Proving no primitive solutions exist for a diophantine equation satisfying the given condition

Question

Suppose $a+b$ is a perfect square divisible by 3. Then show that all solutions of the diophantine equation $a^3+b^3=c^2$ are non primitive. So I factored $a^3+b^3$ into $(a+b)(a^2-ab+b^2)$ and also proved c is divisible by 9 and $a^2-ab+b^2$ must be a perfect square. But i can't seem to move much further.

Answer 1

Assume $(a,b,c)$ is a primitive solution.

Since $3\,{\mid}\,(a+b)$, if either of $a,b\;$is a multiple of $3,\;$they must both be multiples of $3,\;$but then, since $c\;$is a multiple of $3,\;$the triple $(a,b,c)\;$is not primitive.

Thus, neither of $a,b\;$is a multiple of $3$.

Then, since $$a^2 -ab + b^2 = (a+b)^2-3ab$$ it follows that $a^2 -ab + b^2\;$is a multiple of $3,\;$but not a multiple of $9$.

But $a+b\;$is a perfect square, hence the LHS of the equation $$(a+b)(a^2-ab+b^2)=c^2$$ has an odd exponent for the prime factor $3,\;$while the RHS has an even exponent, contradiction.