Mathematics - 2024-01-21 (page 2 of 2)

« first day (4918 days earlier) ← previous day next day → last day (400 days later) »

Jakobian

20:00

yes but its too early to list all solutions

Next, how would you factor $y^2+2y+2$?

Obliv

isn't that complex roots

Jakobian

Yeah. Thats why you need to modify it by some factor of $10$ to make it factor into integers

Obliv

not sure I follow

Jakobian

$y^3+y^2-2 \equiv (y-1)(y^2+2y+2+10k)\pmod{10}$ where $k$ is any integer

while the polynomial doesn't factor as a real polynomial, you can modify it so that it does

Obliv

so I just choose a k that makes it factorable

Jakobian

20:05

One way of doing this is finding a root of $y^2+2y+2\equiv 0\pmod{10}$, which is simple because you only have $10$ possibilities to check

$y = 0, 1, ..., 9$

and indeed, $2^2+2\cdot 2+2 = 10$

Obliv

can't $y=12$ work as well?

Jakobian

Thats a big number and you want to avoid big numbers

either way, $y = 2$ already covers this case

Obliv

so the solutions are $x=1,x=2$?

Jakobian

we're not claiming anything yet we're still factorizing

its too early to solve this

$y^3+y^2-2\equiv (y-1)(y^2+2y-8) = (y-1)(y-2)(y+4)\pmod{10}$

Obliv

wait but $(y-1)(y^2+2y+2) \equiv 0 \pmod{10}$ works for $y = 1,y=2$, no?

Jakobian

20:10

yes but there'll be more solutions

there's no need to list them right now

it will get complicated enough at the end

in fact, trying to list them right now is just going to potentially confuse you

so please don't

Obliv

Oh i didn't try $k = -1$ I was doing positives :\

Jakobian

$2^2+2\cdot 2+2 = 10$ so $2$ is a root of $y^2+2y+2 - 10 = y^2+2y-8$

alright, now you have $(y-1)(y-2)(y+4) \equiv 0 \pmod{10}$

and the real problem is that you can't say that $y \equiv 1, 2$ or $-4 \pmod{10}$

Obliv

ok so there are some tricks to factoring polynomials in modular arithmetic i see

@Jakobian hmm?

Jakobian

Because $xy\equiv 0\pmod{10}$ doesn't mean $x\equiv 0$ or $y\equiv 0\pmod{10}$

because $10$ isn't a prime number

so what can we do? We can reduce it to prime numbers

$(y-1)(y-2)(y+4)\equiv 0 \pmod{2}$ and $(y-1)(y-2)(y+4)\equiv 0 \pmod{5}$

and then you have six different cases

well technically less

it will boil down to four different cases

First equation is $(y-1)y^2\equiv 0\pmod{2}$ and second $(y-1)^2(y-1)\equiv 0\pmod{5}$

Obliv

I don't understand why $(y-1)(y-2)(y+4) \equiv 0 \pmod{10}$ doesn't mean $y = 1,y=2,y=-4$ are solutions. Doesn't $[0]\odot x \equiv 0 \pmod{10}$

Jakobian

20:17

@Obliv they are solutions but not necessarily the only solutions

you want to solve for every $x$ not just the ones you know that are solutions

Obliv

ah ok

@Jakobian that second equation looks off

Jakobian

yeah so we have $y\equiv 1\pmod{2}$ or $y \equiv 0 \pmod{2}$ and $y\equiv 1\pmod{5}$ or $y\equiv 2\pmod{5}$

@Obliv yes it should be $2$ and not $1$

Technically, first two equations don't mean anything, we always have it either congruent to $0$ or $1$

our equation becomes $y\equiv 1\pmod{5}$ or $y\equiv 2 \pmod{5}$

now to list every solution, you just have to check which numbers among $y = 0, 1, 2, ..., 9$ are congruent to $1$ or $2$ modulo $5$

Obliv

i did not expect this problem to be so involved

Jakobian

yeah, so you would miss one solution if you would end at $1, 2, -4$

because $y = -3$ fits as well

(or what amounts to the same solution, $y = 7$)

Obliv

so to break it down, you want to factor and then break the modulus into primes, then find solutions to the equations

Jakobian

20:24

kind of, it will be harder if you have powers of primes in your modulus

because for $4$ for example, you can't do this approach. You need to be more careful about it

Obliv

can I just brute force $x^3 + x^2 = [2]$ by plugging in $x = 0,1,2,...,9$ ?

Jakobian

yep

or $x = [0], ..., [9]$ I guess

Obliv

right, i guess your way is more strategic and important for large $n$.

Jakobian

that will be less involved but equally as valid

Obliv

wait but do I do this for $Z_{10}$ or $Z_{5}$ and $Z_{2}$ separately? (the brute force method as well)

I'm assuming I also have to break apart into primes

Jakobian

20:38

No you don't have to

You can help yourself checking if $y^3+y^2\equiv 2\pmod{10}$ for a particular $y\in \{0, 1, 2, ..., 9\}$ by checking if it equals to $2$ modulo $2$ and $5$, but its not necessary

Obliv

ok, thank you.

Jakobian

its not necessary, but its convenient

since checking something modulo a smaller number means less calculations

Peter

Does anyone know a formula for the content of the $n$-th Faulhaber polynomial , in other words the positive integer with which we have to multiply the $n$-th Faulhaber polynomial to get a primitive polynomial ? Example : For $n=2$ , we have $\frac{x(x+1)(2x+1)}{6}$ , hence the result should be $6$.

Jakobian

$6$ is also the denominator of second Bernoulli number. It might be related

Peter

Mathworld gives a formula where actually Bernoulli-numbers occur. But I cannot derive the content from this formula.

Jakobian

20:50

One can say that the content divides $(n+1)!$, but maybe thats too rough for your purposes

from what I found, the content $c_n$ is always divisible by $n+1$ and the sequence $c_n/(n+1)$ is given by A195441

oeis.org/A195441

Hope that helps @peter

you might also want to look in this article jstor.org/stable/10.4169/amer.math.monthly.124.8.695

@Thorgott I had this funny idea that may be worth pondering about. Let $S$ be a semigroup, and say that $A\subseteq S$ is dense if $S/A$ is the trivial semigroup. Could we give $S$ some natural topology which makes those dense subsets precisely the subsets that are dense in the topological sense?

Peter

21:12

@Jakobian Those donate-appeals over and over again are really annoying !

Jakobian

22:10

@Peter donate-appeals?

if you mean that the article is not public, you can find it in other places