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Math Help - Cubic Equation

  1. #1
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    Cubic Equation

    Find all positive integer roots of the equation $1 + n + n^2 + n^3 = A^2$

    My solution attempt:
    Following Wolframalpha factoring, it can be rewritten as (n+\frac{1}{3})^3+\frac{2}{3}(n+\frac{1}{3})+\frac  {20}{27}=A^2$, then

    $\displaystyle{(3n+1)^3+6(3n+1)+20=27A^2$ and putting $3n+1=k \Rightarrow n=\frac{k-1}{3}}$ and for the

    first positive integer root, $k_1=4 \Rightarrow n_1=1 \Rightarrow A_1=2$
    second positive integer root, $k_2=22 \Rightarrow n_2=7 \Rightarrow A_2=20$

    How can I prove that these $[(n, A)=(1, 2),(7, 20)]$ are the only positive integer solutions? Thanks for any suggestions. Regards.
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  2. #2
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    Re: Cubic Equation

    Is (0,1) not a valid solution?
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  3. #3
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    Re: Cubic Equation

    Thanks for your interest. I guess n=0 is unacceptable because it is not considered positive.
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  4. #4
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    Re: Cubic Equation

    Ok.
    Solution is also:
    n = (u + v)^(1/3) - 2/[9(u + v)^(1/3)] - 1/3
    where:
    u = SQRT(27A^4 - 40A^2 + 16) / [2(3^(3/2)]
    v = (27A^2 - 20) / 54

    (0,1) , (1,2) and (7,20) seem to be the only integer solutions with n=>0.
    I ran a looper with A from 1 to 999999: only above resulted.
    So perhaps above equation can be used to devise the proof...
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  5. #5
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    Re: Cubic Equation

    Thank you very much. I will work on your method. Regards.
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  6. #6
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    Re: Cubic Equation

    Quote Originally Posted by Honore View Post
    Find all positive integer roots of the equation 1 + n + n^2 + n^3 = A^2
    You can factorise it as (1+n)(1+n^2) = A^2. The factor 1+n^2 clearly cannot be a perfect square. So at least one of the squared prime factors in A^2 must be shared between the two factors on the left side. But the only possible common factor of 1+n and 1+n^2 = (1+n)^2-2n is 2. So we can write n+1 = 2s^2 and n^2+1 = 2t^2 for some natural numbers s,t.

    Therefore (2s^2-1)^2 + 1 = 2t^2 which, with a bit of reorganisation and after cancelling the factor 2, can be written as (s^2-1)^2 + s^4 = t^2. This shows that the set (s^2-1, s^2, t) is a Pythagorean twin triple. There are infinitely many Pythagorean twin triples, but we want one in which the middle element is a perfect square. The triples (0,1,1) and (3,4,5) satisfy that condition, leading to the values n=1 and n=7. The next triples are (20,21,29), (119,120,169) and (696,697,985), but none of those has a perfect square as the middle element. After that, the numbers increase very rapidly, as you can see from the listing at A046090 - OEIS. My guess is that there are no more perfect squares in that sequence, but I have to admit that I don't begin to see how to prove that.
    Last edited by Opalg; October 11th 2011 at 12:00 PM.
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  7. #7
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    Re: Cubic Equation

    Thanks a lot. That is much better and efficient than what I have been able to do so far. Regards.
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