1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook