On x² + y³ = z⁶

We show here that the only integer solutions of the equation x² + y³ = z⁶ are of one of the three obvious forms

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 y³ = (z³)² – x², so we have

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

Adding these two gives a³ + b³ = 2z³. 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

Adding these two and dividing by 2 gives z³ + (–a)³ = 2b³. Thus, regardless of whether y is odd or even, we arrive at an equation of the form A³ + B³ = 2C³ 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 three "obvious" solutions of the original equation (listed above).

It only remains to show that an equation of the form a³ + b³ = 2z³ 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 a³ + b³ = 2z³ and simplifying gives

which proves that u and u² + 3v² must both be cubes. Thus, we have integers r,s such that

The right hand equation can be factored as

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

from which it follows that

Recalling that u = r³, the left hand equation factors as

(Notice that if g and h are really half-integers we can clear all fractions by multiplying through by 2³, 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

which imply that n³ + q³ = 2m³. 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 A³ + B³ = 2C³ exist, so the only integer solutions of the original equation are the three obvious solution forms listed at the beginning. In particular, there do not exist any solutions in strictly positive integers.

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