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Of course, what I meant was (p-1)/4 is an integer so long as p=1 or 3(mod 4).Posted by Matt Grime
floor anything is always an integer because that is what the floor function returns.
Of course, what I meant was (p-1)/4 is an integer so long as p=1 or 3(mod 4).Posted by Matt Grime
floor anything is always an integer because that is what the floor function returns.
I think you are losing the meaning behind the floor function. n is an integer, n<n+1/4<n+1 so we must have floor(n+1/4)=n.Oxymoron said:...=n+1/4
Yes. I was guilty of posting before checking.Posted by Shmoe
I think you are losing the meaning behind the floor function. n is an integer, n<n+1/4<n+1 so we must have floor(n+1/4)=n.
By this I mean prove a statement in general.shmoe said:If not, it may have been easier to proceed in a slightly different way. The elements of S less than p/2 are exactly the even integers less than p/2. You can count these directly, right?
Ok that works. You don't have to assume that x is an integer, there are floor(x) positive integers less than x in this case, and so on.Oxymoron said:1) The number of positive integers less than or equal to x is x.
2) The number of even positive integers less than or equal to x is
a) x/2 => x is even
b) floor(x/2) => x is odd
Righty-O. Apply Gauss's lemma and that's what you had set out to prove, minus evaluating this for p=1,3,5, 7 mod 8.Oxymoron said:3) The number of even integers in S less than or equal to p/2 is floor(p/4) since p is odd.
The number of even integers in S strictly larger than p/2 is
(p-1)/2 - floor(p/4)
Yep, so do the cases p=8n+/-1 or 3 seperately and evaluate the exponent in each case. You only care if this is even or odd in the end.Oxymoron said:But now I want to show that 2 is a QR for all primes p congruent to +1,-1 (mod 8) and not for primes congruent to +3,-3 (mod 8).
(8n-2)/2=4n-1 soOxymoron said:2. Let p=-1(mod 8) => p=8n-1
exponent = (8n-2)/2 - floor((8n-1)/4) = 4n - 1 - 2n = 2n-1 (ODD)