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Mathematics Forum
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| The Riemann Hypothesis Quote | Reply | | Quadratic Solutions posted on: 6/4/2006 2:52:39 AM Given that the sqrt(B^2 - 4AC) part of the Q.F. is an integer and A = 1 and C is a known constant, how do I find the values of B. ie, x^2 + Bx + C = 0; C is a known constant What values of B give integer solutions to the quadratic? |
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Euler
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Quadratic Solutions
replied on: 6/4/2006 3:28:31 AM I am not sure if you don't know the answer or if you do and are simply testing people; nevertheless- I shall attempt to do the best I can in my somniferous state. if sqrt(B^2-4AC)=I, where I is some integer such that I=...,1,2,3,... then x=(-B +- I)/2 therefore B= -2x +- I, again where I is a number of the form: I=...,1,2,3,... Hope that is somewhat close, but maybe I will do better after getting some much needed sleep.... |
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The Riemann Hypothesis
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Quadratic Solutions
replied on: 6/4/2006 4:29:30 AM Thanks for replying, but in this problem x is unknow. What I need is to find B in terms of C in: B^2 = I^2 + C Where I is unknown. I don't want to have to go through and test several values of I. I was hoping that there would be a generating function for B in terms of C that didn't involve using I at all. |
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Euler
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Quadratic Solutions
replied on: 6/4/2006 1:00:28 PM Well, what you said: B^2 = I^2 + C is close in that case. If you let sqrt(B^2-4Ac)=I then B^2-4Ac=I^2 and consequently, B^2=I^2 + 4Ac or, B= sqrt(I^2+4Ac) and since A is one and c is a known constant you can say that, 4Ac=C, where C is a known constant; therefore- B=sqrt(I^2+C)... |
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The Riemann Hypothesis
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Quadratic Solutions
replied on: 6/5/2006 3:01:52 AM You're still involving I in the solution. I want a solution that DOESN'T use I. For example: consider x^2 + Bx - 2 = 0 Hence the soutions are: (-B +- sqrt(B^2 + 24)) / 2 Look at sqrt(B^2 + 24) This has integer solutions when B^2 + 24 is a perfect square. By inspection, if B = 1, B^2 + 24 = 25, which is a perfect square. I want a function that will give me B without having to use I as a variable. I want to jump straight from C = -2 to B = 1 |
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Euler
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Quadratic Solutions
replied on: 6/5/2006 11:49:04 AM I know you want B in terms of C, I was simply trying to show that there are infinite solutions for a given C. For any given C you can use any integer value of I and you would have a specific solution out of the ininitely many for that particular C. |
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The Riemann Hypothesis
Quote | Reply | This message was updated on 6/6/2006 8:45:56 AM by The Riemann Hypothesis |
Quadratic Solutions
replied on: 6/6/2006 6:55:44 AM There are an infinite number of solutions, yes. But I = 1,2,3,4,5,...k are not all integer solutions. So I need a generating function that will generate the integer solutions. |
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The Riemann Hypothesis
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Quadratic Solutions
replied on: 6/6/2006 10:17:40 PM Never mind. Thanks for your help though. |
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Euler
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Quadratic Solutions
replied on: 6/14/2006 10:21:29 PM Sorry, I don't think it is possible to achieve what you are asking. The closest thing I can do is what I have already, just you weed out the values of I which satisfy: [I^2+C](mod n)=n; thereby weeding out the perfect squares, but further than that, I don't think there is anything, but I will continue to think about it and get back with you should I find something else. |
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