A Diophantine equation: solve $(3x^2+y^2-4y-17)^3-(2x^2+2y^2-4y-6)^3=(x^2-y^2-11)^3$ (without using Fermat's last theorem)

Question

Solve this Diophantine equation: $(3x^2+y^2-4y-17)^3-(2x^2+2y^2-4y-6)^3=(x^2-y^2-11)^3$

My attempt (use Fermat's last theorem) $$(3x^2+y^2-4y-17)^3-(2x^2+2y^2-4y-6)^3=(x^2-y^2-11)^3$$ $$\Leftrightarrow (3x^2+y^2-4y-17)^3=(x^2-y^2-11)^3+(2x^2+2y^2-4y-6)^3$$ Use Fermat's last theorem, we get:

$$ \left\{ \begin{array}{c} x^2-y^2-11=0 \\ 3x^2+y^2-4y-17=2x^2+2y^2-4y-6 \end{array} \right.$$

or

$$ \left\{ \begin{array}{c} 2x^2+2y^2-4y-6=0\\ 3x^2+y^2-4y-17=x^2-y^2-11 \end{array} \right.$$

Then we continue...

My question is, is there another way to solve that other than Fermat's last theorem? I see that using Fermat's last theorem is like crack a nut by a sledgehammer.

Answer 1

$$(3x^2+y^2-4y-17)^3-(2x^2+2y^2-4y-6)^3=(x^2-y^2-11)^3$$

If we put $A=3x^2+y^2-4y-17$ and $B=2x^2+2y^2-4y-6$, observe that $A-B=x^2-y^2-11$ (the factor on the right).

In other words, we have $a^3-b^3=(a-b)^3$

Expand this to get $a^3-b^3=a^3-3a^2b+3ab^2-b^3$

So $3ab(a-b)=0$ is the simplified equation. You can then show that $a=0, b=0$ or $a-b=0$ has no integer solution.

Answer 2

Let $$3x^2 + y^2 - 4y -17 = a\text{ ,}$$ $$-(2x^2 +2y^2 +4y -6)= b$$ and observe $$x^2-y^2 -11= a+b$$ Your problem may be reexpressed as $$a^3 + b^3=(a+b)^3$$ implying $$3ab(a+b)=0.$$ This gives your proposed solution set, plus the case where $a=0$.