^ k m ^ 2 $, from where $ p $ is the special prime number satisfying $ p equiv k equiv 1 pmod 4 $ and $ gcd (p, m) = 1 $,

Here is my question:

What makes it so difficult to prove that? $ 3 $ does not share an odd perfect number?

MY ATTEMPT

So leave it now $ M = p ^ k m ^ 2 $ be an odd perfect number with a special prime $ p $, Let's assume the opposite $ 3 mid $,

Since $ 3 equiv 3 pmod $ 4, then $ p neq 3 $, (Actually, $ p $ is prime with $ p equiv 1 pmod 4 $ implies that $ p geq $ 5).

So that means that if $ 3 mid $, then $ 3 ^ 2 mid n ^ 2 mid $, Accept that $ 3 ^ 2 || M $ (It means that $ 3 ^ 2 mid $ but $ 3 ^ 4 nmid M $). This implies that $ 13 = 1 + 3 + 3 ^ 2 = sigma (3 ^ 2) mid sigma (M) = 2M $, which means, that $ 13 mid $, This implies that
$$ I (3 ^ 2) I (13) leq I (M), $$
provided that $ p = 13 $, or
$$ 1.60673 approx frac {9507} {5917} = I (3 ^ 2) I ({13} ^ 2) I ({61} ^ 2) I ({97} ^ 2) leq I (M) = 2 $$
assumed $ {13} ^ 2 || M $ implies that
$$ 3 cdot {61} = 183 = 1 + 13 + {13} ^ 2 = sigma ({13 ^ 2}) mid sigma (M) = 2M $$
and
$$ 3 cdot {13} cdot {97} = 3783 = 1 + 61 + {61} ^ 2 = sigma ({61} ^ 2) mid sigma (M) = 2M, $$
where is there no contradiction.

postscript

I dare assume that an odd perfect number is not divisible everything the odd primes below $ 100 (except for $ 3 $. $ 5 $, and $ 7 $ – since we already know that $ 105 nmid M $). However, I have no proof. (Of course, I know that all it takes is a simple calculation.)