Register to reply

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

by ramsey2879
Tags: primes, squares, triangular numbers
Share this thread:
ramsey2879
#1
Nov11-12, 09:02 PM
P: 894
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.
Phys.Org News Partner Science news on Phys.org
New type of solar concentrator desn't block the view
Researchers demonstrate ultra low-field nuclear magnetic resonance using Earth's magnetic field
Asian inventions dominate energy storage systems
phillip1882
#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
#3
Nov14-12, 01:15 PM
P: 894
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
#4
Nov14-12, 10:39 PM
Homework
Sci Advisor
HW Helper
Thanks
P: 9,787
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
#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.


Register to reply

Related Discussions
Proof about a prime between k and 2k. Linear & Abstract Algebra 6
Prove infinitely many prime of the form 6k+5 Calculus & Beyond Homework 3
Prime numbers of given form Linear & Abstract Algebra 7
General form of prime no.s Linear & Abstract Algebra 15
A proof is a proof-says Canadian Prime Minister General Math 0