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Re: a contradiction worth reading (unless a silly mistake was made)
Posted:
Jul 24, 2011 8:52 PM
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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, ...,(n-1)} = {x^(2^j)
>>>>>(mod p),`x = 1,2,3, ...,(n-1)}. 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, ..., (n-1)}
>>>
>>>> equals
>>>
>>>> x^(2^j) (mod p), x = 1,2,3, ..., (n-1)}
>>>
>>> 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 1-1 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
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