Log In

Toy Fermat

Posted on: April 24th, 2013 by
3

Does the equation, x^2 + y^3 = z^4 have solutions in prime numbers? Find at least one if yes, give a nonexistence proof otherwise.

via Math puzzles


3 Responses to Toy Fermat

  1. Bharatram had this to say about that:

    Since we are working over prime numbers, precisely one of x,y,z is 2.

    z=2 trivially gives no solutions.

    If x=2, we need 4+y^3=z^4
    y^3 = z^4-4 = (z^2-2)(z^2+2)
    This limits the possibilities, as y^2 has to be z^2+2 and y has to be z^2-2, giving no solutions, or -y^2 has to be z^2-2 and -y has to be z^2+2, again giving no solutions.

    If y=2, we need x^2+8=z^4
    Again, this gives z^4-x^2=(z^2-x)(z^2+x)=8. Again, either z^2+x=4 and z^2-x=2, giving no solutions, or z^2-x=-4 and z^2+x=-2, again giving no solutions.

    Hence no solutions in primes x,y,z.

  2. Ryan had this to say about that:

    Can you please explain why precisely one of x,y,z = 2? Thanks!

  3. Rarchinio had this to say about that:

    Check the parity of the equation...


{"result":"error", "message":"You can't access this resource as it requires an 'view' access for the website id = 1."}