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Math Help - Geometric Progression

  1. #1
    Junior Member
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    Geometric Progression

    Hello,
    This time I have come up with a geometric progression problem.
    Three natural numbers are known to form a geometric progression. The second member is known to be greater than the first one with +1.
    It is required to find the third member of the geometric progression and this is all that is known about it. So, so far I thought up this: let a, b and c be the members. a = b - 1 as said and from the fact that we are dealing with a geometric progression we get:
    <br />
b^2=(b-1)c<br />
    which is an equation.. with two unknown variables.
    To solve it successfully we need a system of equations, i.e. a second equation, which I do not know where to get from, which is my problem.
    The answer is c = 4.

    I suspect there might be two solutions for b and a, although they are not our concern.

    Thank you.
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hello

    What's important here is that we are trying to solve b^2=(b-1)c for b,c two natural integers.

    If we assume that b is odd then b-1 is even and b^2 is odd so \underbrace{b^2}_{\text{odd}}=\underbrace{(b-1)c}_{\text{even}} can't hold : b can't be odd.

    Now let's assume that b is even. As 2 divides b, 4 divides b^2. As b-1 is odd, 4 doesn't divide b-1 hence from b^2=(b-1)c we get that 4 must divide c : c=4p for some p\in\mathbb{N}. We can now write b^2=(b-1)c as b^2=4(b-1)p\Longleftrightarrow b^2-4pb+4p=0 which is an equation in terms of b.

    Using the discriminant we get that the solutions of this equation are b_1=2p+2\sqrt{p(p-1)} and b_2=2p-2\sqrt{p(p-1)}. We know that b is an integer so we are looking for the values of p such that at least one of the two solutions is an integer. When does this happen ?
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  3. #3
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    This happens when p(p-1) is zero, i.e. p = 0 or p = 1. 0 is not a natural number so p = 1 is a solution and c is, indeed, 4.
    However, I am unsure how to prove that p(p-1)=k^2 can never occur for any integer k bigger or equal to 0.
    I have come up with this:
    Let p(p-1)=k^2, then p^2-p-k^2=0. The discriminant is D=1+4k^2... which should not be an exact square? It can never equal 4, 9, 16, etc., I mean. Is this so?

    I was not expecting this to turn out to be a diophantine equation.
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  4. #4
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by Logic View Post
    This happens when p(p-1) is zero, i.e. p = 0 or p = 1. 0 is not a natural number so p = 1 is a solution and c is, indeed, 4.
    However, I am unsure how to prove that p(p-1)=k^2 can never occur for any integer k bigger or equal to 0.
    I have come up with this:
    Let p(p-1)=k^2, then p^2-p-k^2=0. The discriminant is D=1+4k^2... which should not be an exact square? It can never equal 4, 9, 16, etc., I mean. Is this so?
    Yes, that's it. When I first solved this part of the question I didn't use the discriminant but it works well : The positive root of p^2-p-k^2 is \frac{1+\sqrt{1+4k^2}}{2} which is a natural integer iff 1+4k^2 is a perfect square. Assume 1+4k^2=q^2 for some q\in\mathbb{N}. Rearranging and factoring we get 1=(q-2k)(q+2k). As q-2k and q+2k are integers they must both equal 1 hence the only solution is q=1 and k=0.
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