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February 12th, 2009, 05:39 PM   #1
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Trying to make history, or at least understand primes better

Hi, everybody, Im 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.
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In case it helps, the original way of phrasing the conjecture went something like this:

Let p and n and k_1k_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_1k_n} and a partition of {p_1p_n} such that the construction: abs(A-B) as given above will equal the given prime. Note that the variables here are the ks, 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?
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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.
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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?
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February 18th, 2009, 06:50 PM   #3
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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!
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February 18th, 2009, 07:03 PM   #4
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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...
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February 18th, 2009, 08:52 PM   #5
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Re: Trying to make history, or at least understand primes better

List the squares mod 9.
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February 24th, 2009, 05:23 PM   #6
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disagree

x^2+13x-48=0 has integer roots, yet 9 does not divide 48.
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February 24th, 2009, 06:10 PM   #7
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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.
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