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Homework Help: Rational solutions of equation 3x² + 2 = y²

  1. Oct 11, 2011 #1

    I like Serena

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    1. The problem statement, all variables and given/known data

    Show that the equation 3x2 + 2 = y2 has no integer solutions by calculating modulo 3.
    Proof that the equation has no rational solutions.

    This is a problem from an introductory chapter in algebra that I'm teaching.


    2. Relevant equations



    3. The attempt at a solution

    I've got the first part down, since the left side is always 2 mod 3, while the right side (a square) is always either 0 or 1 mod 3.


    However, I'm stuck on the second part.

    I can write x=a/b and y=c/d, that can not be simplified, yielding:

    3(a/b)² + 2 = (c/d)² ?

    3(ad)² + 2(bd)² = (bc)² ?

    (bd)² is either 0 or 1 mod 3.

    If (bd)² = 1 mod 3, then we can check whether 3(ad)² + 2 = (bc)² mod 3.
    Since this is effectively the same as the original equation for integer numbers, we already know this is not possible.

    So (bd)² = 0 mod 3.
    Consequently from the bold equation: (bc)² = 0 mod 3.
    So b = 0 mod 3, or both c and d are 0 mod 3.

    If both c and d are 0 mod 3, then the fraction c/d could have been simplified.
    This violates the precondition.

    So b = 0 mod 3.
    For the bold equation to hold then a or d must be 0 mod 3.

    If a = 0 mod 3, then a/b could have been simplified.
    Again this violates the precondition.

    So we have:
    a ≠ 0 mod 3, b = 0 mod 3, c ≠ 0 mod 3, d = 0 mod 3.

    As yet I still don't see the end of this.


    This is already quite a long line of reasoning and since it is from an introductory chapter in algebra I'm tentatively drawing the conclusion that I'm missing something.

    Can someone give me a clue?
     
    Last edited: Oct 11, 2011
  2. jcsd
  3. Oct 11, 2011 #2

    Office_Shredder

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    I have an ugly proposal which might lead to some insight. Once you have b=0 mod 3, write it as b=3b'

    3(a/(3b'))2+2=(c/d)2

    1/3*(a/b')2+2=(c/d)2

    clearing denominators again

    (ad)2+6(b'd)2=3(b'c)2

    taking mod 3 again we see that ad=0 mod 3, so a=0 mod 3 or d=0 mod 3. We know a can't be zero mod 3 because of the reduction, so d=0 mod 3 as well. Let's do the same thing again. d=3d'

    1/3*(a/b')2+2=1/9*(a/d')2. Clear denominators

    3(ad')2+18(b'd')=(ab')2

    Taking mod 3 and noting that a can't be divisible by 3, we get that b' is divisible by 3. I think this recurses, i.e. using 3|b' you can get that 3|d', then get that 3|b'', then 3|d'' etc. which is a contradiction because 3 only divides b and d finitely many times
     
  4. Oct 11, 2011 #3

    I like Serena

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    Thanks!
    That works.
    I did see that there was a sort of recursive relationship, but I didn't realize the contradiction that it implies. :)
     
  5. Apr 23, 2013 #4
    my answer is (x^2 + y^2) / -2 = 1

    in other words both x^2 and y^2 = -1 ... see?

    3x^2 + 2 = y^2
    -3 + 2 = -1

    P.j.S .
     
    Last edited: Apr 23, 2013
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