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Math Help - Find two numbers that satisfy these conditions

  1. #1
    Super Member Bacterius's Avatar
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    Find two numbers that satisfy these conditions

    Hi all,
    this is a question I've been doing. Let n (known) be some composite odd integer. Find two numbers b and c such that :
    - x^2 + bx + c + n = 0 is factorable over the real numbers, with integer coefficients in the standard form and factorized form.
    - x^2 + bx + c = 0 is factorable over the real numbers, with integer coefficients in the standard form and factorized form.

    I've been thinking, for a quadratic equation to satisfy these properties, its discriminant must be positive and be a perfect square. Let \Delta_1 and \Delta_2 denote the discriminants of equations one and two. We have :

    \Delta_1 = b^2 - 4(c + n)
    \Delta_2 = b^2 - 4c

    And I also noted that \Delta_1 - \Delta_2 = b^2 - 4(c + n) - b^2 + 4c = 4c - 4(c + n) = 4(c - c + n) = 4n, and this could be useful regarding the difference of two squares.

    And so, if I could find the values of \Delta_1 and \Delta_2, I could easily work out b and c with a simple system of equations. But what I've done beyond this point does not lead to anything interesting, so I'm wondering, do I have enough information about the discriminants to work them out ?

    Does anyone have any idea to put me on the right track again ? Thanks all
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  2. #2
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    Hello, Bacterius!

    I've one observation so far . . .


    Let n (known) be some composite odd integer.
    Find two numbers b and c such that :

    x^2 + bx + c + n \:=\: 0 is factorable over the real numbers, with integer coefficients.

    x^2 + bx + c \:=\: 0 is factorable over the real numbers, with integer coefficients.

    \text{Since }n\text{ is composite, let one factoring be: }\:n \:=\:p\cdot q

    \text{Let }c = 0\,\text{ and }\,b \:=\:p+q


    \text{Then both }\:\begin{Bmatrix}x^2 + (p+q)x + pq \\ \\[-3mm] x^2 + (p+q)x\end{Bmatrix}\:\text{ are factorable over the real numbers.}

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  3. #3
    Super Member Bacterius's Avatar
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    Thank you Soroban, I didn't see this particular case, thanks. Actually, I made a little sign mistake in my original post, it isn't (c + n) but (c - n) in the quadratics and the discriminants, however it does not change your solution. But what if the factorization of n is unknown ?

    For example, let n = 33. I want to find two numbers b and c that satisfy :
    - b^2 - 4c is a perfect square
    - b^2 - 4(c - n) is a perfect square

    After some guess and check, I find that two numbers that work are b = 2 and c = -15, because :

    2^2 - 4 \times (-15) = 64 = 8^2
    2^2 - 4 \times (-15 - 33) = 196 = 14^2

    And thus, x^2 + 2x - 15 is factorable over the reals with integer coefficients, as well as x^2 + 2x - 48.

    But there is a problem : I used guess and check. That's why I was looking for a particular process to find b and c, more efficient than trial.
    Last edited by Bacterius; February 13th 2010 at 09:17 PM.
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  4. #4
    Super Member Bacterius's Avatar
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    Ok, so this is the actual problem, reformulated in a easier to understand yet equivalent way :

    I am looking for two integers, x and y.
    I know the value of some integer n, that can vary.
    I also know that x^2 = b^2 - 4(c - n) and y^2 = b^2 - 4c for some integers b, c.
    I also know that x^2 - y^2 = b^2 - 4(c - n) - (b^2 - 4c) = b^2 - 4c + 4n - b^2 + 4c = 4n.

    Do I have enough information to find x and y ? Or do I need some other information like their sum or product ?
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  5. #5
    Junior Member SuperCalculus's Avatar
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    Argh asdf this is annoying. It seems to be that you need to use some form of substitution, but everything I try seems to go around in circles.
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  6. #6
    Super Member Bacterius's Avatar
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    Quote Originally Posted by SuperCalculus View Post
    Argh asdf this is annoying. It seems to be that you need to use some form of substitution, but everything I try seems to go around in circles.
    I know how you feel, having fiddled with algebra for ten minutes just to find that 0 = 0.
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  7. #7
    Junior Member SuperCalculus's Avatar
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    Quote Originally Posted by Bacterius View Post
    I know how you feel, having fiddled with algebra for ten minutes just to find that 0 = 0.
    Or to find that x^2 = y^2 + 4nfor the fifteenth time.
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