Solve the diophantine equation $3\,{a}^{3}b-13\,{b}^{3}-26\,a-24\,b=0.$ Solve the diophantine equation  $3\,{a}^{3}b-13\,{b}^{3}-26\,a-24\,b=0.$
I have found two obvious solutions $a=b=0$ and $a=b=5.$ Are there another solutions?
 A: Let us try to find a non-zero solutions of Diophantine equation.
Let
$$c=\gcd(a,b),\quad a=cx,\quad b=cy,$$
then
$$x(3c^3x^2y-26)=y(13c^2y^2+24),$$
$$y\,|\,(3c^3x^2y-26),\quad y\,|\, 26,$$
$$3c^3x^3-\dfrac{26}yx = 13c^2y^2 + 24,\quad y\in\{\pm1,\pm2,\pm13,\pm26\}.$$
Taking in account that for any values of $y$ the implicit function $x(c)$ is the asymptote on the coordinate axes and that negative values of $c$ correspond to the opposite values $y,$ we are at each value of $ y $ can get all integer solutions of this equation $(1)$.
All values $(x,y),$ which gives $c\gtrsim 5,$ obtained with using of Mathcad package, are the next:

With the graphic1 for $y=\pm1,$ graphic2 for $y=\pm2,$ graphic3 for $y=\pm13$ and graphic4 for $y=\pm26$ we have all possible integer roots for $x,y,c$.
Among them only $c(1,1) = 5$ is integer. Substituting of $a=b=5$ to the original equation shows that this is the right solution.
So
$$\boxed{(a=0,b=0) \vee (a=5,b=5)}$$ 
are the all integer solutions of the given Diofantine equation.
Note
Of course, in each of these cases, we can show the presence or absence of integer roots using the Rational Root Theorem. However, in this case that looks like a very technical and overload the main logic of proof.
