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Proving that a^2-2b^2-4c^2 = 0 has no positive integer solutions

  1. Oct 17, 2008 #1
    so I have to prove that a^3-2b^3-4c^3 = 0 has no positive integer solutions. I have gotten through most of the proof but now I am stuck, and if you guys could give me a nudge in the right direction that would be great.

    work done so far:

    proof by contradiction, i assumed that a^3-2b^3-4c^3 = 0 does have a positive int solution, therefore by the well ordering principle it has a least solution a = a* , b = b*, c = c*. I then proceeded to show that a*^3 = 2b*^3+4c*^3, therefore a*^3 is an even number, and so a* is also even. I let a* = 2k and plugged that back into the equation to get the following: 8(k^3) = 2b*^3+4c*^3. Now i am stuck. If you guys could give me a nudge in the right direction that would be great, I know that i want to show that k is also a solution to the original equation therefore contradicting that a* is the least solution, thanks for the help.
  2. jcsd
  3. Oct 17, 2008 #2
    First of all, your solution will be an ordered triplet (a, b, c) - so you cannot apply the well ordering principle to your solution set. You CAN however apply it to the set of a's for which there exist corresponding b and c which form solution triples. In other words, your idea is perfectly good, but your wording is a little off. You're going to use the WOP on the least value of a.

    What you've done so far is good. Keep going in that direction. Now that you know a=2k. Try to show that b is also even (b=2m) and then that c must also be even (c=2n). Then plug everything back into the original equation to show that (k, m, n) is another solution where k < a, which is the contradiction you were looking for.
    Last edited: Oct 18, 2008
  4. Oct 17, 2008 #3
    thanks for the help, i have tried this approach, and the come to the point where I have the following expressions for k,m,l:

    k^3 = (1/4)b^3 + (1/2)c^3

    m^3 = (1/16)a^3 - (1/4)c^3

    l^3 = (1/32)a^3 - (1/16)b^3

    and i plug these into the original equation and the a terms do not cancel out, I feel like i am so close but just cannot get it, if u have anymore hints that would be great.
  5. Oct 18, 2008 #4
    I just realized that I had a typo in my original response. I was using the variables k, m, n for most of it and then switched to k, l, m at the very end... I have since edited the post just to keep it consistent.

    Let's back up a bit: We know a is even, a=2k. so [tex]a^3=8k^3=2b^3+4c^3[/tex]. Let's ignore the a^3 bit and look at what's left. The next step, as stated above, is to show that b is even. We're going to use the same strategy you did earlier for a: we're going to isolate b^3.




    What can you say about b? Set b=2m.

    Next step: Show c is even. Go back to your very original statement [tex]a^3=2b^3+4c^3[/tex] and substitute in a=2k and b=2m. Use the same strategy to isolate c^3 and show that c is even (c=2n).

    After you do this, you will be ready for the last step.
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