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mukushla
Find all prime numbers \(p\) such that\[\frac{2^{p-1}-1}{p}\]is a perfect square
@ganeshie8 @sauravshakya
@Neemo this group is a good place for us amigo :)
\[ p=3 ;\, \quad \frac{2^{p-1}-1}{p}=1\\ p=7 ;\, \quad \frac{2^{p-1}-1}{p}=9\\ \]
I think that this is all the primes solutions. I am still working on proving it.
yes sir thats just \(p=3,7\)
@ganeshie8 got any idea?
not yet, \(\frac{2^{p-1}-1}{p} = n^2\) \(2^{p-1} - 1= p*n^2\)
not getting any ideas :S
p=2 is not answer so p is odd\[(2^{\frac{p-1}{2}}-1)(2^{\frac{p-1}{2}}+1)=pn^2\]
very well that makes sense muku.... now we use the factoring thing as we did in prev problems... or this one is hard ?
no this is not about factoring after this
left side factors are co-prime
ahh thats the point gane
so n^2 must equal to one of the factors ?
or if one factor is 1, other factor can equal pn^2
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
ur first guess is right but we need to make that more accurate
we can say n^2 must equal one of the factors ? i dont see hw :|
il grab som coffee brb :)
ok suppose that p is divisor of first factor \[\frac{2^{\frac{p-1}{2}}-1}{p}(2^{\frac{p-1}{2}}+1)=n^2\]
again\[\gcd(\frac{2^{\frac{p-1}{2}}-1}{p}\ , \ 2^{\frac{p-1}{2}}+1)=1\]
so one of the factors is 1 and other is n^2
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 ?
and ur last reply is our start point...so lets work on that
but im sure this is right :) if \(ab=n^2\) and \(\gcd(a,b)=1\) then\[a=m^2\]\[b=k^2\]
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 \(p=7\)
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?
*my last reply\[(2^{\frac{p-1}{2}}-1)\frac{2^{\frac{p-1}{2}}+1}{p}=n^2\]
\[2^{\frac{p-1}{2}}-1=m^2\]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 \(p=3\)
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
these values makes that expression integer ??
the above expression is a perfect square.
between 1000-10000, there is only 1009 1013 1019 1021
exper are u sure? because p=113 dont makes it a perfect square
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
I'm not good with modular arithmetic
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?
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.
what is integerpart function in matlab?
http://www.mathworks.com/matlabcentral/newsreader/view_thread/163080
try viewing with this http://hidemyass.com/
does this work for you http://2.hidemyass.com/ip-1/encoded/Oi8vd3d3Lm1hdGh3b3Jrcy5jb20vbWF0bGFiY2VudHJhbC9uZXdzcmVhZGVyL3ZpZXdfdGhyZWFkLzE2MzA4MA%3D%3D&f=norefer
i think MATLAB cuts the fractional part for big numbers
might ... how do you know that 2^113 - 1/ 113 is not a perfect square.
\[\sqrt{\frac{2^{112}-1}{113}}=6778608243699229.0869912287347921\]
oh .. they are all primes ... didn't note that. BYW what calculator are you using? this is giving me ... e ....
looks like matlab is unable to handle it.
i find numbers hard ... especially big numbers ... lol
\((2^{\frac{p-1}{2}}-1)(2^{\frac{p-1}{2}}+1) = pn^2\) 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 ?
\[2^{\frac{p-1}{2}}+1 = m^2 = (2t+1)^2=4t^2+4t+1\]\[2^{\frac{p-1}{2}} =4t^2+4t=4t(t+1)\]\[2^{\frac{p-1}{2}-2}=t(t+1)\]if \(t>1\) RHS has a an odd factor but LHS is a power of 2 .. so t=1
wow !! so RHS can have prime factorization in 2 only. beautiful proof muku :)
and second case is obvious !!
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