x^3+y^3=z^3⇒(x+y)(x^2-xy+y^2)=z^3. If x+y=z, then x^2-xy+y^2=z^2; but that implies x^2-xy+y^2=x^2+2xy+y^2, and 3xy=0, which is the trivial solution and is disallowed by the theorem. So x+y needs to be an integer factor of z, x+y=az and x^2-xy+y^2=z^2/a; or x+y=az^2 and the quadratic x^2-xy+y^2=z/a. This respectively implies that (1) x^2-xy+y^2=(x^2+2xy+y^2)/a^2; or (2) x^2-xy+y^2=sqrt((x+y)/a)/a.
(1) The equation becomes x^2(a^2-1)-xy(a^2+2)+y^2(a^2-1)=0 for a>1, and, dividing through by a^2-1 we get:
x^2-xy(a^2+2)/(a^2-1)+y^2=0. Solve this by completing the square:
(x-y(a^2+2)/(2(a^2-1)))^2+y^2-y^2((a^2+2)/(2(a^2-1))^2=0.
x-y(a^2+2)/(2(a^2-1))=+ysqrt((a^2+2)^2-4(a^2-1)^2)/(2(a^2-1))=
+ysqrt((a^2+2+2a^2-2)(a^2+2-2a^2+2))/(2(a^2-1))=+ysqrt(3a^2(4-a^2))/(2(a^2-1))=
+aysqrt(3(4-a^2))/(2(a^2-1)).
The square root cannot be negative and a>1, so a=2 and x=y(a^2+2)/(2(a^2-1))=6y/6=y. But this solution is based on the initial assumption that the quadratic could be equal to z/a. Clearly x^3+y^3=2x^3 cannot be z^3 where z is an integer. So the assumption (supposed condition) is false and there are no solutions under (1).
(2) If x^2-xy+y^2=z/a and we write z/a=k, where k is a positive integer, then we can rewrite the quadratic:
(x-y/2)^2-y^2/4+y^2=k, so (x-y/2)^2=k-(3/4)y^2. If y is an even integer y=2w where w is an integer. Then (x-w)^2=k-3w^2. So (x-w)=sqrt(k-3w^2). The left-hand side is an integer so the right-hand side must also be an integer, and this can only happen if k-3w^2=p^2 where p is an integer >0. So k=p^2+3w^2. x=w+p; y=2w; z=a(p^2+3w^2). We have the three variables in terms of three parameters, a, p and w. For the special case of p=0, we can see the effect of substituting the parametrised values into the quadratic: w^2-2w^2+4w^2=3w^2 and into x+y=3w, the product of which is 9w^3=x^3+y^3. This sum cannot be a perfect cube because 9 is not a perfect cube, therefore there are no solutions to the cubic equation when p=0.
When p>0, x+y=3w+p and the quadratic is w^2+2pw+p^2-2w^2T2wp+4w^2=3w^2+p^2. The product is (3w+p)(3w^2+p^2)=9w^3+3wp^2+3pw^2+p^3=(2w)^3+(p+w)^3=y^3+x^3.
If y is odd y=2v-1 where integer v>0 and (x-v+(1/2))^2=k-(3/4)(4v^2-4v+1)=k-3v(v-1)-3/4; (x-v)^2+x-v+1=k-3v(v-1)
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