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mukushla

  • 2 years ago

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

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  1. mukushla
    • 2 years ago
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    @ganeshie8 @sauravshakya

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

  3. eliassaab
    • 2 years ago
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    *

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

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

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

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

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

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

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

  11. ganeshie8
    • 2 years ago
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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 ?

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

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

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

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

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

  17. ganeshie8
    • 2 years ago
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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

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

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

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

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

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

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

  24. ganeshie8
    • 2 years ago
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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 ?

  25. mukushla
    • 2 years ago
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    gane im wrong

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

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

  28. mukushla
    • 2 years ago
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    @ganeshie8

  29. mukushla
    • 2 years ago
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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 \(p=7\)

  30. mukushla
    • 2 years ago
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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?

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

  32. mukushla
    • 2 years ago
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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 \(p=3\)

  33. experimentX
    • 2 years ago
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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

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

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

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

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

  38. experimentX
    • 2 years ago
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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

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

  40. mukushla
    • 2 years ago
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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?

  41. experimentX
    • 2 years ago
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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.

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

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

  44. mukushla
    • 2 years ago
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    Access Denied

  45. experimentX
    • 2 years ago
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    Damn

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

  47. mukushla
    • 2 years ago
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    i got it

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

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

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

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

  52. mukushla
    • 2 years ago
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    lol...my windows calc

  53. experimentX
    • 2 years ago
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    looks like matlab is unable to handle it.

  54. mukushla
    • 2 years ago
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    yeah :/

  55. experimentX
    • 2 years ago
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    i find numbers hard ... especially big numbers ... lol

  56. mukushla
    • 2 years ago
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    :D

  57. ganeshie8
    • 2 years ago
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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 ?

  58. mukushla
    • 2 years ago
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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

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

  60. ganeshie8
    • 2 years ago
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    and second case is obvious !!

  61. mukushla
    • 2 years ago
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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

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