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 February 12th, 2009, 05:39 PM #1 Newbie   Joined: Feb 2009 Posts: 4 Thanks: 0 Trying to make history, or at least understand primes better Hi, everybody, I’m new here but I have been on other math forums. This one looks pretty active, so I thought I would ask if anyone has any ideas about a little conjecture of mine. I have been working for a number of years on this problem, which I have recently re-framed into a question about a type of quadratic equation. If you have seen it on another forum before, it might have changed since I converted it into a quadratic eqn. Please take a look, and contemplate for a minute. Maybe something will "hit" you that I'm just not seeing. Here it is, in brief: Let p be a prime number > 7 and let n be the largest integer such that the nth prime (p_n) is less than root(p). Conjecture: For each such p there exists n positive integers (k_1, k_2, .... k_n), such that the equation x^2+px-c=0 has integer roots, where c = 2^k_1 * 3^k_2 * ... * (p_n)^k_n. ================================================== ======================= In case it helps, the original way of phrasing the conjecture went something like this: Let p and n and k_1…k_n be as given above. Then we can partition the set {p_1….p_n} into two disjoint sets, {p_i…..p_m} and {p_q….p_r}. Let A=(p_i)^(k_i)*…*(p_m)^(k_m) and let B=(p_q)^(k_q)*…*(p_r)^(k_r). In other words, A and B are relatively prime, p_n primorial divides AB, and no prime larger than p_n divides AB. Conjecture 1 (easy to prove) the absolute value of A-B is prime as long it is less than the square of p_(n+1). Conjecture 2 (unproven) for ANY prime less than the square root of p_(n+1), for ANY n, we can find a set of positive integers {k_1…k_n} and a partition of {p_1…p_n} such that the construction: abs(A-B) as given above will equal the given prime. Note that the variables here are the k’s, and the actual partition. In other words, the A-B construction will always produce a prime (given the less-than condition), but WILL it give EVERY prime less than the square of p_(n+1), given that we can change the exponents and the partition? ======================= Hopefully someone can help me see a way to attack this problem. I have looked at the quadratic formula until I'm blue in the face, obviously p^2-4c must be a perfect square. I have looked into matrices, I have tried to bring trig into it even. There seems to be patterns, but nothing that leads to a closed form for the exponents (that would be too much to expect, anyway). Since the options for exponents are unbounded, you might think that there would have to be a way to prove that SOME set of exponents would work for each p, and the quadratic looks like a simple enough approach. I have verified this out past 1000 and there simply must be a way to prove it or disprove it, once and for all. Please help, any ideas are welcome. Thank you.  February 13th, 2009, 09:34 AM   #2
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Re: Trying to make history, or at least understand primes better

Quote:
 Originally Posted by approx obviously p^2-4c must be a perfect square
p^2 + 4c, I think.

Working mod 9, you can show that 9 | c, thus k_2 >= 2. I imagine similar things can be shown mod other prime powers. Have you tried this approach? February 18th, 2009, 06:50 PM #3 Newbie   Joined: Feb 2009 Posts: 4 Thanks: 0 Re: Trying to make history, or at least understand primes better Thank you for the idea, I should have thought of it before (actually I looked at it with 2, but not 3 or the others). Sorry I haven't had time to reply recently. I'll look into this approach, and thanks again!   February 18th, 2009, 07:03 PM #4 Newbie   Joined: Feb 2009 Posts: 4 Thanks: 0 Re: Trying to make history, or at least understand primes better Okay, I can see that p^2, being odd and a square, must be congruent to 1 mod 6, and that p^2+4c, also being an odd square, must be congruent to 1 mod 6, but that just tells me that 4c is a multiple of 6, I'm still working on it but I don't get the "multiple of 9" part... February 18th, 2009, 08:52 PM #5 Global Moderator   Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Trying to make history, or at least understand primes better List the squares mod 9. February 24th, 2009, 05:23 PM #6 Newbie   Joined: Feb 2009 Posts: 4 Thanks: 0 disagree x^2+13x-48=0 has integer roots, yet 9 does not divide 48. February 24th, 2009, 06:10 PM #7 Global Moderator   Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Trying to make history, or at least understand primes better Oops, then? Maybe I switched from working mod 9 to mod 8 at some point. Tags history, make, primes, understand Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post caters Number Theory 67 March 19th, 2014 04:32 PM nikkor180 Calculus 0 April 29th, 2011 09:08 PM slick5657 New Users 2 June 11th, 2009 06:24 AM CRGreathouse Number Theory 0 November 6th, 2008 08:13 AM cknapp New Users 10 January 23rd, 2008 06:01 PM

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