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anonymous
 4 years ago
Find all prime numbers \(p\) such that\[\frac{2^{p1}1}{p}\]is a perfect square
anonymous
 4 years ago
Find all prime numbers \(p\) such that\[\frac{2^{p1}1}{p}\]is a perfect square

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@ganeshie8 @sauravshakya

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@Neemo this group is a good place for us amigo :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[ p=3 ;\, \quad \frac{2^{p1}1}{p}=1\\ p=7 ;\, \quad \frac{2^{p1}1}{p}=9\\ \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I think that this is all the primes solutions. I am still working on proving it.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes sir thats just \(p=3,7\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@ganeshie8 got any idea?

ganeshie8
 4 years ago
Best ResponseYou've already chosen the best response.1not yet, \(\frac{2^{p1}1}{p} = n^2\) \(2^{p1}  1= p*n^2\)

ganeshie8
 4 years ago
Best ResponseYou've already chosen the best response.1not getting any ideas :S

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0p=2 is not answer so p is odd\[(2^{\frac{p1}{2}}1)(2^{\frac{p1}{2}}+1)=pn^2\]

ganeshie8
 4 years ago
Best ResponseYou've already chosen the best response.1very well that makes sense muku.... now we use the factoring thing as we did in prev problems... or this one is hard ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no this is not about factoring after this

ganeshie8
 4 years ago
Best ResponseYou've already chosen the best response.1left side factors are coprime

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ahh thats the point gane

ganeshie8
 4 years ago
Best ResponseYou've already chosen the best response.1so n^2 must equal to one of the factors ?

ganeshie8
 4 years ago
Best ResponseYou've already chosen the best response.1or if one factor is 1, other factor can equal pn^2

ganeshie8
 4 years ago
Best ResponseYou've already chosen the best response.1that looks wrong, cuz if n^2 = k^2m^2l^2... , then few prime factors for ex, k^2m^2 can make first factor, while remaining pl^2 can make the second factor.. hmm

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ur first guess is right but we need to make that more accurate

ganeshie8
 4 years ago
Best ResponseYou've already chosen the best response.1we can say n^2 must equal one of the factors ? i dont see hw :

ganeshie8
 4 years ago
Best ResponseYou've already chosen the best response.1il grab som coffee brb :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok suppose that p is divisor of first factor \[\frac{2^{\frac{p1}{2}}1}{p}(2^{\frac{p1}{2}}+1)=n^2\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0again\[\gcd(\frac{2^{\frac{p1}{2}}1}{p}\ , \ 2^{\frac{p1}{2}}+1)=1\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so one of the factors is 1 and other is n^2

ganeshie8
 4 years ago
Best ResponseYou've already chosen the best response.1i believe u muku but i dont understand how thats possible, \( \frac{2^{\frac{p1}{2}}1}{p}(2^{\frac{p1}{2}}+1)=n^2 \) lets say first factor = k^2, and second factor = m^2 n = km hw can we say one of the factor must equal n^2 ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and ur last reply is our start point...so lets work on that

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but im sure this is right :) if \(ab=n^2\) and \(\gcd(a,b)=1\) then\[a=m^2\]\[b=k^2\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0suppose that \(p\) is a divisor of \(2^{\frac{p1}{2}}1\)\[\frac{2^{\frac{p1}{2}}1}{p}(2^{\frac{p1}{2}}+1)=n^2\]this implies that\[\frac{2^{\frac{p1}{2}}1}{p}=k^2\]\[2^{\frac{p1}{2}}+1=m^2\]\(m,k\) are odd numbers\[2^{\frac{p1}{2}}+1=m^2=4t^2+4t+1\]\[2^{\frac{p1}{2}}=4t^2+4t\]\[2^{\frac{p1}{2}2}=t(t+1)\]so \(t=1\) and \(p=7\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0now suppose that \(p\) is a divisor of \(2^{\frac{p1}{2}}+1\)\[\frac{2^{\frac{p1}{2}}1}{p}(2^{\frac{p1}{2}}+1)=n^2\]this implies that\[\frac{2^{\frac{p1}{2}}+1}{p}=k^2\]\[2^{\frac{p1}{2}}1=m^2\]now what?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0*my last reply\[(2^{\frac{p1}{2}}1)\frac{2^{\frac{p1}{2}}+1}{p}=n^2\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[2^{\frac{p1}{2}}1=m^2\]m is odd\[2^{\frac{p1}{2}}1=4t^2+4t+1\]\[2^{\frac{p1}{2}}2=4t(t+1)\]RHS is divisible by 4 but LHS not this implies that \(t=0\) so \(p=3\)

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0here a numeric solution 11000 3 7 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541 547 557 563 569 571 577 587 593 599 601 607 613 617 619 631 641 643 647 653 659 661 673 677 683 691 701 709 719 727 733 739 743 751 757 761 769 773 787 797 809 811 821 823 827 829 839 853 857 859 863 877 881 883 887 907 911 919 929 937 941 947 953 967 971 977 983 991 997

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0these values makes that expression integer ??

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0the above expression is a perfect square.

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0between 100010000, there is only 1009 1013 1019 1021

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0exper are u sure? because p=113 dont makes it a perfect square

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0i used this piece of code on matlab clc; p = primes(100000); for i=1:length(p) if rem(sqrt((2^(p(i)1)1)/p(i)),1) == 0; disp(p(i)); end; end

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0I'm not good with modular arithmetic

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ahh...my guess is right we want this expression\[\frac{2^{p1}1}{p}\]to be a perfect square not an integer...can u write a little code for it?

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0IDK how matlab works with big numbers... but that code should give you perfect square. for every prime p, the expression should be an integer by fermat last theorem.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what is integerpart function in matlab?

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0http://www.mathworks.com/matlabcentral/newsreader/view_thread/163080

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0try viewing with this http://hidemyass.com/

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0does this work for you http://2.hidemyass.com/ip1/encoded/Oi8vd3d3Lm1hdGh3b3Jrcy5jb20vbWF0bGFiY2VudHJhbC9uZXdzcmVhZGVyL3ZpZXdfdGhyZWFkLzE2MzA4MA%3D%3D&f=norefer

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i think MATLAB cuts the fractional part for big numbers

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0might ... how do you know that 2^113  1/ 113 is not a perfect square.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\sqrt{\frac{2^{112}1}{113}}=6778608243699229.0869912287347921\]

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0oh .. they are all primes ... didn't note that. BYW what calculator are you using? this is giving me ... e ....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lol...my windows calc

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0looks like matlab is unable to handle it.

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0i find numbers hard ... especially big numbers ... lol

ganeshie8
 4 years ago
Best ResponseYou've already chosen the best response.1\((2^{\frac{p1}{2}}1)(2^{\frac{p1}{2}}+1) = pn^2\) two cases : case I : \(p \ \ (2^{\frac{p1}{2}}1) \) => \(2^{\frac{p1}{2}}+1 = m^2\) \(m\) is an odd number \(2^{\frac{p1}{2}}+1 = m^2 = (2t+1)^2\) \(2^{\frac{p1}{2}2} = m^2 = t(t+1)\) now what ? we do trial and error is it ? and hw do we knw only one solution exist for this case ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[2^{\frac{p1}{2}}+1 = m^2 = (2t+1)^2=4t^2+4t+1\]\[2^{\frac{p1}{2}} =4t^2+4t=4t(t+1)\]\[2^{\frac{p1}{2}2}=t(t+1)\]if \(t>1\) RHS has a an odd factor but LHS is a power of 2 .. so t=1

ganeshie8
 4 years ago
Best ResponseYou've already chosen the best response.1wow !! so RHS can have prime factorization in 2 only. beautiful proof muku :)

ganeshie8
 4 years ago
Best ResponseYou've already chosen the best response.1and second case is obvious !!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes u made a good way...and finally we got it completely this posted regarding ur point http://openstudy.com/study#/updates/504857ade4b003bc12040f69 see sir @eliassaab reply
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