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mukushla Group Title

Find all prime numbers \(p\) such that\[\frac{2^{p-1}-1}{p}\]is a perfect square

  • 2 years ago
  • 2 years ago

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  1. mukushla Group Title
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    @ganeshie8 @sauravshakya

    • 2 years ago
  2. mukushla Group Title
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    @Neemo this group is a good place for us amigo :)

    • 2 years ago
  3. eliassaab Group Title
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    *

    • 2 years ago
  4. eliassaab Group Title
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    \[ p=3 ;\, \quad \frac{2^{p-1}-1}{p}=1\\ p=7 ;\, \quad \frac{2^{p-1}-1}{p}=9\\ \]

    • 2 years ago
  5. eliassaab Group Title
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    I think that this is all the primes solutions. I am still working on proving it.

    • 2 years ago
  6. mukushla Group Title
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    yes sir thats just \(p=3,7\)

    • 2 years ago
  7. sauravshakya Group Title
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    @ganeshie8 got any idea?

    • 2 years ago
  8. ganeshie8 Group Title
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    not yet, \(\frac{2^{p-1}-1}{p} = n^2\) \(2^{p-1} - 1= p*n^2\)

    • 2 years ago
  9. ganeshie8 Group Title
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    not getting any ideas :S

    • 2 years ago
  10. mukushla Group Title
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    p=2 is not answer so p is odd\[(2^{\frac{p-1}{2}}-1)(2^{\frac{p-1}{2}}+1)=pn^2\]

    • 2 years ago
  11. ganeshie8 Group Title
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    very well that makes sense muku.... now we use the factoring thing as we did in prev problems... or this one is hard ?

    • 2 years ago
  12. mukushla Group Title
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    no this is not about factoring after this

    • 2 years ago
  13. ganeshie8 Group Title
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    left side factors are co-prime

    • 2 years ago
  14. mukushla Group Title
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    ahh thats the point gane

    • 2 years ago
  15. ganeshie8 Group Title
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    so n^2 must equal to one of the factors ?

    • 2 years ago
  16. ganeshie8 Group Title
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    or if one factor is 1, other factor can equal pn^2

    • 2 years ago
  17. ganeshie8 Group Title
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    that 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

    • 2 years ago
  18. mukushla Group Title
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    ur first guess is right but we need to make that more accurate

    • 2 years ago
  19. ganeshie8 Group Title
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    we can say n^2 must equal one of the factors ? i dont see hw :|

    • 2 years ago
  20. ganeshie8 Group Title
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    il grab som coffee brb :)

    • 2 years ago
  21. mukushla Group Title
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    ok suppose that p is divisor of first factor \[\frac{2^{\frac{p-1}{2}}-1}{p}(2^{\frac{p-1}{2}}+1)=n^2\]

    • 2 years ago
  22. mukushla Group Title
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    again\[\gcd(\frac{2^{\frac{p-1}{2}}-1}{p}\ , \ 2^{\frac{p-1}{2}}+1)=1\]

    • 2 years ago
  23. mukushla Group Title
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    so one of the factors is 1 and other is n^2

    • 2 years ago
  24. ganeshie8 Group Title
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    i believe u muku but i dont understand how thats possible, \( \frac{2^{\frac{p-1}{2}}-1}{p}(2^{\frac{p-1}{2}}+1)=n^2 \) 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 ?

    • 2 years ago
  25. mukushla Group Title
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    gane im wrong

    • 2 years ago
  26. mukushla Group Title
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    and ur last reply is our start point...so lets work on that

    • 2 years ago
  27. mukushla Group Title
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    but im sure this is right :) if \(ab=n^2\) and \(\gcd(a,b)=1\) then\[a=m^2\]\[b=k^2\]

    • 2 years ago
  28. mukushla Group Title
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    @ganeshie8

    • 2 years ago
  29. mukushla Group Title
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    suppose that \(p\) is a divisor of \(2^{\frac{p-1}{2}}-1\)\[\frac{2^{\frac{p-1}{2}}-1}{p}(2^{\frac{p-1}{2}}+1)=n^2\]this implies that\[\frac{2^{\frac{p-1}{2}}-1}{p}=k^2\]\[2^{\frac{p-1}{2}}+1=m^2\]\(m,k\) are odd numbers\[2^{\frac{p-1}{2}}+1=m^2=4t^2+4t+1\]\[2^{\frac{p-1}{2}}=4t^2+4t\]\[2^{\frac{p-1}{2}-2}=t(t+1)\]so \(t=1\) and \(p=7\)

    • 2 years ago
  30. mukushla Group Title
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    now suppose that \(p\) is a divisor of \(2^{\frac{p-1}{2}}+1\)\[\frac{2^{\frac{p-1}{2}}-1}{p}(2^{\frac{p-1}{2}}+1)=n^2\]this implies that\[\frac{2^{\frac{p-1}{2}}+1}{p}=k^2\]\[2^{\frac{p-1}{2}}-1=m^2\]now what?

    • 2 years ago
  31. mukushla Group Title
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    *my last reply\[(2^{\frac{p-1}{2}}-1)\frac{2^{\frac{p-1}{2}}+1}{p}=n^2\]

    • 2 years ago
  32. mukushla Group Title
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    \[2^{\frac{p-1}{2}}-1=m^2\]m is odd\[2^{\frac{p-1}{2}}-1=4t^2+4t+1\]\[2^{\frac{p-1}{2}}-2=4t(t+1)\]RHS is divisible by 4 but LHS not this implies that \(t=0\) so \(p=3\)

    • 2 years ago
  33. experimentX Group Title
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    here a numeric solution 1-1000 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

    • 2 years ago
  34. mukushla Group Title
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    these values makes that expression integer ??

    • 2 years ago
  35. experimentX Group Title
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    the above expression is a perfect square.

    • 2 years ago
  36. experimentX Group Title
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    between 1000-10000, there is only 1009 1013 1019 1021

    • 2 years ago
  37. mukushla Group Title
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    exper are u sure? because p=113 dont makes it a perfect square

    • 2 years ago
  38. experimentX Group Title
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    i 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

    • 2 years ago
  39. experimentX Group Title
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    I'm not good with modular arithmetic

    • 2 years ago
  40. mukushla Group Title
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    ahh...my guess is right we want this expression\[\frac{2^{p-1}-1}{p}\]to be a perfect square not an integer...can u write a little code for it?

    • 2 years ago
  41. experimentX Group Title
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    IDK 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.

    • 2 years ago
  42. mukushla Group Title
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    what is integerpart function in matlab?

    • 2 years ago
  43. experimentX Group Title
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    http://www.mathworks.com/matlabcentral/newsreader/view_thread/163080

    • 2 years ago
  44. mukushla Group Title
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    Access Denied

    • 2 years ago
  45. experimentX Group Title
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    Damn

    • 2 years ago
  46. experimentX Group Title
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    try viewing with this http://hidemyass.com/

    • 2 years ago
  47. experimentX Group Title
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    does this work for you http://2.hidemyass.com/ip-1/encoded/Oi8vd3d3Lm1hdGh3b3Jrcy5jb20vbWF0bGFiY2VudHJhbC9uZXdzcmVhZGVyL3ZpZXdfdGhyZWFkLzE2MzA4MA%3D%3D&f=norefer

    • 2 years ago
  48. mukushla Group Title
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    i got it

    • 2 years ago
  49. mukushla Group Title
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    i think MATLAB cuts the fractional part for big numbers

    • 2 years ago
  50. experimentX Group Title
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    might ... how do you know that 2^113 - 1/ 113 is not a perfect square.

    • 2 years ago
  51. mukushla Group Title
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    \[\sqrt{\frac{2^{112}-1}{113}}=6778608243699229.0869912287347921\]

    • 2 years ago
  52. experimentX Group Title
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    oh .. they are all primes ... didn't note that. BYW what calculator are you using? this is giving me ... e ....

    • 2 years ago
  53. mukushla Group Title
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    lol...my windows calc

    • 2 years ago
  54. experimentX Group Title
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    looks like matlab is unable to handle it.

    • 2 years ago
  55. mukushla Group Title
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    yeah :/

    • 2 years ago
  56. experimentX Group Title
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    i find numbers hard ... especially big numbers ... lol

    • 2 years ago
  57. mukushla Group Title
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    :D

    • 2 years ago
  58. ganeshie8 Group Title
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    \((2^{\frac{p-1}{2}}-1)(2^{\frac{p-1}{2}}+1) = pn^2\) two cases : case I : \(p \ |\ (2^{\frac{p-1}{2}}-1) \) => \(2^{\frac{p-1}{2}}+1 = m^2\) \(m\) is an odd number \(2^{\frac{p-1}{2}}+1 = m^2 = (2t+1)^2\) \(2^{\frac{p-1}{2}-2} = m^2 = t(t+1)\) now what ? we do trial and error is it ? and hw do we knw only one solution exist for this case ?

    • 2 years ago
  59. mukushla Group Title
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    \[2^{\frac{p-1}{2}}+1 = m^2 = (2t+1)^2=4t^2+4t+1\]\[2^{\frac{p-1}{2}} =4t^2+4t=4t(t+1)\]\[2^{\frac{p-1}{2}-2}=t(t+1)\]if \(t>1\) RHS has a an odd factor but LHS is a power of 2 .. so t=1

    • 2 years ago
  60. ganeshie8 Group Title
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    wow !! so RHS can have prime factorization in 2 only. beautiful proof muku :)

    • 2 years ago
  61. ganeshie8 Group Title
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    and second case is obvious !!

    • 2 years ago
  62. mukushla Group Title
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    yes 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

    • 2 years ago
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