Need proof re prime of the form 8N +/-1


by ramsey2879
Tags: primes, squares, triangular numbers
ramsey2879
ramsey2879 is offline
#1
Nov11-12, 09:02 PM
P: 891
I need help or direction on how to prove that if A = S^2 - (T^2 + T)/2 Then 8A-1 can not be factored into the form B*C where B and C are coprime and each of the form 8N+/-3. For instance -4*8-1 = -33 can be factored as -3*(8+3) and 5*8-1 = 39 = 3*(8*2-3). Thus neither -4 or 5 can be expressed as S^2 -(T^2+T)/2 where S and T are integers.

So far I have proven that if A = f(S,T) = S^2 - (T^2+T)/2 then A = f(S',T') where S' = 3S + 2T +1 and T' = 4S + 3T + 1, but I don't know where to go from there.

Any ideas.
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phillip1882
phillip1882 is offline
#2
Nov14-12, 10:56 AM
P: 39
S^2 = 1, 4, 9, 16, 25 ...
(T^2 +T)/2 = 1, 3, 6, 10, 15, 21...

9-1 = 8; 8*8-1 = 63; 63 = 3*(8*3-3).
therefore your statement is false.
edit: nm, missed the co prime part.
okay; i programmed a check up to many values, as far as i can tell this is true.
how to prove ti is beyond me though.
ramsey2879
ramsey2879 is offline
#3
Nov14-12, 01:15 PM
P: 891
Quote Quote by phillip1882 View Post
S^2 = 1, 4, 9, 16, 25 ...
(T^2 +T)/2 = 1, 3, 6, 10, 15, 21...

9-1 = 8; 8*8-1 = 63; 63 = 3*(8*3-3).
therefore your statement is false.
edit: nm, missed the co prime part.
okay; i programmed a check up to many values, as far as i can tell this is true.
how to prove ti is beyond me though.
Thankyou for your post. Glad to see that someone was interested enough to check my apparent finding.

haruspex
haruspex is offline
#4
Nov14-12, 10:39 PM
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Need proof re prime of the form 8N +/-1


Suppose p is prime, 3 or 5 mod 8.
Easy to show that p cannot be expressed as 2*a2-b2.
Also seems to be true that if p|2*a2-b2 then so does p2. That looks like it might be associated with your observation.
Norwegian
Norwegian is offline
#5
Nov15-12, 07:41 AM
P: 144
Hi Ramsey, your observation is a consequence of the following:

Lemma: Let N=2x2-y2 with x and y integers. Let p|N be a prime of the form 8ką3. Then ordp(N) is even. (By ordp(N) we mean the exponent of p in the factorization of N.)

Proof: First recall that 2 is a quadratic residue modulo a prime q if and only if q is of the form 8ką1. Since p|N we have 2x2 = y2(mod p). Since 2 is a quadratic nonresidue, it follows that y=x=0 (mod p), and all the terms of the equation N=2x2-y2 can be divided by p2. Repeat as long as N has prime factors of the form 8ką3, and qed.

Your observation follows immediately from this by setting N=8A-1, x=2S, y=2T+1, and by observing that in a coprime factorization N=bc, all factors pa are of the form 8ką1.

I assume the lemma is well known, but I couldn't immediately find a reference. It is analogous to the celebrated theorem about sums of two squares, one version being: A positive integer N can be written as a sum of two squares if and only if for all primes p of the form 4k+3, ordp(N) is even.


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