Andy wrote: << [I wrote]: << Find the set of prime numbers p such that p^2 == 1 (mod 24). If (n, 24) = 1, then n^2 == 1 (mod 24). So that would be all primes > 3.
This statement "If (n, 24) = 1, then n^2 == 1 (mod 24)" is true and easy to verify, but it is also profound. There is no number > 24 for which the corresponding statement is true. (This is also not hard to prove, but I wonder if there is a deep underlying reason it is true.) --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele
It's just that 2^3 and 3 are the only prime powers that have a solitary non-zero square. The next best is 5 which has two, namely 1 and 4, so that if n is prime to 120, there aren't many possibilities for n^2 mod 120. R. On Fri, 11 Apr 2008, Dan Asimov wrote:
Andy wrote:
<< [I wrote]:
<< Find the set of prime numbers p such that p^2 == 1 (mod 24).
If (n, 24) = 1, then n^2 == 1 (mod 24). So that would be all primes > 3.
This statement "If (n, 24) = 1, then n^2 == 1 (mod 24)"
is true and easy to verify, but it is also profound.
There is no number > 24 for which the corresponding statement is true.
(This is also not hard to prove, but I wonder if there is a deep underlying reason it is true.)
--Dan
_____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele
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On 4/11/08, Dan Asimov <dasimov@earthlink.net> wrote:
There is no number > 24 for which the corresponding statement is true.
(This is also not hard to prove, but I wonder if there is a deep underlying reason it is true.)
I think the reason is that for p > 2, exactly half the elements of Zp are quadratic residues. 3 is the largets prime for which (p-1)/2 = 1.
--Dan
_____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele
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