
Re: a contradiction worth reading (unless a silly mistake was made)
Posted:
Jul 24, 2011 8:52 PM


quasi <quasi@null.set> wrote: >Simon Roberts <2assume@gmail.com> wrote: >>On Jul 23, 4:53 m, Simon Roberts <2ass...@gmail.com> wrote: >>> On Jul 23, 11:30am, quasi <qu...@null.set> wrote: >>>> Simon Roberts <2ass...@gmail.com> wrote: >>>>> >>>>>This is my proposistion, (i tried to mimic the terms or >>>>>style you used and probably all mathematicians use) >>>>> >>>>>p is prime and p=3 (mod 2). >>>>>Then I propose: {x^(2^i) (mod p), x = 1,2,3, ...,(n1)} = {x^(2^j) >>>>>(mod p),`x = 1,2,3, ...,(n1)}. i can be fixed to any, i is equal to >>>>>an, integer greater than or equal to one and j can be fixed to any, j >>>>>is equal to an, integer greater than or equal to one. >>> >>>> There are already several issues. >>> >>>> Firstly, you surely don't mean p = 3 (mod 2), since every odd >>> >>> yes p=3 (mod 4) >>> >>>> Also, you surely don't mean >>> >>>> x^(2^i) (mod p), x = 1,2,3, ..., (n1)} >>> >>>> equals >>> >>>> x^(2^j) (mod p), x = 1,2,3, ..., (n1)} >>> >>> no, this is EXACTLY what I mean. I sure do mean. What I have >>> meant all along. j!=i necessarily j>=1 and i>=1 they are both >>> fixed in these sets. >> >>If it's trivial, for goodness sakes, just tell me it's trivial. > > Well, maybe not trivial, but certainly an easy exercise. > In outline form, the argument is as follows ... > > Since p = 3 (mod 4), 1 is not a square (mod p), hence the > map f(x) = x^2 (mod p) is bijective on the set of nonzero > squares (mod p). Thus the set of nonzero squares (mod p) is > the same as the set of nonzero 4'th powers (mod p), hence is > also the same as the set of nonzero (2^i)'th powers (mod p) > for any positive integer i, which proves your claim.
SIMPLER and more generally: in a finite group G of order 4k+2 every square a^2 is the square of some other square b^2 since
a^2 = a^(4k+4) = (b^2)^2, for b = a^(k+1)
Hence the map x > x^2 is onto so 11 on the squares G^2.
In any group H of odd order 2k+1 the equation x^2 = c has solution x = c^(k+1) (a result due to Lagrange in 1769).
See also my post here http://math.stackexchange.com/q/4872/242
Bill Dubuque

