Question

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

Answer 1

@ganeshie8 @sauravshakya

Answer 2

@Neemo this group is a good place for us amigo :)

Answer 3

*

Answer 4

\[
p=3 ;\, \quad \frac{2^{p-1}-1}{p}=1\\
p=7 ;\, \quad \frac{2^{p-1}-1}{p}=9\\

\]

Answer 5

I think that this is all the primes solutions. I am still working on proving it.

Answer 6

yes sir thats just \(p=3,7\)

Answer 7

@ganeshie8 got any idea?

Answer 8

not yet,

\(\frac{2^{p-1}-1}{p} = n^2\)

\(2^{p-1} - 1= p*n^2\)

Answer 9

not getting any ideas :S

Answer 10

p=2 is not answer so p is odd\[(2^{\frac{p-1}{2}}-1)(2^{\frac{p-1}{2}}+1)=pn^2\]

Answer 11

very well that makes sense muku.... now we use the factoring thing as we did in prev problems... or this one is hard ?

Answer 12

no this is not about factoring after this

Answer 13

left side factors are co-prime

Answer 14

ahh thats the point gane

Answer 15

so n^2 must equal to one of the factors ?

Answer 16

or if one factor is 1, other factor can equal pn^2

Answer 17

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

Answer 18

ur first guess is right but we need to make that more accurate

Answer 19

we can say n^2 must equal one of the factors ? i dont see hw :|

Answer 20

il grab som coffee brb :)

Answer 21

ok

suppose that p is divisor of first factor \[\frac{2^{\frac{p-1}{2}}-1}{p}(2^{\frac{p-1}{2}}+1)=n^2\]

Answer 22

again\[\gcd(\frac{2^{\frac{p-1}{2}}-1}{p}\ , \ 2^{\frac{p-1}{2}}+1)=1\]

Answer 23

so one of the factors is 1 and other is n^2

Answer 24

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 ?

Answer 25

gane im wrong

Answer 26

and ur last reply is our start point...so lets work on that

Answer 27

but im sure this is right :)

if \(ab=n^2\) and \(\gcd(a,b)=1\) then\[a=m^2\]\[b=k^2\]

Answer 28

@ganeshie8

Answer 29

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 \(p=7\)

Answer 30

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?

Answer 31

*my last reply\[(2^{\frac{p-1}{2}}-1)\frac{2^{\frac{p-1}{2}}+1}{p}=n^2\]

Answer 32

\[2^{\frac{p-1}{2}}-1=m^2\]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 \(p=3\)

Answer 33

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

Answer 34

these values makes that expression integer ??

Answer 35

the above expression is a perfect square.

Answer 36

between 1000-10000, there is only
1009 1013 1019 1021

Answer 37

exper are u sure? because p=113 dont makes it a perfect square

Answer 38

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

Answer 39

I'm not good with modular arithmetic

Answer 40

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?

Answer 41

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.

Answer 42

what is integerpart function in matlab?

Answer 43

http://www.mathworks.com/matlabcentral/newsreader/view_thread/163080

Answer 44

Access Denied

Answer 45

Damn

Answer 46

try viewing with this
http://hidemyass.com/

Answer 47

does this work for you
http://2.hidemyass.com/ip-1/encoded/Oi8vd3d3Lm1hdGh3b3Jrcy5jb20vbWF0bGFiY2VudHJhbC9uZXdzcmVhZGVyL3ZpZXdfdGhyZWFkLzE2MzA4MA%3D%3D&f=norefer

Answer 48

i got it

Answer 49

i think MATLAB cuts the fractional part for big numbers

Answer 50

might ... how do you know that 2^113 - 1/ 113 is not a perfect square.

Answer 51

\[\sqrt{\frac{2^{112}-1}{113}}=6778608243699229.0869912287347921\]

Answer 52

oh .. they are all primes ... didn't note that. BYW what calculator are you using? this is giving me ... e ....

Answer 53

lol...my windows calc

Answer 54

looks like matlab is unable to handle it.

Answer 55

yeah :/

Answer 56

i find numbers hard ... especially big numbers ... lol

Answer 57

:D

Answer 58

\((2^{\frac{p-1}{2}}-1)(2^{\frac{p-1}{2}}+1) = pn^2\)

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 ?

Answer 59

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

Answer 60

wow !! so RHS can have prime factorization in 2 only. beautiful proof muku :)

Answer 61

and second case is obvious !!

Answer 62

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