no, exactly
#elementary-number-theory
wild zinc
shadow saffron
ah ok
wild zinc
you don't need to necessarily use the prime factorisation, any factorisation will do
and sometimes some other factorisation will give you a smaller answer than just using the prime factorisation
shadow saffron
oh yeah, so there is no definite method for finding the smallest number with n factors?
wild zinc
other than just trying all the different factorisations I don't think so
shadow saffron
shame
wild zinc
well
I don't know of one
I know there are some more things you can apply to get rid of some cases (I think due to ramanujan?)
I'll have a look
shadow saffron
shadow saffron
ah ok
wild zinc
the reason it's a product to infinity is that it includes all the primes that don't divide it as well
it's just that a_k in those cases is 0
shadow saffron
thats a shame it doesnt work all the time
wild zinc
yeah
for a small number of factors though there shouldn't be too many combinations to try (at least for a computer)
frank shoal
Hello. I'm not sure if this question belongs here, but number theory seems like the closest thing to it. I'm studying RSA cryptography and I'm somewhat confused on the idea behind the public key being strictly made up of prime numbers, and more specifically, co prime numbers. Wouldn't any particularly large integer be enough to make the decryption process incredibly complex?
sacred junco
@frank shoal, no, because you can factor it one at a time. Take 2097152. This is just 2²¹ and you can deduce this quickly by just dividing out a single 2 at a time. You can't do this with semiprimes because they only have 1 valid pair of factors (the coprime factors).
frank shoal
@sacred junco Ah, that does make sense. Thank you 😃
clever remnant
f(a, b) = a/b + b/a + 1/ab where a,b are integers. If f(a, b) is an integer, show that it's a multiple of 3. Can one of you number theory geniuses help me out here?
wild zinc
f(a,b) = (a^2 + b^2 + 1)/ab
I suspect it's vieta jumping tbh
but there may be a nicer solution
I have ab | (a^2+1)(b^2+1)
oh wait
if 3 | ab then 3 | (a^2+1)*(b^2+1) which is impossible because -1 isn't a quadratic residue modulo 3
so 3 doesn't divide ab
so a^2, b^2 and 1 are all 1 mod 3
so a^2+b^2+1 is 1 mod 3
and since 3 doesn't divide ab, 3 divides f(a,b)
waxen kernel
gilded tinsel
owo already on wikipedia
wild zinc
@clever remnant
clever remnant
@wild zinc thx man. I appreciate it
wild zinc
where'd you get the problem from?
low jolt
how do yall become so good at nt
wild zinc
practice
low jolt
auuuuh so damn hard to keep motivated
nocturne token
Do lots of number theory. You have to make it through all the little pieces of riff raff, which can be hard as it seems boring, but it gets better.
clever remnant
@wild zinc that problem was actually on the Euclid math contest (not sure if you've ever heard of it)
low jolt
@nocturne token i have the tendency to want to prove every theorem i stumble upon in elementary number theory, but i am totally new to proofs, have never taken calculus or real analysis or any proof based subject, so proving them are also intimidating. zzz
clever remnant
@wild zinc this year
nocturne token
You may want to do some discrete structures or calculus first.
wild zinc
@clever remnant hopefully the contest was over before you asked that :^)
nocturne token
If you are not familiar with proof based mathematics it will become a stumbling block.
clever remnant
yup it was
low jolt
@nocturne token i see. i'm gonna try studying them in my spare time. thanks.
nocturne token
Excellent. And remember there are people here to help you if you have questions.
low jolt
@wild zinc i mean any subject with high level of rigor, calculus and real analysis to name a few.
wild zinc
right, but I don't think calculus will really help you with proofs because there aren't many in it, and if you want to learn elementary number theory it's not really necessary
however
low jolt
it's interestting though
clever remnant
I need to get started on number theory but don't know where. I have experience doing inductive proofs and proofs in geometry but not in number theory. Plus I barely know any modular arithmetic - should I start looking into that?
nocturne token
Yeah, maybe that wasn’t the best suggestion. I do a lot of analytic number theory so my view on what’s important may be slanted. But I still feel that being able to go through long formal manipulations with special functions/series helps build your ability to go through a long proof. Maybe not.
low jolt
never taken Real Analysis before. does that solidify my basics on proofs?
nocturne token
Yeah, it will.
low jolt
thanks on your advice @nocturne token @wild zinc
swift shard
Welp tough question
woeful ravine
Yeah that's why I posted it
swift shard
Although taking mod 9 might give some answers
woeful ravine
how
swift shard
If we take mod 9, the pattern becomes
1,4,7,1,4,7,1...
Luckily enough, 1×4×7 ≡ 1 (mod 9)
woeful ravine
I like your mod 9 solution
swift shard
But how much does that give us lol
woeful ravine
Yeah it seems a bit lengthy there has to be something
Ok
I'll try from here
I only tried mod 10 for some reason
swift shard
The first digit will be 0 since there's a multiplication by 10
There will probably be a lot of zeroes
swift shard
That's actually pretty cool
That's very helpful
@woeful ravine
Are you allowed to use a trick like the above?
woeful ravine
Yes
swift shard
Hmm, nvm I can't think of a way to bend it to our will. I'll keep thinking on it
woeful ravine
(1x4x7x0x3x6x9x5x8) mod 10?
swift shard
≡ 0 (mod 10) because of the 0
woeful ravine
yeah
wait
nvm
Time to ping Daniel
@wild zinc
Tbh
I feel like its getting close
wild zinc
uhh
find the highest power of 5 that divides it
and the highest power of 2 that divides it
woeful ravine
what?
wild zinc
to find the highest power of 10 that divides it, divide A by that
swift shard
Prime factorization duh
wild zinc
then find that number mod 10
woeful ravine
Tf, you want to do prime factorisation with these numbers?
Yeah
So what you're saying is
A/10^x (mod 10)
wild zinc
don't need full prime factorization
just cancel appropriate 2s/5s to remove trailing zeroes
so uhh
0 mod 5, 2 mod 3 -> 5 mod 15
0 mod 25, 2 mod 3 -> 50 mod 75
0 mod 125, 2 mod 3 -> 125 mod 375
0 mod 625, 2 mod 3 -> 1250 mod 1875
none are divisible by 3125 so not a problem there
...
lmao it's 1 mod 3
woeful ravine
uh
wild zinc
0 mod 5, 1 mod 3 -> 10 mod 15
0 mod 25, 1 mod 3 -> 25 mod 75
0 mod 125, 1 mod 3 -> 250 mod 375
0 mod 625, 1 mod 3 -> 625 mod 1875
wild zinc
count how many satisfy each
134 are 10 mod 15
27 are 25 mod 75
5 are 250 mod 375
and 1 is 625 mod 1875
so highest power of 5 that divides it is 134+27+5+1 = 167
woeful ravine
yes?
wild zinc
5^167 I mean ofc :^)
woeful ravine
I still don't see where this is going
wild zinc
well 2^167 also divides it
so it has 167 trailing zeroes
so we know we cancel 167 5s and 167 2s
hmm slightly different idea
we know 2^168 also divides it so A/10^167 is even
if we can also find A/10^167 mod 5
woeful ravine
Why doesn't mod 10 work
Ok
Thanks @wild zinc @swift shard Imma sleep on it
If that's fine
wild zinc
well mod 10 works
but it's probably easier to do mod 5
I mean it's essentially the same because of how many excess factors of 2 there are
but anyway that's fine yeah
sacred junco
any good books on
elementary number theory?
Like, actually elementary? Modulus, prime numbers, etc.
sacred junco
If x | y and x | y ± z, then x | z;
Does it have to be both ± or just one?
Im guessing it must divide both + and - right?
ivory patrol
oh my god
this problem is so tedious unless there's a clean way to do it that im not seeing
nocturne token
@sacred junco Number Theory by George Andrews
It’s cheap, starts basic and ends strong.
And you could probably find a pdf of it online.
amber harness
Need help Find integers a and b such that a+b≡a- b (mod 5)
static sparrow
there's this\
$x\equiv y\mod n\iff n$ divides $x-y$
sterile pumiceBOT
Tuong:
amber harness
if $a=10,b=15$ then $$10+15\equiv (10-15) \mod 5\Rightarrow 25\equiv -5 \mod 5\Rightarrow -5-25\equiv -30\mod 5=0$$
sterile pumiceBOT
⨋ρ:
sacred junco
you also could've done a = b = 0
vague fable
How do I prove Bezout's identity in a GCD domain (which may not have a total ordering?)
dim ocean
Find the unique positive integer n with the property that both n and n + 77 have exactly nine positive
divisors.
how do u sth like this
ivory patrol
Since n has an odd number of factors, it is a square. Let n=q^2
q^2 + 77 = r^2
77 = r^2 - q^2
77 = (r+q)(r-q)
r+q can equal 77 or 11.
If r+q = 77 then r=34 q=33
If r+q=11 then r=9 q=2
9^2 only has 5 positive divisors, whereas 34^2 has 9.
Checking, we see that 33^2 + 77 = 34^2 and both 33^2 and 34^2 have 9 divisors. n=33^2
dim ocean
@ivory patrol thanks so much but gomez showed me how to do it a while ago
unborn oriole
So, the diophantine eqn x^3+y^3+z^3-d=0 was just solved for d=33, as many of you know
but it was done using only 512 CPU cores and a couple days
Theoretically, couldn't the algorithm be easily parallelised to solve ones orders of magnitude higher, given enough compute power?
Maybe this is a question for computer science, but I feel like it wasn't exactly the most computer-intensive task
unborn oriole
Like, I understand that it's orders of magnitude harder to crack 42. But my university allows access to orders of magnitude more compute power than what was used to solve 33
ivory patrol
If there was an easy way then i'm sure 42 would have been solved already
mossy folio
If it ran on 512 cores, then I'm sure it was parallelized. @unborn oriole
I just rewatched the video, and it seems to me it is very easily parallelizeable.
unborn oriole
Right, but has anyone tried it?
mossy folio
who knows?
unborn oriole
he said he tested through 10^16. I suppose I'll apply for access to a few servers, see if I can't get through 10^18 by May
mossy folio
Even if someone was working on it, it's not probable that anything would be published before any concrete results.
Do you have access to that kind of computation power?
unborn oriole
Certainly. Of course, there's little mathematical implication to going through and finding solutions
yes, my university has a lot of compute power
mossy folio
Well, then good luck. Don't forget to optimize your code.
vast vessel
It's also very easy that the solutions grow much faster than you'd expect
for all you know, that jump might make the smallest solutions to be larger than 10^50 or something
sacred junco
unborn oriole
I'm cautiously optimistic that it's under 10^25, but I'll know if I'm wrong soon enough. Thanks
exotic raptor
np
unborn oriole
=p
sacred junco
lol
quick furnace
#famouslastwords
noble jay
@unborn oriole oh damn are you trying to find another solution?
unborn oriole
Yeah, with the compute power I have access to it'd be a sin not to
dense scroll
Huh ukdl, nice seeing you here
unborn oriole
Huh, neat
sacred junco
This may be the wrong channel but anyone know a good text book for introductory number theory
unborn oriole
Oh, I have just the textbook for you
@sacred junco let me get home real quick and I'll get you the name, it's a springer text that my professor uses for undergrad NT
unborn oriole
this one. Elementary Number Theory by Gareth A. Jones and J. Mary Jones. https://www.amazon.com/Elementary-Number-Springer-Undergraduate-Mathematics/dp/3540761977?SubscriptionId=AKIAILSHYYTFIVPWUY6Q&tag=duckduckgo-d-20&linkCode=xm2&camp=2025&creative=165953&creativeASIN=3540761977
noble jay
@sacred junco also if you're looking for good starting exercises i heard mit ocw has some good ones
twilit locust
The higher arithmetic is a really nice one
fierce flame
didn't know where to ask this, but one question, how are prime numbers used in encryption?
like i understand the general idea of easy to multiply, hard to divide
but how is that used to encrypt messages and prevent hackers from intercepting and breaking the message?
swift shard
RSA uses public key encryption. Note the very important implication of "public key".
For many types of encryption, you need to give people a way to encrypt and decrypt a message. How do you safely get that to someone? What if it gets intercepted? Now some hacker has your "key" and can break your messages.
RSA is asymetric. The method by which you encrypt a message is different from the process by which you decrypt. So, you give everybody (even the hacker) the encryption key, but keep the decryption key for yourself. Now, you don't need to worry about sending it.
How do RSA do? Your private key is two large prime numbers. Your public key is their product.
@fierce flame
Someone can get the private key if they are very good at factoring, but those numbers can get very large
fierce flame
ok so lets say i want to send information to bank
i have two private keys, and one public key
what tf do i send to the bank
or how does message sending work then?
swift shard
You take the bank's public key, and use that to encrypt a message. You then send it to the bank, who can decrypt it
Note that you are not capable of decrypting the message you just encrypted (with the obvious exception of having the original message lel). You need the bank's private key to do that
fierce flame
oooOOOOOOOo
so the bank sends u its public key
and then what do you do to that public key before sending it back
swift shard
@fierce flame
Bank gives everybody the public key. You can get it whenever you want. This allows you to make encrypted messages that ONLY the bank can decrypt
fierce flame
where do the prime numbers come in
im trying to give a presentation on how prime numbers is used in encryption
mossy folio
They are randomly selected?
From a prime pool of sorts?
But this is rather an implementation detail.
swift shard
@fierce flame
A simpler version is the Diffie-Hellman exchange. Two hippies in the 70s came up with the first idea that started public key methods. It's still used a lot today
Definitely worth presenting on, there's a lot there and it's possible to explain via a presentation
fierce flame
Hmmmm ok
crystal pond
I'm not sure if this goes into number theory
But say like if $n=m^3-m$, where $n,m {\in}\mathbb{Z}$, then $6 | n$
sterile pumiceBOT
Itadakimasu!:
crystal pond
Yeah so like I wrote a proof but I'm not sure if it's valid it goes like this
sterile pumiceBOT
Ann:
crystal pond
Notice that $6=3\times 2$
\Meaning $3|m^3-m$ and $2|m^3-m$
\Say m was odd, then we can rewrite m as $m=2k+1, k{\in}{\mathbb{Z}}$
\$m^3-m=(2k+1)^3-(2k+1)=8k^3+12k^2+4k$ which is even
\Say m was even, then m can be rewritten as $m=2k, k{\in}{\mathbb{Z}}$
\$m^3-m=8k^3-2k$ which is also even
\Hence $m^3-m$ must be divisible by 2
\To prove $3|m^3-m$, we factor the expression giving
\$m^3-m=m(m+1)(m-1)$
\If $3|m {\implies} 3|m(m+1)(m-1)$
\If $3|m+1 {\implies} 3|m(m+1)(m-1)$
\If $3|m-1 {\implies} 3|m(m+1)(m-1)$
\If $3|m+2$ we can rewrite $m=3x-2, x{\in}{\mathbb{Z}}$
\Then $(m-1)(m+1)=(3x-3)(3x-1)=9x^2+12x+3$ which is divisible by 3
\If $3|m-2$we can rewrite $m=3x+2 x{\in}{\mathbb{Z}}$
\Then $(m-1)(m+1)=(3x+3)(3x+1)=9x^2-12x+3$ which is divisible by 3
\Hence, $m^3-m$ is divisible by 6
Yeah the proof is a bit long
quick furnace
okay so like
crystal pond
anbd thanks for point tha out^
quick furnace
can i make some tex criticism before reading the proof
crystal pond
HAHAH okay
quick furnace
and you have some mathmoded text, too
sterile pumiceBOT
Itadakimasu!:
quick furnace
okay, so like.
crystal pond
Ok I think it's uh okay now
quick furnace
one does not simply assume X in order to prove X
quick furnace
you can say "To show that 6 | n, it suffices to show 3 | n and 2 | n, because 3 and 2 are coprime"
crystal pond
Ah okay
quick furnace
now cutting more to the essence of the proof...
you could have proved 2 | n by factoring as well
crystal pond
Ah okay
quick furnace
of three consecutive numbers, at least one is guaranteed to be even
and at least one is guaranteed to be a multiple of 3
crystal pond
oh
oh nkd jksdh fjkdsh
Okay that's actually
quite smart
Sigh
So that's all there was to it, okay thanks
sharp pagoda
0³=0=0 mod6
1³=1=1 mod6
2³=8=2 mod6
3³=27=3 mod6
4³=64=4 mod6
5³=125=5 mod6
Therefore for all m in Z
m³=m mod6
m³-m=0 mod6
6 | m³-m for all m in Z
upbeat wren
Never give examples? Just show a proof of the theorem?
unborn oriole
You don't need the word technically... It literally is a proof.
Also, on another note, I've verified computationally that the solution for the diophantine equation x^3+y^3+z^3-42=0 has no solutions for x,y,z <= 10^19. The servers will be done with 10^20 by the end of the day.
Specifically, There are no solutions when all of them are less than 10^19. Wanted to clarify.
mossy folio
how does the program work?
unborn oriole
Very much the same as the graduate of my university Andrew Brooker used for 33. If you've seen the numberphile video, then you'll understand the basics of how it works. Essentially, iterating through possible values of x+y to determine the possible values of 42-z^3
Except, unlike his setup, mine has much more computational power. So I expect that I'll be finding solutions soon enough
I'm renting this hardware from the university until Monday, so I'll optimistically get all the way through 10^21 before I have to rent again.
limber crow
Anyone have suggestions for a second number theory book
I'm just finishing the introductory parts of apostol, and I haven't touched on the analytic number theory stuff since I don't have a background in complex analysis yet
sharp pagoda
I have not read apostol
Is it easy to understand?
unborn oriole
Another update, had a runtime error and couldn't get through 10^21 this weekend. Unfortunate, but I'll soon be renting out servers again and starting from 10^20
mossy folio
What language do you use? @unborn oriole
unborn oriole
C++
mossy folio
with GMP or sth?
unborn oriole
GMP
but I don't think I'm using it efficiently enough. I haven't used C++ in a brick and I think my algorithm is a little slow
tbh given my programming habits over the past 2 years, I'd probably do better to make it in JS, ruby, or even python
but my friends in the CS department insisted I use C++. So I did
sterile pumiceBOT
Linguist = Trash:
oak mica
It works for n = 2.
After I set n = k, I had difficulty proving that this works for n = k+1
$k^{1/k} < 2 - 1/k$
sterile pumiceBOT
Linguist = Trash:
oak mica
$(k+1)^{1/(k+1)} < 2 - 1/{(k+1)}$
sterile pumiceBOT
Linguist = Trash:
oak mica
What I notice is that 1/n in the exponent matches the 1/n being subtracted from 2
I think I have to take advantage of this fact, but i am unable to
Please ping me if you can make any contribution, it will help
(proof by induction btw)
slim agate
@oak mica do you know how to show that (n+1)^(1/(n+1) ) < n^(1/n) \forall n>2?
oak mica
no sir
slim agate
try showing that first :o
oak mica
.... i spent my last 30 mins gaming
@slim agate is this true?
i graphed both functions
here let me take a screenshot
blue line is f(x) = (x+1)^1/(x+1)
slim agate
yep
slim agate
basically, you get maxima for n^(1/n) at n=e (n>1)
slim agate
have fun :o
the inequality you gave initially is true for all n>1
but we can simply show for n=2 and n=3 and then use induction
oak mica
i did that, but not really
slim agate
essentially, you have: n^(1/n) + 1/n > (n+1)^(1/(n+1)) + 1/(n+1) for all n >2
oak mica
n^(1/n) > (n+1)^(1/(n+1))
from which i got that
(n+1)^(1/(n+1)) > (n+2)^(1/(n+2))
from which i got that
(n+2)^(1/(n+2)) > (n+3)^(1/(n+3))
continues on and on
using my flawed logic
slim agate
and the flawed logic was?
oak mica
i thought the right side would eventually equal (infinity)^(1/infinity)
which is 1
which would be greater than the left side
what is your approach?
slim agate
you raise both side to the power of n(n+1)
Then divide throughout by n^n
You'll then just have to show that (1+1/n)^(n) < n
this follows from binomial theorem
slim agate
ye
slim agate
can you see that each term in the binomial expansion is less than or equal to 1?
oak mica
(1+1/(n+1))^(n+1) < (1+1/n)^(n+1)
(1+1/n)^(n+1) = (1+1/n)(1+1/n)^n
(1+1/n)(1+1/n)^n < n * (1+1/n)
slim agate
(1+1/n)^(n) < n you can directly show this for n>2 (without induction)
oak mica
hmm, i showed it for n>=3
slim agate
how? :o
oak mica
the ratio of 4^3:3^3 is less than 3
it only decreases after
5^3/4^3 is less than 4^3:3^3
yeah im stumpede
i think the previous approach was better
oak mica
2^3>3^2
oak mica
$$(1+1/(n+1))^{n+1} < (1+1/n)^{n+1}$$ and
$$(1+1/n)^{n+1}= (1+1/n)(1+1/n)^n < n(1+1/n) = n+1$$ by assumption
sterile pumiceBOT
Linguist = Trash:
oak mica
yep i got it, thank you and i am sorry for wasting your time
good night
i can sleep in peace
sacred junco
also for the last part of the question i got $\frac{p^{m-l}-1}{p-1}$ for the exponent of p is that correct?
sterile pumiceBOT
same:
sacred junco
<@&286206848099549185>
sacred junco
Also
For the last part I have $(x,n)=1 \implies x^{\phi{(n)}}\equiv 1 \mod n$ by Euler's theorem. So $(x^{\prod_{i=1}^k(q_i-1)})^{\prod_{j=1}^k\frac{n-1}{q_j-1}}=x^{(n-1)^k}\equiv 1 \mod n$
sterile pumiceBOT
same:
sacred junco
But from that how can I get that $x^{n-1}\equiv 1 \mod n$ which is equivalent to showing $k$ is odd I guess
sterile pumiceBOT
same:
sacred junco
The positive digits a, b, c each have 3 digits, and for eaxh integer the first digit is the same as its last digit. Also, b = 2a+1 and c=2b+1. How many possibilities are there for integer a?
The answer could be 0, I suppose... hahaha
slow swift
First we see (b-1)/2=a which means b-1 is even so b=odd
From second we see (c-1)/2 = b => c-1 is even so c is odd
sacred junco
Hmm, but does this restriction necessarily mean that the first digit != last digit
slow swift
No no i was wrong about somethinf
It means first and last digit of the number b and c are odd
c=4a+3 by both formulas
i got a=151
😂
slow swift
Oh wait i missed somethings
Ok i got the general formula
b=3a3 Where a is an integer from 0 to 9
It is the only solution that works for b
slow swift
I used restriction method to derive this
So we know the 3 numbers are 3 digits
So they must be inside 100 and 999
Solving that by putting 2b+1 we get 100<=b<=499
We know first and last integer of b is the same
And its also odd
So 1x1 and 3x3 where x is an integer
But 1x1 wont work bcs then a is less than 100
So 3x3 is the only b value
"How many possibilities are there for integer a?" Answer = 10 imo
slow swift
I actually proved that a<250 too
I just didnt send the photo of my work
@sacred junco what i didnt do is test the numbers
If we had done that we wouldve gotten it wouldnt making ur friend troubled
UwU tell Star thank u from me too
sacred junco
Hahaha yeah, he solved it in like 2minutes @slow swift
while he was playing a video game
Guy enjoys math, somewhat so he was happy.to oblige. Lmk if u find any i teresting problems like this so I can give them a go, too!
slow swift
I will say the same
I love maths
It was pretty easy to get the numbers and stuff
Im new to number theory so
sacred junco
@ivory patrol whats that
Oh- youve mgiht confused it with:
How many three-digit positive integers ABC are there such that (A +B) C is a three-digit integer and an integer power of 2?
ivory patrol
nah that's a completely different problem
sacred junco
Any idea why the 4 is subtracted?
mfw
I know why; we are solving for n
That was my issue...gotta be more careful smh
sterile pumiceBOT
Requested by rudy#0446
sacred junco
Any number higher than 3^6 would divide this
pretty sure
nvm
It'd be exactly
3^6
sacred junco
7! + 8! + 9! = 7!(1 + 8 + 72)
sharp pagoda
Yeah
sharp pagoda
$3^4(7!) =3^4(2^4\cdot3^2\cdot 5\cdot 7)=3^6(2^4\cdot 5\cdot 7)$
sterile pumiceBOT
Nick(Smiling):
sacred junco
lol did you actually factorise
sharp pagoda
Factorization difficult 😿
But also can do without factorization
shut raven
sharp pagoda
sharp pagoda
If
x²(p-1) = 0 ( mod x³+x²)
Can we write either x²=0 or p-1=0 mod (x³+x²)
sharp pagoda
Oic
exotic raptor
combined they have all the necessary factors but individually they may not
exotic raptor
yes
sharp pagoda
Kk thx
exotic raptor
np
sacred junco
np
warm sedge
Show that given any increasing sequence of positive integers a_n, the set of all rationals of the form k/a_n in lowest form for some integers k, n is dense in [0, 1].
upbeat wren
what is k?
upbeat wren
okay this isn't too bad.
Let r be in [0, 1]
Let epsilon > 0 be given. Show that k/(a_n) can be less than epsilon away from r.
for some k.
any thoughts about that?
Let M > 1/epsilon. For all a_n > M, 1/(a_n) < epsilon ... ? 😃
just chime in when you want to try a bit on your own or don't understand something.
So now we know 1/(a_n) < epsilon, and we "inch up" by 1/(a_n) or a value less than epsilon by letting K be 1, 2, 3, ... Since k/(a_n) is an arithmetic sequence, we know it must eventually surpass 1. This means there is a k1/(a_n) below r and a k2/(a_n) above r.
warm sedge
Err
If k and a_n share a factor you’re in trouble
It has to be k/a_n in lowest form
upbeat wren
um, sorry?
warm sedge
Err no
For example
A_n = 4
Or let’s say 6
Then the admissible values of k are only 1 and 5
upbeat wren
I'm trying to make the point that there will be a k where k/(a_n) is out of range.
sorry greater than r
upbeat wren
er why?
warm sedge
Because question says so
Show that given any increasing sequence of positive integers a_n, the set of all rationals of the form k/a_n in lowest form for some integers k, n is dense in [0, 1].
Note the in lowest form
upbeat wren
okay.
wow ... I might need a lemma here. Thanks for that. I totally read it wrong.
Um, so now I'm thinking on my feet.
warm sedge
The fact that it’s in the number theory channel should be a hint hehe
upbeat wren
Okay. Let's try this. 2, 3, 5, 7, 11, 13, 17, 19 😃
warm sedge
Yeah primes xD
upbeat wren
Um, let f(x) be the product of the first x primes.
warm sedge
That works cause every k is admissible
upbeat wren
yes!
warm sedge
But the question asks for any a_n
upbeat wren
I now 😃 I'm getting there.
What I'm thinking here is if M > (f(x))^n for some n then, there exists a prime, p, up to the x'th where p^n is a factor of M OR a prime q > x'th prime where q is a factor of M.
warm sedge
Yes that’s correct
upbeat wren
well there it is 😃 you have a p^n which is relatively prime to everything except multiples of p OR you have a lower bound for a prime that divides M. Suppose q divides M, ....
warm sedge
How does that help with the original question tho?
upbeat wren
qt = M. (tn + 1) /q shouldn't reduce except once every q turns.
upbeat wren
My goal is to say there exists an N for which any number above N, call it D, will either have a high power of some prime as a factor, or be a multiple of a sufficiently big prime.
warm sedge
Okay..
I still don’t see how it links to the original question
2*59 has a big prime
But it reduces every 2 steps
upbeat wren
but 59 is the bigger prime ...
warm sedge
And?
upbeat wren
so 1, 3, 5, 7, 9 will be fine except for every 59th step
upbeat wren
I'm saying beyond a certain N, all numbers will have sufficiently large powers of primes or have a large prime factor. This is good ... separately:
p^n | D --> zp^n = D and (1 + hz) / (zp^n) is close to h/(p^n)
q | D --> xq = D and (1 + gx)/(xq) is close to g/(q)
warm sedge
Right
upbeat wren
and (1 + hz) / (zp^n) is reduced except every p^n
and (1 + gx) / (xq) is reduced except every q
setting epsilon to 1/(2p^n) or 1/(2q) would be enough
warm sedge
Yeah I believe this works
upbeat wren
I hope so 😃
warm sedge
Nicely done
upbeat wren
lol 😃
upbeat wren
it's 3:45 AM. I'll take you up on that offer possibly some other day. But again, thanks. 😃
upbeat wren
cya
warm sedge
Eh this is a problem
I don’t think you can say anything about the factors of 1 + hz
It may contain p, we have no way of knowing
Same with 1 + gx and q..
@upbeat wren
upbeat wren
well, D = zp^n and if you let (z, p) = 1 , 1 + hz will have a factor of p only once every p h's.
Similarly, D = xq^m and q^m | (1 + gx) once every q g's
@warm sedge
warm sedge
Oh is that true?
Ooh right
Some number theory.. thingy
But then what if p is smal
It’s once every “p” rounds
upbeat wren
we have to hope that z is big then.
upbeat wren
p^n | D --> zp^n = D and (1 + hp) / (zp^n) is close to h/(zp^(n-1))
Or either of:
q | D --> xq^r = D and (1 + gq)/(xq^r) is close to g/(xq^(r-1))
q | D --> xq^r = D and (1 + gx)/(xq^r) is close to g/(q^r)
ok ok 😃
noble anvil
so ... for some reason, i'm struggling with Number Theory
I don't really know where I'm lacking, but , while I understand the lesson and all, I'm having trouble with applying it in exercises ...
versed storm
Look at numbers
noble anvil
lmao
it's just that i don't know how to approach the exercises for it
err, lemme figure out LaTeX to type one out ... hehe
let $$ n \in \mathbb{N}\setminus {0,1} $$
prove that if $$ 5^{n} - 3^{n} $$ is prime then n is prime
sterile pumiceBOT
阿良々木暦:
noble anvil
yeah
exotic raptor
vacuously true?
noble anvil
i actually have the solution and
in the end it said : that 5^n - 3^n isn't prime
and i'm like : wut
wild zinc
Prove that if p and p^2+2 are prime, then p^3+2 is also prime
noble anvil
hmm
just a thought, we assume "if p and p^2+2 are prime"
and try to prove " p^3+2 is also prime"
by assuming by contradiction that : " p^3+2 isn't prime"
and then proving that contradicts with the first statement, thus false ?
idk, just a guess (i guess ? )
sharp pagoda
p=0,1 or 2 (mod 3)
if p=1 or 2 (mod 3)
p²+2= 1+2=3=0 (mod 3) or
p²+2=4+2=6= 0(mod 3)
p²+2 not prime
If p=0 (mod 3)
p not prime except p=3
wild zinc
interestingly p^4+2 is also prime :^)
and by interestingly I mean
completely coincidental
p^5+2 is not tho
exotic raptor
neither is p^6+2 
versed storm
neither is p^69 + 2
ivory patrol
for all prime $p > 3$, $p \equiv \pm 1 \pmod 6$, $p^2 + 2 \equiv 3 \pmod 6$. Hence $p\leq 3$. $p=2$ or $3$. $2|2^2+2$ but $3^2+2$ is prime. $3^3 +2 = 29$ QED
Rip latex
sterile pumiceBOT
Plum:
exotic raptor
69=4*17+1 so p^69+2 isn't yeah
upbeat wren
hehe. (a^2 + 3z - 9)/(3b^2 + 5b + 8) = 3c^2 + 4c - 3 has no solutions in the integers.
exotic raptor
a=3, z=2,b=-1,c=-2 wants to know your location
quick furnace
limber crow
if g is a primitive root of an odd prime p, prove that -g = g^((p-1)/2) mod p
not quite sure where to start with this one
versed storm
shouldn't it be -1 instead of -g
wild zinc
it should be -1
I'll just remind you of the definition of a primitive root modulo p:
the smallest k such that g^k = 1 (mod p) is p-1
wild zinc
and then hopefully that will be enough to show you where to start
limber crow
Oops I meant plus in the exponent that was my issue
wild zinc
well it's equivalent to -1 = g^((p-1)/2) mod p
wild zinc
Solution (don't check without trying for a reasonable amount of time pls thanks):
||Note (g^((p-1)/2))^2 = g^(p-1), which is 1 (mod p) as g is a primitive root modulo p||
||(g^((p-1)/2))^2 - 1 factorises as (g^((p-1)/2)-1)(g^((p-1)/2)+1).||
||Notably from the first line (g^((p-1)/2))^2 - 1 = 0 (mod p), so as p is prime, by Euclid's lemma either g^((p-1)/2)-1) = 0 (mod p) or g^((p-1)/2)+1, which means that g^((p-1)/2) is either 1 or -1 (mod p)||
||Noting (p-1)/2 < p-1, and that since g was a primitive root p-1 was the smallest exponent k such that g^k = 1 (mod p), we cannot have g^((p-1)/2) = 1 (mod p), and hence we must have g^((p-1)/2) = -1 (mod p).||
sharp pagoda
" the smallest k such that g^k = 1 (mod p) is p-1 "
I didn't understand this
Can you give an example
ivory patrol
fermats last theorem states that g^(p-1) = 1 (mod p) for all p, but the power of p-1 isn't necessarily the smallest positive power where this is true
sharp pagoda
I mean someone told the definition of primitive root mod p
" the smallest k such that g^k = 1 (mod p) is p-1 "
But I didn't understand
upbeat crypt
if (m1, n1) and (m2, n2) are integer points on x^2 - 2y^2 = 1 show that point (m, n) is also on the curve, where m and n are integers such that: m - n root(2)=(m1 - n1 root(2))(m1 - n2 root(2))
does anyonbe know how to derive the forumla from the qurstion. (i think the question involves pells equation)
i know (given pells equation) that if (a,b) is a solution to x^2 - 2y^2 = 1 then the general equation for finding solutions are infinity^n=(a + b root(2))^2 (if that helps)
sharp pagoda
In m-n√2 in the second bracket is it
m1-√2n2 or m2-√2n2?
sacred junco
@sharp pagoda do you know some group theory
sharp pagoda
Heard of it when friend was teaching me NT
sacred junco
do you know the concept of generator
sharp pagoda
Not sure but I think some elements can generate whole group
sacred junco
right
so if you have the group of integers mod n under multiplication (set of numbers in {0,..,n-1} coprime to n), and there is an element that generates the entire group
that element is known as a primitive root mod n
sharp pagoda
Is it unique
sharp pagoda
I forgot how element generates group
Can you give example
sacred junco
an element $g$ of a group $G$ is said to generate $G$ if for all $x \in G$, $x=g^k$ for some integer $k$
sterile pumiceBOT
sacred junco
like you can express every element of the group as some amount of applications of g to itself
sharp pagoda
So powers of g in mod should generate every element of G?
sacred junco
yes
sharp pagoda
"if g is a primitive root of an odd prime p, prove that -g = g^((p-1)/2) mod p"
So here how do you know which group they are talking about
sacred junco
(Z/pZ)^*
sharp pagoda
Oic thx
devout trail
Hopefully it's the right place for that..
p-adic analysis:
Let $\pi \in K$ where $K$ is an extension field of $\mathbb Q_p$ (the p-adic rationals) and $ord_p(\pi)=\frac 1 e$ where $e$ is the index of ramification.
Then any $x \in K$ can be written uniquely in the form $\pi^mu$ where $|u|_p=1$ and $m \in \mathbb Z$.
Can you please help me understand why that is right?
sterile pumiceBOT
idan:
upbeat crypt
@sharp pagoda SORRY it was m2-√2n2 my bad
do you have any idea on how to solve it...
sharp pagoda
Oic
I can try
Let's say (a,b) (c,d) instead of (m1,n1) and (m2,n2)
$a^2-2b^2=1$ and $c^2-2d^2=1$
sterile pumiceBOT
Nick (HS):
sharp pagoda
$(a^2-2b^2)(c^2-2d^2)=1\a^2c^2-2a^2d^2-2b^2c^2+4b^2d^2=1$ let's call it equation (1)
sterile pumiceBOT
Nick (HS):
sharp pagoda
$m-\sqrt{2}n = (a-\sqrt{2}b)(c-\sqrt{2}d)\=(ac+2bd)-\sqrt{2}(ad+bc)$
sterile pumiceBOT
Nick (HS):
sharp pagoda
Comparing $m=ac+2bd$ and $n=ad+bc$
sterile pumiceBOT
Nick (HS):
sharp pagoda
We want to show that $m^2-2n^2=1$
sterile pumiceBOT
Nick (HS):
sharp pagoda
$(ac+2bd)^2-2(ad+bc)^2\=a^2c^2+\cancel{4abcd}+4b^2d^2-2a^2d^2-\cancel{4abcd}-2b^2c^2$
sterile pumiceBOT
Nick (HS):
sharp pagoda
By equation $ (1) $ this equals 1
sterile pumiceBOT
Nick (HS):
sharp pagoda
@upbeat crypt
upbeat crypt
omg bro thank you i skipped the question because none of my friends could answer it
you are a life saver it was really well explained too 😃
upbeat crypt
@sharp pagoda The only question I have is: is the question asked us to prove m,n on this curve or to prove that root2 equation is right? .I cannot prove it any other way around(knowing mn on the curve and prove root2 equation is right)
sharp pagoda
The way you asked the question means we have to show that (m,n) is on the curve
Show the printed question if there is any difference
wild zinc
@sacred junco only unique if p <= 3: there are phi(p-1) generators
sacred junco
ah ok
wild zinc
if g is a generator and x = g^k, then x is a generator iff gcd(k,p-1) = 1
versed storm
primitive root
wild zinc
works in a general cyclic group of order p
sharp pagoda
what is this question
If (m1,n1) is on the curve then show that (m1,n1) is on the curve?
Printing mistakes ig
upbeat crypt
yea it was a printing mistake
we were told to say (m1, n1) and (m2, n2) are inteeger
and show (m, n)
forgot to mention that
sharp pagoda
I think if we can find two solutions then we can find more using them
sharp pagoda
If $(a,b)$ and $(c,d)$ are solutions then $m=ac+2bd$ and $n=ad+bc$ is also solution
sterile pumiceBOT
Nick (HS):
upbeat crypt
isnt (m, nroot(2))^n = (3 + 2root(2))^n a general solution? and you just state n can be a whole integer
for exmaple (m - nroot(2))^2 =17 + 12root(2)
therefor (17, 12) is a solution
sharp pagoda
You talking about general solutions of Pell's equation ig
upbeat crypt
yea.. i thought thats what the question was asking
sharp pagoda
But using this exercise idk how would you get this general solution
That's why I said using two solutions we can find more
upbeat crypt
oh okay i should be stemming of question 8
not using another equation
is what you are trying to say?
sharp pagoda
I mean they use some advanced NT to find that general solution ig but I don't know advanced NT well enough
(3,2) and (17,12)
m=99 n=70 check if this is solution
upbeat crypt
yes, that is a solution
did you find those number by subbing into
m = ac + 2bd and n = ad +bc
as stated above by you
sharp pagoda
$m=ac+2bd$ and $n=ad+bc$
sterile pumiceBOT
Nick (HS):
sharp pagoda
Yeah
upbeat crypt
but one thing still confuses me what does it mean by approx to the irrational number root(2)
inner crest
$\frac xy=\sqrt{2+\frac1{y^2}}$
sterile pumiceBOT
EpicGuy4227:
inner crest
when x and y have the same sign
upbeat crypt
oh so you rearragned the equation in terms of x/y and as x and y approach infinity
1/y^2 get exponentially smaller
making it basically nothing and gives an approximation of root(2)
omg thank you so much
@sharp pagoda thank you for helping me much appreciated
and @inner crest thank you for the equation
sharp pagoda
@upbeat crypt
I was thinking about it
Actually we can find solutions using one solution (3,2)
Say (3,2) and (3,2) is solution
Then
m=9+2(4) and n=6+6
upbeat crypt
@sharp pagoda shit that is true then that means we can simplify our general solution
If $(a,b)$ is a solutiona then $m=a^2+2b^2$ and $n=a^2+b^2$ is also solution
sterile pumiceBOT
Devo:
sharp pagoda
,w (17^2+2*(12)^2)^2-2(2(17)*12)^2
sterile pumiceBOT
Requested by Nick (HS)#3263
upbeat crypt
sorry i just realised
If $(a,b)$ is a solutiona then $m=a^2+2b^2$ and $n=2ab$ is also solution
sterile pumiceBOT
Devo:
upbeat crypt
@sharp pagoda
isnt this also pythagoreans formula from memory?
ohhh no its not pythag is a^2-b^2, 2ab, a^2 + b^2
but its awfully similar
sharp pagoda
Yeah 2ab
desert rose
oh sorry, it says. prove interchangably a and b for all primitive pythagorean triples a, b and c can be obtained from (a,b,c)=(2uv,u^2-v^2,u^2+v^2) where u and v are positive integers where u>v such that u times v have a common factor greater than 1 where either u and v are odd or even.
wild zinc
Prove all primitive Pythagorean triples (a,b,c) are of the form (2uv, u^2-v^2, u^2 + v^2), where u and v are positive integers such that u > v and gcd(u,v) = 1.
dusty granite
Wow I'm actually reading the same proof @desert rose
wild zinc
my advice for how to prove it is ||if a^2 + b^2 = c^2, then (a/c)^2 + (b/c)^2 = 1, and vice versa, so it's a case of finding all rational points on the unit circle. You should be able to find at least one, and then you can generate the rest from that||
desert rose
@dusty granite lol where'd u get the problem from?
dusty granite
i am reading book called topology of numbers
first topic it talks about is pythagoream triplets
desert rose
@wild zinc the proof shouldnt be too confusing its the way the question is worded i am confused at what they are asking me to do
oh nice if u happen to find any discoveries be sure to let me know
@wild zinc would u just create a linear equation y= -rx + r where r=u/v
and sub it into x^2 + y^2 = 1
therefor x^2 + (-(u/v)x + (u/v))^2 = 1
solve for x therefor, x= (u^2-v^2) / (u^2 + v^2)
and y = (2uv)/(u^2+v^2)
therefor because x and y are rational points x = a/c and y= b/c
therefor a = u^2 - v^2 b=2uv and c= u^2+v^2
but i am not sure if thats what the question wants?
desert rose
could you just state that ka^2 + kb^2 = kc^2
dusty granite
u know wht primitive triplets r right
dusty granite
it is enough when 2 r just co prime
once 2 hv a comon factor, third one wul be automatically hv the factor
upbeat wren
Huge Oopsies:
Corrected: (a^2 + 3a- 9)/(3b^2 + 5b + 8) = 3c^2 + 4c - 3 has no solutions in the integers.
sacred junco
@sacred junco do you know the thing for a finite geometric sum
$\frac{5^{125} - 1}{5^{25} - 1} = 1 + 5^{25} + \br{5^{25}}^2 + \br{5^{25}}^3 + \br{5^{25}}^4$
prolly replace 5 with x and then factor that expr
ivory patrol
I think you meant 1 + 5^25 + 5^50 + 5^75 + 5^100
ivory patrol
🤔
sacred junco
hm
sterile pumiceBOT
sacred junco
ok
autumn violet
versed storm
non negative
grim thicket
What does the negative mean in this? a = -1 mod n
sacred junco
a is 1 less than a multiple of n
grim thicket
So 5 mod 3 is congruent to 2 mod 3 and -1 mod 3?
sacred junco
yes
grim thicket
thank you
sacred junco
np
magic egret
Which theorem in number theory says that $ \ f(x)\ mod\ g(x)=f(x)-g(x)\left \lfloor \frac{f(x)}{g(x)} \right \rfloor$
sterile pumiceBOT
boilhats:
foggy shadow
Hey I was suggested to ask the following here: is there a pain-free way of calculating 444^101 (mod 1961) or did my book just assume that one uses a computer for that?
mossy folio
are you sure that's what you have to calculate?
foggy shadow
Yep, it is part of a bigger problem though
mossy folio
and are you sure that you've made no mistakes in the bigger problem?
foggy shadow
Yeah, like in the answer it says 777 which is 444^101 (mod 1961).
as 101 was another part of the problem
mossy folio
what theorems did you use?
foggy shadow
literally multiplying and finding a pattern (as it is modular)
mossy folio
that was all?
foggy shadow
They also had one where the exponent is converted into binary and then rewritten to multiplications of the base raised to powers of two. For our example: 444^64 * 444^32 * 444^2 * 444^1. Still way too extreme to do by hand unless I'm missing something
like, they find the congruency of each smaller number instead and multiply them in the end
mossy folio
well
the cyclic pattern repeats at every 53rd power
so that's probably not what they're looking for
foggy shadow
Its on decrypting an RSA-encrypted message so they're literally looking for the number for the expression I gave. I guess I'l justl hope that the exam will be with numbers I can brute force ^^
sharp pagoda
@foggy shadow
You just have to use CRT
444^(101)=0 mod (37)
444^(101) =
20^(101)=find this number (mod 53)
By Fermat's theorem
20^(52)=1 (mod 53)
20^(104)=1 (mod 53)
Inverse of 20 is 8 in mod 53
Therefore 20^(101)=8^3 (mod 53)
20^(101)=35 (mod 53)
mossy folio
wow the math checks out
how do you reach the conclusion that 444^101 = 20^101 (mod 53) ?
sharp pagoda
Cos 444=20 (mod 53)
sharp pagoda
Let me check my calculations for 20^(101)
sharp pagoda
Oic
Now using CRT will give required value
444^(101)=0 mod (37)
444^(101) =35 (mod 53)
foggy shadow
@sharp pagoda thanks a lot man but I don't really understand what you're doing. Like, where did 53 come from? Also, what is CRT?
foggy shadow
Ah! ok that makes sense
Would you mind going through the other steps as well?
I don't get where cos and 20 came from
sharp pagoda
cos=because
We wanted to find what is 444^(101) in (mod 53)
This is equivalent to what is 20^(101) in (mod 53) if you know modular arithmetic
Idk how to explain that in english
foggy shadow
Aha hahah ok yeah I get that, thought you meant cosine and was super confused
But I don't get it, we wanted to find 444^(101) (mod 1961) I don't get the same result from factorising it like you did in wolfram alpha
sharp pagoda
I didn't find 444^(101) mod (1961)
We can find that after using CRT on
444^(101)=0 mod (37)
444^(101) =35 (mod 53)
Using CRT
444^(101)=35* 43* 37 (mod 1961)
,w 35* 43 * 37 (mod 1961)
sterile pumiceBOT
Requested by ♫ick (HS)#3263
sharp pagoda
,w 444^(101) (mod 1961)
sterile pumiceBOT
Requested by ♫ick (HS)#3263
foggy shadow
Ok, thanks a lot. I was confused as I wasn't familiar with CRT. I'll read up on it
thank you for your help 😄
sharp pagoda
Oh okay
west cedar
@swift shard sorry didnt see your response until now, thanks I'll check that out
versed storm
Simultaneously
light flicker
It's a normal system of equations, just use elimination or something
sacred junco
but what does the answer even look like?
sacred junco
how do iknow that x,y are unique mod 53 (which im assuming they are)
vast vessel
Is elementary numbre theory really uni level tho
vast vessel
crystal pond
Or it could be high school hmm
sacred junco
its defo early uni in uk
obviously there are people who learn it before
but not required except for competitions
north pagoda
early uni is definitely a better fit
though the "pre-uni" category does seem fairly small, i think it works
sacred junco
do u learn shit like this pre uni in america?
like CRT, FLT, Fermat-Euler theorem, euclid's algorithm, bezouts lemma and stuff like that
crystal pond
I’m quite impressed
I thought the Singapore education system was good
I guess noT
north pagoda
meh, i dont think elementary number theory should be a HS topic anyway
more useful things to teach kids
light flicker
@sacred junco you know they're unique through usual linear algebra ideas
Since the determinant is not 0 mod 53
sacred junco
ah ty
wild zinc
@vast vessel hmm didn't think I'd see you talking about number theory
vast vessel
wild zinc
ping who I like
wild zinc
nope
vast vessel
wild zinc
ah well
vast vessel
wild zinc
can't expect you to understand the nuances of language
vast vessel
what's that
wild zinc
I knew it
vast vessel
sacred junco
Oh gecko guy and simple art are here
sterile pumiceBOT
same:
sacred junco
or i guess that $\sum_{k=1}^{p^i-1} \binom{p^i}{k} x^k \equiv 0$
sterile pumiceBOT
same:
light flicker
So consider the coefficients of those two polynomials
sacred junco
which 2 polynomials
can i just say that I can set x=1,2,....,p^i-1 and use gaussian elimination to show that they're all 0
also for the last part by considering the coefficient of $x^k$ in the expansion of $(1+x)^n$ I have that $\binom{n}{k}=\prod_{i=0}^l \binom{p^i n_i}{p^i k_i}$
sterile pumiceBOT
same:
versed storm
Lucas theorem
unkempt void
@sacred junco bit late but isn't bezout's lemma in the new FP2 stuff?
i have a feeling some of the elementary number theory has been moved into optional modules of the new further maths a level
wild zinc
there is a little elementary number theory in one of the optional modules in the new further maths a level in one of the exam boards, yes
sacred junco
whoa ours got downgraded to user friendly on the other hand :/
versed storm
Which exam board
wild zinc
I think edexcel
light flicker
What do you mean by increases?
sterile pumiceBOT
agentnola:
crimson lagoon
Or $/leq$
sterile pumiceBOT
agentnola:
light flicker
That's not true? $\varphi(5) = 4 > 2 = \varphi(6)$
sterile pumiceBOT
Zopherus:
wild zinc
what's the original question? @crimson lagoon
sacred junco
@unkempt void wtf number theory in fp2
sacred junco
ah
unkempt void
it's an optional module very few do
sacred junco
arent u y12
sacred junco
fuking nerds
sacred junco
number theory is trivial anyway
wild zinc
it's true
versed storm
true like a loo
woven phoenix
Hi everyone, I'm current going through Elementary Number Theory by Underwood Dudley. Any other books you would recommend to really get a grasp on integers, factorization, congruences, and prime numbers? (will use it in the context of cryptography and elliptic curves)
granite compass
Try Hardy and wright
I think it's called elementary number theory
There's also Davenport, but it's more advanced @woven phoenix
woven phoenix
@granite compass Wow, thanks a lot! hardy & wright looks to be accessible at my current level.
raven carbon
Is (b) limited to the natural numbers found in the sequence?
raven carbon
Isn't the sequence stopping at 58?
sacred junco
I tried applying the formula but it is giving weird answerz
It is said 2 digit numbers divisible by 13
raven carbon
2 digit natural numbers
sacred junco
Ohshit
I messed up
Ol
Lol
No
I'm doing correct
I guess
Help me :'(
@raven carbon
raven carbon
To be clear, the sequence starts at -12 and ends at 58?
raven carbon
Haha good job
sacred junco
@raven carbon you still here?