ganeshie8
  • ganeshie8
\(\gcd((p+1)/2,~p-1) = ?\) \(p\) is an odd prime
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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dan815
  • dan815
what is the GCD algorithm
ganeshie8
  • ganeshie8
wolfram says the answer is always either 2 or 1 http://www.wolframalpha.com/input/?i=Table%5Bgcd%28%28p%2B1%29%2F2%2C+p-1%29%2C+%7Bp%2C3%2C13%7D%5D
ganeshie8
  • ganeshie8
gcd(a, b) = gcd(a-b, b) = gcd(a, b-a)

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ganeshie8
  • ganeshie8
i have been messing with that but no success..
dan815
  • dan815
1/2, 2, 1/2, 1
dan815
  • dan815
okay (p+1)/2 = p/2 + 1/2, and p-1 = 2*(p/2 - 1/2)
ganeshie8
  • ganeshie8
|dw:1440466619213:dw|
dan815
  • dan815
i think showing this pattern recurring is better
dan815
  • dan815
we can show that all oodds will be 1 and 2 repeating
dan815
  • dan815
let p=2n+1
dan815
  • dan815
GCD((2n+1)+1)/2,2n-1)
ganeshie8
  • ganeshie8
\(p\) is given as an odd prime so..
dan815
  • dan815
if we show this happens to all odd numbers then we can atleast say it happens to all odd primes, the patternn of the 1s and 2s might be random
dan815
  • dan815
GCD((2n+1)+1)/2,2n-1) = 1 or 2 in fact we can evn do 1 step better by changing n=2k and 2k+1
dan815
  • dan815
to show every 2nd add is 2 and every 1st odd is 1
ganeshie8
  • ganeshie8
Oh! for odd numbers also it is giving 1 or 2 didn't notice before.. interesting..
dan815
  • dan815
right
dan815
  • dan815
oops i mean (2n-1)-1
dan815
  • dan815
|dw:1440467089588:dw|
dan815
  • dan815
or 1
dan815
  • dan815
lets see every 4th odd is 1 when odd of form 4k+1 when odd of form 4k+3 we have 2s
ganeshie8
  • ganeshie8
Nice! it works perfectly! GCD((2n+1)+1)/2,2n-1) = GCD(n+1, 2n) = GCD(n+1, n-1) = GCD(2, n-1)
dan815
  • dan815
ooo okay that looks neater
dan815
  • dan815
when odd of form 4k+1 we have 1s as GCD when odd of form 4k+3 we have 2s
ganeshie8
  • ganeshie8
if p = 4k+1, then (p+1)/2 = (4k+1 + 1)/2 = 2k+1 clearly 2 doesn't divide 2k+1, so gcd = 1
dan815
  • dan815
yep what u solved for already shows
dan815
  • dan815
(2,n-1)
dan815
  • dan815
every 2nd n has 2, every 1st n has 1 gcd
ganeshie8
  • ganeshie8
if p = 4k+3, then (p+1)/2 = (4k+3 + 1)/2 = 2k+2 = 2(k+1) so gcd = 2 this form looks better to me as it says when exactly is gcd 1 or 2
dan815
  • dan815
and since ur looking at the numers of form 2n+1 that shows 4k+1 and 4k+4 have 1 and 2 GCD respectively
dan815
  • dan815
4k+3*
dan815
  • dan815
http://prntscr.com/88jluw this line u did here pretty much is 100% completed work
ganeshie8
  • ganeshie8
\[\gcd((p+1)/2,~p-1) = \left\{\begin{array}{}1&:&p\equiv 1\pmod{4}\\2&:&p\equiv 3\pmod{4}\\ \end{array}\right.\]
dan815
  • dan815
okay now to figure out something how the order of 1s and 2s follow for primes that will be something
dan815
  • dan815
thats just like asking if there are an even number of oodds or an odd number of odds between each prime number
dan815
  • dan815
feels almost as complex as knowing the next prime number itself
ganeshie8
  • ganeshie8
the same applies to primes as well right.. if it works for odd numbers, then it works for odd primes too...
dan815
  • dan815
yeah
dan815
  • dan815
but u dont know if its a 1 or a 2
dan815
  • dan815
it would be interesting if it was like always alternating perfectly
dan815
  • dan815
i am pretty sure its quite random, just like primes
ganeshie8
  • ganeshie8
ohk you're thinking of sorting them..
dan815
  • dan815
it would be interesting if u could prove something based on density of primes and stuff
ganeshie8
  • ganeshie8
yea the primes don't alternate between : {4k+1, 4k+3, 4k+1, 4k+3,...} so there should not exist any pattern like : {1, 2, 1, 2, ...}
ganeshie8
  • ganeshie8
|dw:1440468566168:dw|
dan815
  • dan815
ya im saying it would be interseting if we can prove something abot the 1s and 2s if like there should be more 2s or more 1s, or if its truly random or some pattern exist and such
dan815
  • dan815
the mysteriess
dan815
  • dan815
lol
ganeshie8
  • ganeshie8
so basically you're asking, which set is bigger : \(\{p:\text{p is an odd prime of form 4k+1}\}\) or \(\{p:\text{p is an odd prime of form 4k+3}\}\)
dan815
  • dan815
oh true
ganeshie8
  • ganeshie8
they both are infinite sets and have same cardinality, but it might look intresting to see the density graph for each of them..
ganeshie8
  • ganeshie8
wow! you're not the first person to observe that, this looks interesting In number theory, Chebyshev's bias is the phenomenon that most of the time, there are more primes of the form 4k + 3 than of the form 4k + 1, up to the same limit. This phenomenon was first observed by Chebyshev in 1853.
dan815
  • dan815
oh cool
Empty
  • Empty
I wonder if there are more primes of the form 6n+1 or 6n+5 while we're at it.
Empty
  • Empty
Also I wonder if we can help sort the 4n+1 and 4n+3 primes by acknowledging they are really both 2n+1 primes. So what I mean is this is part of a larger tree that will sort further: 8n+1, 8n+3, 8n+5, 8n+7 So we can see that we are further cutting apart 4n+1 --> 8n+1, 8n+5 4n+3 --> 8n+3, 8n+7 If that makes sense, since this gives us a way to try to understand them better?
ganeshie8
  • ganeshie8
just completed reading first page of this http://www.jstor.org/stable/27641834?&seq=1#page_scan_tab_contents looks interesting.. i wish i had access to the full doc
dan815
  • dan815
its kind of cool the difference looks like its increasing constantly but around some binomial distribution increase
Empty
  • Empty
Fascinating, the difference between them in the "race" isn't very far all the way up to 100,000 cool
Empty
  • Empty
Might be time to make some programs :P
dan815
  • dan815
we should consider log sets
Empty
  • Empty
Fun fact: All Mersenne primes are of the form 4n+3
dan815
  • dan815
like instead of considering sets of 10,20,30 how about sets of length approximately e and e^2 , e^3
dan815
  • dan815
i always wonder how these numbers that are discovered more purely hold up against these random things
dan815
  • dan815
it could result in a linear growth of the different between team 1 and team 3 primes
dan815
  • dan815
difference*
dan815
  • dan815
e,pi,phi lets gooo lol
dan815
  • dan815
then e*pi*phi
ganeshie8
  • ganeshie8
it seems the prime number theorem also works independently for each of the forms : 1 mod4 and 3 mod4 so the asymptotic behavior is same for each of them https://gyazo.com/3a056dd0d8a7c6c07dbae6b805c379df
ganeshie8
  • ganeshie8
extension of prime number theorem to arithmetic progressions is something beyong my grasp..
dan815
  • dan815
u know something knowing someone like chebyshev was the one to discover it, he probably showed something like the difference is getting less and less proportionally and maybe show how the standard deviation could be changing approximately too
dan815
  • dan815
i was working on something similiar to this the other day on here, i found out it was something very similar to the central limit theorem
ganeshie8
  • ganeshie8
interesting, central limit theorem relates sampling distributions to sample distribution right ?
dan815
  • dan815
yeah
dan815
  • dan815
it comes in trying to answer this question " if u notice some standard deviation for n samples, what standard deviation will you notice for k*n samples"
ganeshie8
  • ganeshie8
the more samples u take, the more steep the normal curve of the distribution of samples becomes ?
dan815
  • dan815
i didnt really finish it lol i just got some integral expression
dan815
  • dan815
http://prntscr.com/88k0av http://prntscr.com/88k0hm i got upto there and forgot about it xD
dan815
  • dan815
but i think the steps i took there is pretty logical its how someone naive would try to approach it
dan815
  • dan815
like if u flip 2 coins heads and tails, you will expect quite a bit of variation often
ganeshie8
  • ganeshie8
Ahh okay, i remember it from some highschool level statistics course.. i don't think they took time to prove CLT but i remember doing many problems that use CLT. .
dan815
  • dan815
and how the variation changes can be shown with some combinatorics
dan815
  • dan815
by looking at hte prob of 2 heads and so on, then u can do the same thing with 10 coins and see how the prob changes then you try to relate the 2 prbabilities as a function of the grown in sample space
dan815
  • dan815
and u end up with the central limit theorem i think
dan815
  • dan815
u wud expect bigger fluctuations around the mean for smaller sample spaces because theres higher probabilties for the values around the mean there
dan815
  • dan815
|dw:1440471255706:dw|
Empty
  • Empty
``` 8n+1 counter = 2384 8n+3 counter = 2409 8n+5 counter = 2399 8n+7 counter = 2399 ``` So here's the output of looking at the number of primes up to 100,000 similar to how they did it here where @ganeshie8 showed us earlier http://www.jstor.org/stable/27641834?&seq=1#page_scan_tab_contents
Empty
  • Empty
You can see that 8n+1 counter and 8n+3 counter's total will add up to 4783 which is the value of 4n+1 like we'd expect, listed on his site. Just thought it'd be interesting to show, I can put my Java code if you're curious.
ganeshie8
  • ganeshie8
``` 4n+1 --> 8n+1, 8n+5 4n+3 --> 8n+3, 8n+7 ``` kai just a basic question, you're splitting 4n+1 into two subtrees by considering n=even/odd right ?
Empty
  • Empty
That's one way to look at it, yeah
Empty
  • Empty
Here's my code, check it out and play around: https://repl.it/BD31/1
ganeshie8
  • ganeshie8
it is very interesting, how evenly they distribute themselves among 16a+b progressions!
Empty
  • Empty
Yeah I'm not sure if this is to be expected or not for them to be so uniformly distributed like this, very interesting
Empty
  • Empty
Also if you see any ways to make my code faster, I think my prime checking method is kind of poorly made but it gets the job done at least haha
Empty
  • Empty
Uh oh I think I made a tiny bug left in there
ganeshie8
  • ganeshie8
i came across this witty limit just some time back \[\lim\limits_{n\to\infty} \dfrac{n\log n}{p_n} = 1\] \(p_n\) is nth prime not gonna share the proof as i don't wanto spoil the fun incase if you haven't seen it before..
Empty
  • Empty
``` 16n+1 counter = 1188 16n+3 counter = 1208 16n+5 counter = 1200 16n+7 counter = 1200 16n+9 counter = 1196 16n+11 counter = 1201 16n+13 counter = 1199 16n+15 counter = 1199 ``` That looks better I think
Empty
  • Empty
Oh that looks fun like a modification on the prime number theorem thingy maybe? I'll try playing around.
ganeshie8
  • ganeshie8
haha all i can think of is putting sqrt for that n ``` for (int i = 2; i < Math.sqrt(n); i++) ```
ganeshie8
  • ganeshie8
yes it has to do with prime number thm..
Empty
  • Empty
Oh right of course! XD
Empty
  • Empty
Ahhh also it looks like \(p_n\)~ n log n
Empty
  • Empty
Improving the code slightly https://repl.it/BD31/2 I got some fun results up to a million ``` max = 1000000 16n+1 counter = 9841 16n+3 counter = 9838 16n+5 counter = 9816 16n+7 counter = 9832 16n+9 counter = 9878 16n+11 counter = 9815 16n+13 counter = 9807 16n+15 counter = 9837 max = 1000000 4n+1 counter = 39342 4n+3 counter = 39322 ``` Interesting how close it stays, it is still only 20 higher, very peculiar!
Empty
  • Empty
I think I'll start making some methods to calculate the standard deviation and mean and things like that
Empty
  • Empty
Although maybe have something like an ordered dependent mean so we can see when one surpasses the other or look at the stability of these things with each new number added
Empty
  • Empty
Or any other kind of fun ideas we should try looking at?
ganeshie8
  • ganeshie8
sometimes i use lookup tables for prime numbers instead of generating them everytime http://primes.utm.edu/lists/small/millions/ just to make it run faster..
ganeshie8
  • ganeshie8
i never had to wry about primes beyond 50 million using code before.. i can wget them and upload the file in JSON format if you want..
Empty
  • Empty
I think I can play with them, but I think before I worry about optimizing my code to look at some 50 million crazy primes I think playing around and generating the only 100,000 primes every time doesn't really slow me down too much to look at stufff.
Empty
  • Empty
I guess what I'm saying is I just wanna see if we can find anything cool, and if we think it's worth looking at then we can look past the small stuff and see if this extends to the millions of primes or whatever. Right now I'm not sure what to think about all this so I don't want to put too much effort into something I might not use tomorrow XD
ganeshie8
  • ganeshie8
``` for (int i = 3; i < Math.sqrt(n); i += 2) ``` this returns true for numbers of form \(p^2\) too.. need an equality there when i ran ur program, it didnt change the counter for 4n+3 for obvious reasons :)

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