On x^2 + y^3 = z^6

The only integer solutions of the equation x^2 + y^3 = z^6 are of 
the form
                {x=0, y=k^2, z=+-k}
                {x=+-k^3, y=-k^2, z=0}
                {x=+-3k^3, y=-2k^2, z=+-k}

for any integer k.  Clearly if any two of x,y,z have a common factor 
then they also share a common factor with the third, so we can divide 
until the three numbers x,y,z are coprime.  Of all such solutions, 
let's consider the one with the smallest absolute value of z.

Now, the equation implies  y^3 = (z^3)^2 - x^2, so we have

                  y^3  =  (z^3 + x)(z^3 - x)                     (1)

If y is odd then the two factors on the right side are mutually 
coprime, so they must each be cubes, meaning there are coprime 
integers a,b such that

        z^3 + x  =  a^3      and        z^3 - x  =  b^3

Adding these two gives  a^3 + b^3 = 2z^3.  On the other hand, if y is
even, then the two factors on the right side of (1) share a common 
factor of 2, so one of them is twice a cube and the other is four times
a cube, meaning there exist coprime integers a,b such that

        z^3 + x  =  2a^3      and        z^3 - x  =  4b^3

Adding these two and dividing by 2 gives z^3 + (-a)^3 = 2b^3.  Thus,
regardless of whether y is odd or even, we arrive at an equation of 
the form A^3 + B^3 = 2C^3 in coprime integers A,B,C, and if we take 
the trivial solutions {A=B=C} and {A=-B,C=0} we get one of the 
"obvious" solutions of the original equation
 
                {x=0, y=k^2, z=+-k}
                {x=+-k^3, y=-k^2, z=0}
                {x=+-3k^3, y=-2k^2, z=+-k}

It only remains to show that an equation of the form a^3 + b^3 = 2z^3 
has no solutions other than those with a=+-b.  To prove this, assume 
a,b,z are the solution with the absolutely smallest value of z.

Since a,b are coprime with unequal magnitudes, and the sum of their 
cubes is even, they must both be odd.  This implies the existence of 
coprime integers u,v such that a=u+v and b=u-v.  Substituting into 
a^3 + b^3 = 2z^3 and simplifying gives

                      u (u^2 + 3v^2)  =  z^3

which proves that  u  and  u^2 + 3v^2  must both be cubes.  Thus, we
have integers r,s such that

             u = r^3        and           u^2 + 3v^2 = s^3

The right hand equation can be factored as

              (u + v sqr[-3]) (u - v sqr[-3])  =  s^3

Following Euler's somewhat reckless path, we can take advantage 
of the fact that the ring of numbers of the form  a + b sqr(-3)  
possesses unique factorization (where a,b are actually half-integers), 
and assert that each of the (coprime) factors on the left is a cube.  
Thus, we have coprime (half)integers g,h such that

     u + v sqr[-3]  =  (g + h sqr[-3])^3

                    =  g (g^2 - 9h^2)  +  h (3g^2 - 3h^2) sqr[-3]

from which it follows that

         u = g (g^2 - 9h^2)           v = 3h (g^2 - h^2)

Recalling that u = r^3, the left hand equation factors as

               r^3  =  g (g+3h) (g-3h)

(Notice that if g and h are really half-integers we can clear all
fractions by multiplying through by 2^3, which still gives a cube on
the left side.)  Since g,h are coprime, each of the above three factors 
are coprime and therefore each must be cube.  Thus, we have integers 
m,n,q such that

        g = m^3        g+3h = n^3       g-3h = q^3

which imply that  n^3 + q^3 = 2m^3.  But clearly m < g < r < z, so
m < z, which contradicts our assumption that z is the smallest integer
such that the double of its cube equals the sum of two cubes.  Therefore,
no non-trivial integer solutions of A^3 + B^3 = 2C^3 exist, so the
only integer solutions of the original equation are

                {x=0, y=k^2, z=+-k}
                {x=+-k^3, y=-k^2, z=0}
                {x=+-3k^3, y=-2k^2, z=+-k}

In particular, this shows that there do not exist any solutions in 
strictly positive integers.

Return to MathPages Main Menu