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.

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