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Number Theory Problem

  1. Jul 7, 2009 #1
    I've been stuck on this for a while now, and I was wondering if anyone could help me out. The problem is:

    If ax[tex]^{2}[/tex]+bx+c=0, prove that all integer roots divide b

    I'm fairly new to number theory, but this is the one problem that's been really tough for me. If someone could even give me someplace to start I'd appreciate it
     
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  3. Jul 7, 2009 #2
    Have you tried looking for counterexamples?
     
  4. Jul 7, 2009 #3

    HallsofIvy

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    1. This is not "Linear & Abstract Algebra" so I am moving it to "Number Theory"

    2. Are we to assume that a, b, and c are integers?

    As Moo Of Doom suggested- look for a counterexample- because this is NOT true!
     
  5. Jul 7, 2009 #4
    Sorry, I was looking at something in linear algebra and I must have posted this there by mistake.

    Also, yes, the problem says that a, b, and c are all integers
     
  6. Jul 8, 2009 #5

    HallsofIvy

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    Good, have you found your counterexample yet? (Almost any quadratic with integer roots will do!) Once again, the reason it is "tough" for you is probably because it is not true.
     
  7. Jul 8, 2009 #6
    Try to write b in terms of the roots.
     
  8. Jul 8, 2009 #7
    Got it! Thanks for the help. I also found a related problem which is giving me trouble:

    If x[tex]^{2}[/tex]+ax+b=0 has an integer root, show that it divides b.

    It's in a chapter on integers, but I'm not sure if a and b are necessarily integers. I'm assuming they are. I was thinking that I should factor it and write it as two polynomials in terms of a and b, which will then show that an integer x divides b. Am I on the right track?
     
  9. Jul 8, 2009 #8
    Actually, let me just show the proof I came up with, and if anyone has any suggestion on improving it, let me know.

    For x[tex]^{2}[/tex]+ax+b=0 to have integer roots, several conditions must apply. I rewrote it as (x+a/2)[tex]^{2}[/tex]=0. This means that b must be equal to a[tex]^{2}[/tex]/4. Since x needs to be an integer, a must be of the form a=2n, n[tex]\geq1[/tex]. With these conditions in place, x=-a/2, which divides b for b=a[tex]^{2}[/tex]/4.

    How did I do?
     
  10. Jul 8, 2009 #9

    Office_Shredder

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    THere's no reason to believe that the polynomial is the square of a linear function... for example, (x-2)(x+3) can't be written in the proposed form.

    Use this: If a polynomial p(x) has a root a, then (x-a) divides p(x), i.e. there exists q(x) such that p(x) = (x-a)q(x). If p is a quadratic polynomial, then q must be another linear polynomial. What can you conclude about the shape of q(x)?
     
  11. Jul 9, 2009 #10
    q(x) is linear and has the same slope (1) as (x-a), with its y-intercept also being negative. It will be in the form (x-n). Is this what you're asking? I'm not sure I follow you completely.
     
  12. Jul 10, 2009 #11

    Office_Shredder

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    That's correct, except you don't have that much information on the y-intercept really. I realized calling the root a was a poor choice, so let's assume that c is a root So now we have
    x2+ax+b = (x-c)(x-n) for some n. Multiply out the right hand side
     
  13. Jul 10, 2009 #12
    yeah, I dont think the set of all roots of the equation divide b. But there is a set of integer roots that will divide b but the roots will have to satisfy a couple of conditions, concerning the certain values of a, b, and a new variable ns that has to be introduced to avoid roots with square-root terms.

    The conditions for x|b, or algebraically, [tex]{b \over x} = k[/tex] where k is any number in the set of integers, are:

    [tex]x = {\pm n_s \over 2a + k}[/tex]

    where [tex]n^2 = k^2x^2 - 4ac[/tex] and ns is the set that makes (kx)2 - 4ac equal to a number n2 that the square root, n, is also an integer.

    and thats it.

    as for the formula for finding the set of k's that satisfy x|b,

    [tex] k_s = {2ab \over \pm n_s - b}[/tex] where ks is the set that makes the values of a, b, and ns in [tex]{2ab \over \pm n_s - b}[/tex] a subset of the set of all integers.
     
    Last edited: Jul 10, 2009
  14. Jul 26, 2009 #13
    I think I've got it, albeit in a simpler form than what I feel like it should be.

    For an integer root n, the polynomial can be factored to (p(x))(x-n). p(x) must be of the form (x+m) for some number m. Using the foil definition, nm=b. Thus n|b for all integers n.

    Is this proof correct/complete?
     
  15. Jul 26, 2009 #14
    Well mostly. Given your own definition we have:
    (x+m)(x-n) = x^2 + ax + b
    Comparing the constant terms we get:
    b = -nm
    not b = nm, but for the sake of this proof it doesn't matter.

    Also it's incorrect to conclude "Thus n|b for all integers n." since it's not true for all integers, only all integer roots of the polynomial (I'm assuming this is what you meant). As you have already stated that n is an arbitrary integral root it would suffice to state "Thus n|b" or "Thus n|b for all integer roots n of the polynomial".

    I have no idea what this foil thing is. I don't really see where you use it though? You simply seem to expand and equate the constant terms (which is the correct thing to do).

    EDIT: Interesting aside: you only showed the result for integer roots, but in fact all rational roots of such quadratics are integers so it's possible to expand the theorem to "All rational roots of an arbitrary monic quadratic x^2 + ax + b are integers that divide b". It's even true that for all monic polynomials P(x) the rational roots are integers that divide the constant term, and in very general terms: If a reduced rational number p/q is root of a degree polynomial P(x) with constant term b and leading coefficient a, then p | b and q | a. The generalizations to nth degree polynomials may be a bit hard to prove if you are not used to manipulating summations or doing inductive arguments, but you may very well be able to show that all rational roots of a monic quadratic are integral (HINT: simply let p/q be an arbitrary root in reduced form and show that q must be 1 or -1 by showing q | p and since p/q is reduced we must have |q| = 1).
     
    Last edited: Jul 26, 2009
  16. Jul 27, 2009 #15
    Yes, I meant to say that n|b for all integer roots n of the polynomial. When I said foil I was talking about the FOIL method of multiplying (x+m)(x-n), the last part of which says to multiply m by -n. I couldn't think of a better way to state this; I think it might be superfluous to the proof anyway.

    I think I understand what you're saying about proving that all rational roots are integers, but I'm not very good with summations, which you mentioned might be needed. Either way, I'm pretty sure I understand how this would be true for most p/q.
     
  17. Jul 27, 2009 #16

    HallsofIvy

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    What do you mean "rewrote it as"? That is not at all the same. It as [itex]x= \pm a/2[/itex] as solution while the roots of the first equation could be anything.

    Unfortunately, you have no reason to think you can "rewrite" the equation in that way. Assuming you mean "a" there to be the same as "a" in [itex]x^2+ ax+ b[/itex], completing the square gives [itex]x^2+ ax+ b= x^2- ax+ (a/2)^2- (a/2)^2+ b[/itex][itex]= (x- a/2)^2+ b- (a/2)^2= 0[/itex].

    Try this: suppose the two roots are [itex]\alpha[/itex] and [itex]\beta[/itex]. Do you understand that that means [itex]x^2+ ax+ b= (x-\alpha)(x-\beta)[/itex]?
     
    Last edited: Jul 27, 2009
  18. Jul 27, 2009 #17
    You're right, I wrote that assuming (wrongly) that the quadratic was a perfect square and could be written as such. Later I realized that I was making it more complicated than it needed to be, so I came up with my second proof, which as far as I can tell is very similar to what you suggested, except I used n and m instead of [tex]\alpha[/tex] and [tex]\beta[/tex].
     
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