help-28

[forest trail]

There's probably more advanced ways to construct it, but it's just something similar to this

We say that R[x] is the vector space of all polynomials with real coefficients.

I.e every polynomial in the form a1x^n + a2x^(n - 1) + ... + a0 where a0, a1 .. an are real numbered.

Then we define R[x]_n to specifically be the set of all polynomials with degree less than or equal to n, following the above rules.

[torn jolt]

that notation looks like the notation for the coordinate matrix column vector

[forest trail]

Yep

Well, it was the notation that was introduced to me

So its the one I've stuck with :)

[torn jolt]

😂

What math did you take so far bro

[forest trail]

Currently 2nd year (going into 3rd year) undergrad, I've done Linear Algebra, Multivariable Calculus, Vector Calculus, and this semester I'm doing Real and Complex Analysis as well as PDEs

So far, enjoying the ride :)

[torn jolt]

sounds good bro

you definitely seem to like it considering how much you know

vs just getting the material over with like most ppl

[forest trail]

The trick is you gotta find stuff that makes the subject interesting

Once you see the connections within LA (i.e like this one, where its somehow related to polynomials - something you expect wouldn't be linear), then it gets a bit more interesting

If maths was just about memorising formulas and random facts, it'd be pretty pretty boring ngl

So you have to make the most of it.

[torn jolt]

of course

whats ur major

[forest trail]

Maths and Education :)

Yours?

[full forum]

@torn jolt Has your question been resolved?

[full forum]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[glossy valve]

Renato

[torn jolt]

😭

[narrow glacier]

Um... the hell is that? number theorists have gone too far!

2 ⋅ 17^(p+1) + 2 ⋅ 45^(p^2 + p - 1) - 2 ⋅ 187 + 2 ⋅ 6p

So it's such that 3p | that/2

Also, final 12p is irrelevant since it'll necessarily be divisible.

So... 3p | 17^(p+1) + 45^(p^2 + p - 1) - 187

Surely that helps...

[torpid perch]

yes

[torpid perch]

how to continue

[void plume]

okay uhh..... hmm... take the modulo of 3 on both sideeess...???

cause both sides are divisible by 3....

not sure though

cause what im seeing is that the 45^(p^2 + p -1) is divisible by 3

and the remainder should be 0

[narrow glacier]

I suppose that should reduce the search somewhat

[void plume]

17 is congruent to (-1) mod 3

and -187 is uhh

2 mod 3

[hidden belfry]

bro I gave you half the solution earlier 💀

[void plume]

seriously

[hidden belfry]

yea

[void plume]

rip

[hidden belfry]

I was helping this person and renato never replied to the solution 🥀

[void plume]

you did take the modulo of 3 on both sides right...

[hidden belfry]

I just wanted to know what i had so far made sense to the person

[hidden belfry]

ye

[void plume]

yee same here xD

[torpid perch]

you replied 5 hours later

you said you needed to attend a call and never came back

[hidden belfry]

I replied 30 mins later

[void plume]

uhh...

[hidden belfry]

and I pinged you 2 times

after I wrote half of the solution on the channel

[torpid perch]

yes

[hidden belfry]

you never replied

[hidden belfry]

nice

[torpid perch]

you left the groupchat for 5 hours

[void plume]

im sure you got this interaction in the bag 🙂

[hidden belfry]

I didnt

[void plume]

did you go invisible

[hidden belfry]

I can search up the chat?

[hidden belfry]

no

i never do that

[buoyant ravine]

was it in this server? i suggest maybe linking the message if so

[hidden belfry]

yes

Idk how to link a message tho ;-;

[void plume]

uhh

[buoyant ravine]

click more, then copy message link

[void plume]

i only know how to forwaaard and theres the answer

[hidden belfry]

ok

[torpid perch]

so instead of using divisibility we use modular arithmetic is what you are saying

[buoyant ravine]

you should see this on desktop btw

[ancient solar]

#help-23 message

[hidden belfry]

Yea it was this chat

[void plume]

yeah cause we know its divisible by 3

[torpid perch]

sure

[hidden belfry]

thx man

[void plume]

pretty sure Ryxo elaborates more on it in the chat linked above but not sure

[torpid perch]

do you mind we start from scratch

i am getting confused by so many people talking about different things

[hidden belfry]

ok 😭

[void plume]

uhh ryxo you got this right

[hidden belfry]

ye ig

[glossy valve]

Renato

[torn jolt]

very nice

[torn jolt]

undecided tbh

[torpid perch]

this is my channel btw

[torn jolt]

oh my bad lol

[hidden belfry]

aight so we have 6p is divisible by that equation

[torn jolt]

hope i can give you some help back

sadly i might not be able to

[hidden belfry]

we first check if the equation is even divisible by 2 and 3 since 6= 2x3

this gives us valuable info as well

to do this we use modulus bc its easier and divisibility rules vary with primes

so first we check if equation is divisble by 2

we set it to modulo two and i mean its pretty obvious its divisible by two

bc the coeffients and constands are all even so the equation is divisble by 2

Next we check modulo 3

and we get that 45 is obviously 0 mod 3 and so is 12 p

17 is congruent to 2 mod 3 and -374 is congruent to 2 mod 3 since we get -1 mod 3 and that equals 2 mod three

[torpid perch]

yes, so?

what do we conclude from mod 3

[hidden belfry]

so this simplifies to 2*2^(p+1) +1

we move that over which gives 2^2(p+1) is congruent to -1 mod 3 which is also conrguent to 2 mod 3

[torpid perch]

can you latex what you are saying

[hidden belfry]

Uhhhhhhhhhhhhhhhhhhhhhhhhhh

lemme try?

[torpid perch]

do you have paper at hand?

[hidden belfry]

yea

imma try latex tho

[torpid perch]

ok

[hidden belfry]

the above equation simplified down too

$2^(p+2) ≡ 2 (mod 3)$

[glossy valve]

Ryxo
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[hidden belfry]

:/

lemme try again

$2^(p+2) congruent to 2 (mod 3)$

[glossy valve]

Ryxo

[torpid perch]

$2^{p+2} \equiv 2 \pmod 3$

[hidden belfry]

yes this

its congruent to it tho

[glossy valve]

Renato

[hidden belfry]

yep

now look at the powers of 2

[torpid perch]

i dont think I follow

[hidden belfry]

in mod 3 ofc

[hidden belfry]

where are you stuck?

[torpid perch]

well

[hidden belfry]

...

we are testing the equation to see if it is even divisible by 2 and 3

if the above is false then no primes will divide the equation whatsoever

[torpid perch]

can we start feom scratch

because we have 6p

[hidden belfry]

bro 😭

[torpid perch]

suddenly everything got complicated

[void plume]

conceptually, if you want the number to be divisible by 6, it has to be divisible by both 2 and 3

[hidden belfry]

because we want to find what value of P lets 6P divide the equation

Its not just P

6P must divide it

so 2 and 3 must divide the equation

[torpid perch]

sure

[hidden belfry]

do you get that?

[torpid perch]

2 divides it

[hidden belfry]

yep

[torpid perch]

3 on the other hand

[hidden belfry]

now we are checking 3

[torpid perch]

shit gets messy

[hidden belfry]

its an elegant solution once you muddle throught the mucky part

[hidden belfry]

so we this rn

we need to look into powers of 2 in mod 3

$when

2^1 = 2 (mod 3)

2^2 = 4 = 1 (mod 3)

2^3 = 8 = 2 (mod 3)

2^4 = 16 = 1 (mod 3)$

[glossy valve]

Ryxo
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[hidden belfry]

finally

do you get what this means?

[torpid perch]

its either 1 or 2

[hidden belfry]

Uhm yea

[void plume]

couldn't you just change the 2 to -1 mod 3 and that'd make more sense since -1^odd = -1

[hidden belfry]

yea this works too

[hidden belfry]

ok so the main thing is that for any odd power we get 2 and for any even power we get 1

do u understand what this means?

[void plume]

$(-1)^{p+2} \equiv -1 \pmod 3$

[glossy valve]

ERROR

[void plume]

Essentially this

[torpid perch]

how

is this fermat?

[hidden belfry]

no

[void plume]

no

[hidden belfry]

that ones gonna come in use soon

[torpid perch]

can we start from scratch

[hidden belfry]

this is correct but IDK if it actually works like that

[void plume]

pretty sure it does since

$17 \equiv -1 \pmod 3$

[glossy valve]

ERROR

[hidden belfry]

you can try expanding it to other numbers

[hidden belfry]

wait ur right

[void plume]

and $-187 \equiv -1 \pmod 3$

[glossy valve]

ERROR

[torpid perch]

you guys send too many unrelated messages but like I have to read through the messages and the math in between them, there is no structure in this

[hidden belfry]

Bro there is no structure in this already 🥀

[torpid perch]

we were evaluating the mod 3

[hidden belfry]

you wanted to restart from scratch again 😭

[torpid perch]

then, something happened

[hidden belfry]

atp you just want the answer???

[torpid perch]

!noans

[full forum]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[hidden belfry]

i mean ur not understanding what we are saying

[void plume]

what about i take it from the mod 3 part

[hidden belfry]

???

[torpid perch]

all am I saying is that I already lost track of where we were

[hidden belfry]

Ok so do you understand the math we are doing rn?

I dont mean it in a rude way

[glossy valve]

Renato

[hidden belfry]

Im just asking if you are following

[torpid perch]

so, we need to find which primes p have the condition such that 6p divides that

for 6p to divide that, we need 3 and 2 to divide this

[hidden belfry]

yep

[torpid perch]

2 clearly divides it

[hidden belfry]

yep and we are doing 3

and if we put the equation into mod 3 it simplifies down to this

The latex was deleted :/

[hidden belfry]

we get this

[torpid perch]

how

[hidden belfry]

now we have to find what powers of 2 is equal to 2 mod three

[hidden belfry]

bro you put simplify it with the properties of modular equations

do you not get what we did?

[torpid perch]

i am trying to do that

[hidden belfry]

if you don't get mods then I can't explain it to you

divisibility is basically all about modulus

[torpid perch]

dont be rude

I know the basics of modular arithmetic

[hidden belfry]

Im not trying to be its just that IDK if you actually understand mods

[torpid perch]

is just that we went too fast and the messages got lost in between the chit chat you were doing with the other guys

[hidden belfry]

we were talking abt useful stuff related to the problem 😔

so do you understand how to simplify an equation with mods?

[torpid perch]

yes but jumping to different parts of the problem

[torpid perch]

12p = 0 (mod 3) for example

[hidden belfry]

yes

do you understand how to simplify powers with mods?

[torpid perch]

no

[hidden belfry]

like numbers raised to powers

[torpid perch]

maybe fermat

[hidden belfry]

mmmm

ok

so do you know multiplication rule for mods?

[torpid perch]

no

[hidden belfry]

what do you know about mods then?

[torpid perch]

maybe i know it but not as a 'multiplication rule', again, dont be rude

[hidden belfry]

im not trying to be

im just genuinly trying to get what you understand on the problem

so what happens when you multiple two mods together?

[torpid perch]

what?

[hidden belfry]

for example we have a is congruent to 2 (mod 5) and b is congruent to 3 (mod 5)

what happens when you have ab

in mod 5

[torpid perch]

a + b = 0 (mod 5)

[hidden belfry]

No I mean a*b

like a times b

[torpid perch]

who knows

[hidden belfry]

I gave you what a and b is congruent to in mod 5

thats all you need to know it

[torpid perch]

i guess you can say that ab = 1 (mod 5)

[hidden belfry]

yes

[torpid perch]

but im not sure

like, i am not sure if this holds

[hidden belfry]

dude you need to understand the rules of modulus in order to understand the problem

[hidden belfry]

it does

its the multiplication rule in modulus

[torpid perch]

if I would know everything I wouldn't be asking for help, dont be rude

[hidden belfry]

In this problem if you understand the rules of modulus you can still mess up because its not just solve a mod congruency

[torpid perch]

the non linear modular arithmetic complicates me

generally speaking

[hidden belfry]

yea

hey man

Imma not be rude

but you NEED to learn the rules of modulus bf you tackle this problem

[void plume]

or i can try to teach you stuff

[hidden belfry]

youll also need it to take the class that gave you this problem

[hidden belfry]

That would take a while tho

they are easy anyways

[void plume]

since 17 is congruent to -1 mod 3, and its 17^n, its essentially (17)(17)(17).....(17) n times, which is same to (-1)(-1)(-1)....(-1) n times

[hidden belfry]

yea but i don't think renato understands this stuff yet?

@torpid perch I think the best I can do today is just give you the answer cause I'm not going to teach modulus to you. Its very easy and you'll understand the rules quickly

[torpid perch]

dont give me the answer

[hidden belfry]

would you like me to give you a website or vid on it so you can learn all the rules?

then maybe tommorow we can try this again

cause I don't have time today

its getting late for me

[torpid perch]

I would prefer if you dont help me again because you make me feel bad about me taking this class and you are always in a hurry

[hidden belfry]

Bro how do I make you feel bad?

I'm just genuinely asking what you know

its what I need to understand the situation

and Im not being overly rude abt it like Im calling you stupid or saying that the solution is obvious

Also I took my time to try to help you with this problem

I left and came back

and spent another 30 mins on this same problem

and all I get in return is Your being rude and I dont want tou to help me again?

[subtle verge]

hi!
took a look at the q just wanted to ask why (mod 3) both sides was what you decided to do?
guessing its an intuition thing that I just don't have yet?

[hidden belfry]

You mod 3 the equation

[void plume]

actually just ping a helper or i can help

[hidden belfry]

Fr 😔

I reccomend ANN

see if she can help

[hidden belfry]

you should help

[white karma]

lol

no need to get heated up like this guys

its alright

[hidden belfry]

bro

[white karma]

imo both sides made some mistakes in communication

but thats not a problem

[torpid perch]

me aswell(?)

[subtle verge]

ah
well I thought to (mod 6p) the equation altho mayb theres something there that I haven't learned yet hmmm... interesting
(mod 3) does sound easier
sorry bout my interjection gonna dip

[white karma]

all good, just stop fighting

[void plume]

alr bye

[white karma]

all of you are just wasting your time by fighting over stuff like this

[hidden belfry]

its alright
I think you can do that too its just that seperating into 2 and 3 is easier to deal with

[torpid perch]

nobody is fighting, i am trying to solve my homework problem

[void plume]

here comes another person into the conversation.... time for me to leave... i just couldn't bear to see the interaction

[hidden belfry]

NOOOOO

[white karma]

sure, but it happened along the way

[hidden belfry]

come back

[white karma]

anyways

@torpid perch

time to solve the problem

[torpid perch]

how

[hidden belfry]

Bro I tried my best aight 😔

[white karma]

thats what we will get into rn

[white karma]

np bro one can always try again

(tho you are not obliged lol)

[hidden belfry]

yea

[white karma]

ok so this is the question

we want to find all such primes p

since we are already requiring 6 to divide the RHS (right hand side)

lets assume first that p is different than 2 and 3

[hidden belfry]

Btw you might wanna explain to renato some more modulus concepts
Not being rude tho renato didnt understand numbers raised to powers and stuff so you may need to explain that

aight imma dip

[white karma]

because if not for these cases we can live in peace and apply nice theorems like fermat's little theorem

[white karma]

have a great day/night

[torpid perch]

well since 6p divides that then necessarily 6 AND p divides that

[white karma]

right

[torpid perch]

alright, since we can take that 6 divides this by granted and so 2 and 3 come as granted

[white karma]

so we first rule out the prime factors of 6 for now

and then deal with them in separate cases

[torpid perch]

6 = 2x3

[white karma]

right

so we assume p is different than 2 and 3

we see what we get for p

and after that we see if p=2 or p=3 (or both) work or no

all good?

[torpid perch]

good so far

[white karma]

alright

so assuming p neq 2 and p neq 3. lets reduce the RHS mod p

[torpid perch]

alright okay

[white karma]

so we want to see what happens to $2\cdot 17^{p+1}+2\cdot 45^{p^2+p-1}-374+12p$ mod $p$

[torpid perch]

12p = 0 (mod p)

[white karma]

since we have p's in the exponents, and p is a prime, what comes to mind?

[white karma]

first there is this

[glossy valve]

ali yassine

[white karma]

right mb

[torpid perch]

something something fermat is little

[white karma]

nice

so what does fermat's little theorem say?

[torpid perch]

i don't remember

[white karma]

alright np i will remind you

fermat's little theorem says the follow:\Let $p$ be a prime and $a\in\mathbb{Z}$ be any integer. Then $a^p\equiv a\pmod p$

[glossy valve]

ali yassine

[white karma]

you may also see it in another form

[torpid perch]

okay thats actually understandable

no idea about the proof but everything good so far

[white karma]

which is : let $p$ be a prime and $a\in\mathbb{Z}$ be an integer such that $(a,p)=1$, then $a^{p-1}\equiv 1\pmod p$

[glossy valve]

ali yassine

[white karma]

here (a,p) denotes the gcd of a and p

[torpid perch]

yikes

[white karma]

so this says that if p does not divide a then this congruence follows

[torpid perch]

a,p is gcd?

[white karma]

(a,p) yea

[torpid perch]

oof

[white karma]

at least some texts use that

others just say gcd(a,p)

so it depends on the author etc..

ok, so far so good?

[white karma]

alright then, now we want to use that info

[white karma]

lets try to use this

[torpid perch]

wait

whats the difference between the first theorem and the second one

[white karma]

alright let me tell you

[torpid perch]

because the first theorem is simple but second looks daunting

[white karma]

no its actually more or less the same

there is only one difference

so i have a question for you

if p|a, then what can you say about: a mod p?

[torpid perch]

yes

p | a <=> a = 0 (mod p)

[white karma]

nice

what about a^p mod p in this case?

[torpid perch]

well

using lil fermat

[white karma]

no

if a\equiv 0 mod p

[torpid perch]

it doesnt apply?

[white karma]

what does that say about a^p mod p?

[torpid perch]

a^p = 0 (mod p) still

[white karma]

i want you to analyze this without resorting to fermat because the reason i am doing this is to explain the difference between the 2 "versions" i presented

[white karma]

why?

[torpid perch]

because

if p | a then p | a^n

[white karma]

and in general p|a^n for any n

nice

[torpid perch]

that, is very fundamental yeah, idk why I said n = 2

[white karma]

ok so if p|a then from the above we can conclude that a\equiv a^p mod p right

[torpid perch]

a^p equiv a (mod p) , yes

the other way around aswell, but is stronger implication

i think (?)

[white karma]

ok so now if p does not divide a

then (a,p)=?

right

[torpid perch]

wait

if p does not divide a what can we say about their gcd?

they dont have any common divisors

other than 1

[white karma]

so whats their gcd?

[torpid perch]

1 of course

[white karma]

right

this property is special to prime numbers

a prime number p either divides a natural number or has no common divisor with it other than 1

since p doesnt divide a, (a,p)=1

[white karma]

so assuming this is true

we have a^p\equiv a mod p in this case too

but this time a^p and a are not 0 mod p

[white karma]

since the (a,p)=1, we can divide both sides of this congruence by a just like with normal equations

did you know this before?

[torpid perch]

wdym

ahh because a and p are coprime

[white karma]

right

[torpid perch]

gcd(a,p) = 1 so p does not divide a

[white karma]

nice

so a is not 0 mod p

[torpid perch]

yes

[white karma]

and if we take this for granted, we conclude that a^p is not 0 mod p too

[torpid perch]

because the multiplicative inverse exist?

[white karma]

exactly

[torpid perch]

kind of, i am newbie to NT

[white karma]

np, you will get used to it with practice

[torpid perch]

but i encountered multiplicative inverse when covering China Theorem

[white karma]

ohhhh i see

[white karma]

so multiply by the inverse a^{-1} of a

[torpid perch]

so?

[white karma]

what do you get?

[torpid perch]

I SEE

dude 😎

[white karma]

but here comes the question

[white karma]

does this work if p|a?

[torpid perch]

idk why nobody explained it that before, its just the same theorem but multiplying by the multiplicative inverse, saying gcd = 1 and so the inverse exists

[white karma]

right

[torpid perch]

of course not

because then their gcd is fucked

[white karma]

lmao

yea well p|a so a^{p-1} equiv 0 mod p

it cant be equiv 1 mod p

ok so now lets tackle the problem

$2\cdot 17^{p+1}+2\cdot 45^{p^2+p-1}-374+12p$ mod $p$

[glossy valve]

ali yassine

[white karma]

we will use fermat's little theorem to simplify quite a bit here

[torpid perch]

this is when it gets nasty

[white karma]

sotrue

[torpid perch]

12p = 0 (mod p)

[white karma]

right

[torpid perch]

17^p = 17 (mod p)

[white karma]

what else

[torpid perch]

17^{p+1} = 17 x 17 (mod p)

[white karma]

no be careful

the first congruence is right

the second is not

now it looks good

[torpid perch]

well I get stuck

45^p = 45 (mod p)

then

45^(p^2) = 45^p = 45 (mod p)?

[white karma]

well its not good to leave the -1 alone

because we dont even know if 45^{-1} exists or no right

we know that p is not 2 and not 3

[torpid perch]

oh my

[white karma]

but 45=3^2.5

so if p is 5 then there is no 45^{-1}

so we should keep the -1 with some other thing

[white karma]

this is a good step

[white karma]

and this too

you only have to do one thing now

which is the following: $45^{p^2+p-1}=45^p45^{p^2-1}$

[glossy valve]

ali yassine

[white karma]

now 45^p\equiv 45 mod p

so now you have 45.45^{p^2-1}

now there is a way to get rid of the -1 in the exponent

what is it?

[torpid perch]

wait

[torpid perch]

unsure

we multiply by 45?

[white karma]

you already have this 45 somewhere right?

[white karma]

from this

[torpid perch]

so we have (45^2)(45^{p^2})

[white karma]

and this

[white karma]

why 45^2?

[torpid perch]

I'm not sure

$45^{p^2+p-1}=45^p45^{p^2-1}$

[glossy valve]

Renato

[white karma]

ok so is everything good till here?

[torpid perch]

yes but

I have 45 . 45^{p^2 - 1}

[white karma]

right and now 45^p\equiv ?? mod p

[white karma]

nice

[torpid perch]

45

[white karma]

so?

you are multiplying 2 things with same base

what do you do with the exponents?

[torpid perch]

qh right

right right

45 . 45^{p^2 - 1} = 45^{p^2}

[white karma]

nice

[white karma]

now this

[torpid perch]

and 45^{p^2} = 45 (mod p)

[white karma]

great

[torpid perch]

2x17x17 + 2x45 - 374 (mod p)

i have

[white karma]

so now we have \$2\cdot 17^{p+1}+2\cdot 45^{p^2+p-1}-374+12p\equiv 2\cdot 17^2+2\cdot 45-374\pmod p$

[glossy valve]

ali yassine

[white karma]

nice now lets calculate this

,calc 2(17^2)+2(45)-374

[glossy valve]

Result:

294

[torpid perch]

so 294 = 0 (mod p)?

[white karma]

right

[torpid perch]

is 294 prime?

[white karma]

no, its even and >2

[torpid perch]

,w 294 prime factorization

[glossy valve]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by renatoxbz

[white karma]

and now what did we assume about p?

[torpid perch]

so p is 7

[white karma]

ok does 6 also divide this?

we need that too right

[torpid perch]

6 | 294

[white karma]

so we are good to go

so we found p=7

we still have 2 cases

p=2 and p=3

[torpid perch]

ah this is when p > 3

[white karma]

if p=2 then 6p=12

so we need to check whether $6p\mid 2\cdot 17^{p+1}+2\cdot 45^{p^2+p-1}-374+12p$ or not after we replace every $p$ with $2$

[glossy valve]

ali yassine

[torpid perch]

alright okay

[white karma]

as usual 6p|12p so no need for the term 12p

[torpid perch]

,calc 2(17^3) + 2(45^(4+2-1))-374+12(2)

[glossy valve]

Result:

3.69065726e+8

[white karma]

no dont do this

in the exam you are probably not allowed to use calculator right?

[torpid perch]

fair enough yeah right

12 does not divide 2x17^3

[white karma]

do you know euler's generalization of fermat's little theorem?

[torpid perch]

not really no

[white karma]

it has to do with euler's totient function φ

[torpid perch]

not really no

[white karma]

alright i see

ok so i think its good if you use chinese remainder theorem here

to make this congruence into a system of congruences

[torpid perch]

how

care to elaborate on that?

[white karma]

wait wait nvm no need for that

ok so what is 17 mod 12?

[torpid perch]

is 5

[white karma]

nice

so 17^3 is 5^3

which is 125

and 125 mod 12 is?

[torpid perch]

is 5

[white karma]

nice

so 2.17^3 is 2.5=10 mod 12

thats the first term

next

2.45^{2^2+2-1}=2.45^5

45 mod 12 is?

[torpid perch]

well

12 36 48

[torpid perch]

9

[white karma]

right but 9 is a big number especially since we want to raise to an exponent of 5 later

so lets write it as -3

[torpid perch]

2x9^5 = 2x9 = 18 = 6 (mod 12)

[white karma]

wait how did you get that

thats correct lol but how did you get it

[torpid perch]

idk

[white karma]

hahaha np

that essentially follows by euler's theorem that i told you about earlier

[white karma]

you can also reach this by calculating 9^2 mod 12 and then repeating the process

[white karma]

you can also do this

2.(-3)^5 mod 12

(-3)^5=-3^5=-243\equiv -3 mod 12

[white karma]

so this is -6 mod 12

or in other words 6 mod 12 (since -6\equiv 6 mod 12)

[torpid perch]

how so?

[white karma]

Start by calculating 9² mod 12

[torpid perch]

if I am working with single number exponents in a congruence I just ignore them

idk if its good or not

[white karma]

so you discard the exponents immediately?

[torpid perch]

i mean

[white karma]

Like you write 9⁷\equiv 9 mod 12 for example ?

[torpid perch]

say for example you have

a = x (mod p)
a^n = ? (mod p)
summing n times 'a' will get me a^n
and that is
a^n = nx (mod p)
but then idk how to complete the proof

what i said was a mistake @white karma

[torpid perch]

here

[white karma]

I see yeah that's wrong

[torpid perch]

this is a mistake

its a mistake from my part

[white karma]

yea np, the result is correct although from false reasoning

Just next time verify something before you write it (in the exam for example)

Because here I can correct you if you are wrong

But in the exam if you are wrong then you will lose some grades

[white karma]

So you were asking this question

what is 9² mod 12

[torpid perch]

idk

81 mod 12

but that intuition only works when the n in 9^n is small

[white karma]

Write. Because higher powers of 9 are hard to compute by hand

Here it works nicely thankfully

[white karma]

But you can do this

This makes the calculations easier

Because it's not too hard to find that 3⁵=243

[torpid perch]

3^10000

[white karma]

Whereas 9⁵ is much harder to compute by hand

[torpid perch]

what if the exponent is big

[white karma]

So $2.17^3\equiv 10\pmod {12}$ and $2.45^5\equiv 6\pmod {12}$ so $2.17^3+2.45^5-374\equiv 10+6-374\equiv-358\pmod {12}$

[torpid perch]

pmod{12}

it's the exponents

[glossy valve]

ali yassine

[white karma]

You always try to convert the base to a smaller base if possible like writing 9 as -3 here

But if the exponent is big also on 3

Then you first get rid of a chunk of the exponent then proceed

[torpid perch]

wdym

1000^999

mod 3

1^999 = 1

okay that was easy, but let me think something harder

[white karma]

For example, 3^{45}=(3³)¹⁵=(27)¹⁵\equiv 3¹⁵\equiv 3⁵\equiv 3 mod 12

[torpid perch]

you can take mod in the exponent?

[white karma]

No

You can only use Fermat (or Euler if you can but you didn't take that)

[white karma]

I didn't do that

3⁴⁵=(3³)¹⁵=27¹⁵\equiv 3¹⁵ mod 12 because 27\equiv 3 mod 12

[torpid perch]

ok

[white karma]

Then 3¹⁵=(3³)⁵=27⁵\equiv 3⁵=243\equiv 3 mod 12

[torpid perch]

ok good enough for me

where were we

?

[white karma]

Here

[torpid perch]

So $2.17^3\equiv 10\pmod {12}$ and $2.45^5\equiv 6\pmod {12}$ so $2.17^3+2.45^5-374\equiv 10+6-374\equiv-358\pmod {12}$

[glossy valve]

Renato

[white karma]

I just put everything we got together

now we have to see if this is 0 or no

-358=-360+2

[torpid perch]

how is 245 = 6?

[white karma]

No not 245

[torpid perch]

it's not

[white karma]

2.45

[torpid perch]

yes how is it 6

[white karma]

2x45

[torpid perch]

45 = 9

[white karma]

Well you just found it a few moments ago

[white karma]

Ok

2.9=18

Which is 6 mod 12

[torpid perch]

no no

2.9^5

[white karma]

Yea that too is 6 mod 12

[torpid perch]

how

[white karma]

Let's do it again

[torpid perch]

is what i am asking

[white karma]

2.9⁵\equiv 2(-3)⁵ mod 12

All good ?

[torpid perch]

okay

[white karma]

(-3)⁵=-243

[torpid perch]

suure okay

[white karma]

And -243\equiv -3 mod 12

[torpid perch]

or we can use fermat

[white karma]

No

[torpid perch]

since exponent is prime

[white karma]

Fermat is only for primes

[white karma]

What you need is prime modulus

[torpid perch]

5 is prime

ah

[white karma]

the modulus here is 12 which isn't a prime

[torpid perch]

imo this is almost never the case

[white karma]

You can use Euler but you didn't cover that

[white karma]

lmao

[torpid perch]

when fermat really shines is when using china

i think

[white karma]

Right

you can use Chinese remainder theorem

[torpid perch]

no no, it's fine

we can, you want to?

[white karma]

To make any congruence a system of congruences with prime moduli

[white karma]

No we are near the end

I would've done it with you

But after this I should sleep

Since I have to wake up for uni

[torpid perch]

ok let's finish this bad boy

[white karma]

Right

[white karma]

So all good with this?

[torpid perch]

sure

[white karma]

Nice

Now -358=-360+2\equiv 2 mod 12

[torpid perch]

so we get a contradiction

[white karma]

Not really a contradiction but rather that p=2 doesn't work

Well you can probably write it in a way that makes it a contradiction

I get what you mean tho

The last case to check is when p=3

[torpid perch]

mod 18

[white karma]

Right

Tho I gtg sleep now

I will leave that to you

[torpid perch]

?

[white karma]

I trust that you can solve it with no problems

[white karma]

The thing is that I have a bit less than 3 hours to sleep before waking up for uni

[torpid perch]

alright have a good one buddy

[white karma]

So if I stay awake any longer I will sleep in the lecture or oversleep and skip the lecture

[torpid perch]

i will see if i manage to finish this crap

[white karma]

It was fun doing the exercise with you

[white karma]

You too

[white karma]

You can, just trust yourself

It's essentially the same thing we did with p=2

Just different things to calculate

Find each term mod 18 separately then join them all together

At the end you should reach that ||p=3 doesn't work too so the only p is p=7||

gn, have a great day/night!

[torpid perch]

The other guy was having trouble with me not knowing the basics

[torpid perch]

Why is that?

[white karma]

np don't mind everything that happens lol. Forget about it and dw

[torpid perch]

I just think patience comes when you get more mature

[white karma]

Tough question

I don't really know why immediately

[torpid perch]

The problems are nice, innit?

[white karma]

It's good, but I don't like these problems too much

number theory has much nicer problems

If you study analytic and/or algebraic number theory some day you will know what I mean

[torpid perch]

Well is my first course on intro to proofs, this is not exactly nt course

[torpid perch]

I see

[white karma]

Ohhhh I see

I gtg sleep now, cyaaa!

[torpid perch]

Cya gn

Ty for the help

[white karma]

gn

[white karma]

np, you did the work!

[torpid perch]

.close

[full forum]

Channel closed

Closed by @torpid perch

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[celest bay]

I need help

Plz

[slate violet]

where are you stuck?

[celest bay]

It’s not like all the other vector questions I have done

It’s different

[slate violet]

okay, ignore the 15 km/h current for now

you have a right triangle where the height is 10m

[celest bay]

Hmm ok

[slate violet]

and then the 85 degrees gives you one of the angles inside the triangle

draw it out first

[slate violet]

nope, that angle is not 85 degrees

[celest bay]

Oh

[void nova]

Otherwise that would be almost a right angle

[celest bay]

There?

[slate violet]

yep, and the height is 10m again

[celest bay]

Yes got that

[slate violet]

right, so what's the length of the hypotenuse?

[celest bay]

Well it’s 3km/h

[slate violet]

no, that's not a length

[celest bay]

But that’s not a length

[void nova]

Use trig definitions

Such as sine

[celest bay]

So it would be 10/sin85

That’s basically 10m

[slate violet]

but you shouldn't round early

[celest bay]

Yeah ok

[slate violet]

right, so you know the length of the hypotenuse now

do you see how you can use distance = speed * time here?

[celest bay]

To get which distance?

[slate violet]

let me draw it out

the idea is when you combine the horizontal and vertical components, you get this diagram

but you want to convert the 15 km/h to a distance

[celest bay]

Oh ok

But wouldn’t that depend on the time

[slate violet]

exactly!

[celest bay]

So what would the time be then?

[slate violet]

that's what I'm asking you

[celest bay]

Ok

Well time = distance/speed

[slate violet]

yep

[celest bay]

But we don’t know the distance of the 15km/h

[slate violet]

but you know the distance of the 3 km/h!

[celest bay]

Ohh

So if 3km/h = 10/sin85

Then 15km/h = 50/sin85

Just times it by 5

[slate violet]

yep, that's some pretty slick reasoning actually

(in 1 hour, or in the same time period in general, there is 5 times more distance)

[celest bay]

Ok so now

Time = distance / speed

[slate violet]

great, you're almost done
there's just one more length you need to find here

actually, you don't need to find the time anymore

[celest bay]

Well now I have two vectors

[slate violet]

there are 3 distances in this picture you needed to find

so far you've found 2 of them

[celest bay]

I just need to find the resultant vector

And the x coordinate is how far down streak the fish traveled

[slate violet]

yeah, so that would be the base length of your right triangle here

[celest bay]

Would it tho

Hm ok

It equals 10cos85/sin85

[slate violet]

yep, that's correct

[celest bay]

That doesn’t seem right

[slate violet]

think about it, your triangle is very, very thin

cause the other angle in the right triangle is 5 degrees

so the hypotenuse will be nearly the adjacent

and the opposite side to the 5 degree angle will be very small

[celest bay]

Ohh ok

[celest bay]

So is this the answer?

Or do I need to account for the 15km/h stream

[slate violet]

now you have to put everything together

the red hypotenuse is what we want to find

[celest bay]

So putting together

Would give: (10cos(85) + 50)/sin(85)

Around 50.1m

[slate violet]

you didn't use Pythagoras

remember, we don't know any of the angles in this big triangle
other than the right angle

[celest bay]

We know 85 don’t we

Ok well the first little part of the bottom is 10/tan5

That seems wrong

But tan = o/a

[slate violet]

yeah, that's wrong

wait

I labelled the angles wrong, one sec

[slate violet]

you've already found the base length

and you've already found the 15 km/h length as well

[celest bay]

Which one is the 15km/h and which is the 3km/h

[celest bay]

Oh ok

So the base is the 15km/h + the other part

So: 50/sin85

Plus 10/tan85

Which is around 51.1m

[slate violet]

right, that's correct

and then from way in the question where it says that the stream is 10m wide, that's how you know the height is 10

[slate violet]

also, give at least 3 dp before your final answer

[celest bay]

Yeah ok

51.066m

Wow that’s pretty cool

[slate violet]

did you get 52.0 m as the final answer?

[celest bay]

Not really

[slate violet]

you still didn't use Pythagoras!

that's the length of the base

[celest bay]

Isn’t that what you want tho

[slate violet]

no

[celest bay]

You just want the distance down the stream

[slate violet]

oh wait, I misread the question

yeah okay I see now

alright, if you're done for now, type .close

[celest bay]

Ok thank you

?close

.close

[full forum]

Channel closed

Closed by @celest bay

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[lyric narwhal]

Computing the fourier series of a function. I am calculating b_n and the approach i have taken is undefined for n=1 so i solved for the rest of bn excluding when n = 1, i cant just throw it away right, the solution doesnt include that n=1 part tho. I have my working if you need more context

[lyric narwhal]

<@&286206848099549185>

[full forum]

@lyric narwhal Has your question been resolved?

[lyric narwhal]

does my question make sense?

[tired quarry]

Not sure why would you want to omit b1

[lyric narwhal]

well at n=1 its undefined right

[tired quarry]

Your solution isn't, but the integral is

[lyric narwhal]

yeah

[tired quarry]

You just have to make special case for n=1

[lyric narwhal]

yes i did that and i got a nonzero answer but they didnt include that in their solutino

[tired quarry]

Maybe they wrote it in some different form and b1 appears elsewhere

[full forum]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[lyric narwhal]

.reopen

[full forum]

✅ Original question: #help-28 message

[lyric narwhal]

when we are finding bn where n = 1
we find that the odd n is equal to 0, for the general n formula, does that mean b1 is 0 automatically

if we compute b1 case we get integral from 0 to pi/2 of sin^2(x) dx which is equal to pi/4 which evidently isnt 0

[tired quarry]

That's bc, from you calculation, we have integral of cos(x(1-n))

Which for n=1 is cos(0)

Which is constant

Which means that your transformations work only for n>1

[full forum]

@lyric narwhal Has your question been resolved?

[full forum]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[brisk minnow]

is my part a correct?

[brittle sun]

Minor nitpick, ϕ isn't abelian, G is

[brisk minnow]

oh yea sorry 😭

[brittle sun]

Also the order in which you've written the equalities don't make sense

[brisk minnow]

how would it be best to write them

[brittle sun]

Start with h1h2 and end with h2h1

[brisk minnow]

Ah alright

[brittle sun]

It would be the same but every equality would be an obvious step

[brisk minnow]

yea I see what u mean

also yh for part b, I’m a bit confused on where to start, like what do I consider

I want to show g1g2 = g2g1

how do I utilise the fact that phi is injective

[brittle sun]

Every g1 has a unique image h1

[brisk minnow]

Hm alright I’ll see what I can do with that

[queen gull]

Intuitivly (b) means that G is isomorphic to a subgroup of H, and of course a subgroup of an abelian group is abelian

[brittle sun]

I doubt they've done isomorphism theorems yet

But if they have then yes

[brisk minnow]

would this work?

Like using the fact that the pre image is unique

[brisk minnow]

I haven’t done first theorem of isomorphism yet so unsure what this is

[queen gull]

Haven't mentioned it even

And I was trying to give intuition

[brisk minnow]

Sorry

[queen gull]

Its fine dw

[brittle sun]

Technically that is first isomorphism theorem lol

[brittle sun]

You can't use ϕ^-1

Cuz ϕ might not be surjective

But you have the right idea

[brisk minnow]

Oh yea

Hm

[brisk minnow]

does this show

That g1g2 = g2g1

[brittle sun]

Yup

Since ϕ is injective

[brisk minnow]

yey Ty

[brittle sun]

ϕ(a) = ϕ(b) means a = b

[brisk minnow]

I’ll include that ❤️

Ty

.close

[full forum]

Channel closed

Closed by @brisk minnow

Use .reopen if this was a mistake.

[full forum]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[still epoch]

what is d1 and d2

oh alright

i assume the lines are the intersection of two planes?

such that y+3z+1=0 is a plane etc

ok

and alpha and beta are both positve and real (not variables)

R^2 means what?

shouldnt it just be R

never knew that

ok

definitely to do with normals

[sick vortex]

find direction vector of line 1 and line 2

and check whether they're scalar multiples of each other

yes

u can use gaussian elimination ig

solve for a condition on alpha and beta such that the system is consistent

[full forum]

@chrome haven Has your question been resolved?

Channel closed

Closed by @chrome haven

Use .reopen if this was a mistake.

[full forum]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[cold sluice]

Can I please get help with this? When I try to use the quadratic equation I get -60 in the discriminant but they're asking me to graph the points of intersections/solutions to the system of equations, and when I tried to divide 2x^2 -2x +8=0 by 2 and then subtracting 4 from both sides and then taking the common x out of the LHS and solving for that I got the coordinates 4,9 and 5,13 but the graph doesn't extend to 13, what am I doing wrong?

[cold sluice]

It can't be a complex solution right?

I think i see my mistake, the 4 is suppoed to be positive

i made the linear equation equal to y in my head instead of doing it on paper because i thought it was simple

[sonic stratus]

You probably messed up one of your coefficients while solving the 2nd equation.

[cold sluice]

Yea

is this correct now?

[knotty grail]

your equation reads
-4x - 7 = 2x^2 - 6x + 1

yes

thats it

[cold sluice]

Ty!!

.close

[full forum]

Channel closed

Closed by @cold sluice

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[old fox]

why is a complete bipartite graph of K(2,n) non planar?

[wintry burrow]

tomorrow i have math text,i need help please

[old fox]

move the 7 to the other side

|2-x|=4

then solve for x

(2-x)=4 or (x-2)=4

[wintry burrow]

isnt -|2-x|=-4?

[frigid carbon]

!occupied

[full forum]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[wintry burrow]

sorry

[frigid carbon]

it's alright

[full forum]

@old fox Has your question been resolved?

[full forum]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[full forum]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[plucky totem]

Am I using Karnaugh Diagram correctly?
GPT pal says my P is wrong

[onyx glen]

!noai

[full forum]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[onyx glen]

also your $\neg$ symbol is backwards all throughout

[glossy valve]

Ann

[onyx glen]

also what's plus here

[neon basin]

or

[neon basin]

What are you doing on the right?

You've got 1 in the entire row

[plucky totem]

ur right oops

[plucky totem]

• is or

[plucky totem]

What is this supposed to tell me? did I group wrong

[onyx glen]

im pretty sure your light blue thing should be $\neg S_1 \land F$ with an AND not an or.

[glossy valve]

Ann

[onyx glen]

and then the connective between the clauses should be OR not and

so you have those backwards too

but unfortunately those are not the kinds of double backwards to make a forwards

[plucky totem]

I need to replace the "+" I see

does that explain why my solution Y=¬S1 is wrong?

[neon basin]

Yes.