Diophantine Equation $a^3+b^3=22c^3$ Found this equation in a book on the Riemann Hypothesis. I've found solution without proof on MathStack asking for an algorithm for all the solutions. I'm trying to find a proof for the smallest positive integer solutions. I think I'm close, but not sure how to proceed.
Solve $a^3+b^3=22c^3$ over the positive integers. Seeking a smallest solution we need $a$,$b$, and $c$ all to be pairwise relatively prime. If the sum of two numbers is even, then those numbers are either both even or both odd. By the above argument then, $a$ and $b$ must both be odd. It follows that their sum is even.
$a^3+b^3=(a+b)(a^2+b^2-ab)$. By 25 trials, it can be shone that $a^2+b^2-ab$ is never divisible by 11. It follows that $a+b$ must be, since the right hand side of the equation is. So $a+b$ is divisible by 22.
$a+b=22k$ for some integer $k$.
$22c^3=a^3+(22k-a)^3=a^3+22^3k^3-3a(22k)^2+ 3a^2(22k)-a^3$
$3ka^2-66ak^2+(22^2k^3-c^3)=0\implies a= \frac{66k^2\pm\sqrt{66^2k^4-12k(22^2k^3-c^3)}}{6k}$
Introduce $n$ as  $66^2k^4-12k(22^2k^3-c^3)=36k^2n^2$
It follows that $c^3=k(11^2k^2+3n^2), a=11k+n, b=11k-n$. For $a$ and $b$ to be relatively prime then so are $k$ and $n$. If $z^3=xy$ and $gcd(x,y)=1$, then $x$ and $y$ are also perfect cubes.
So if $3\nmid k$, both factors of $c^3$ are relatively prime. This means $\exists m$ so that $k=m^3$. It then follows that $c^3=m^3(11^2m^6+3n^2)$ and both factors are perfect cubes.
On the other hand, if $3|k$, then $9|k$ and $3|c$. After $c=3p, k=9q$ we have:
$27p^3=11^2\cdot 9^3q^3+3n^2\cdot 9q\implies p^3=q(27\cdot11^2q^2+n^2)$ By previous arguments $q$ and $n$ are relatively prime. It follows that :
$p^3=m^3(3\cdot(33)^2m^6+n^2)$ where both factors are perfect cubes.
This might be a red-herring, but whether $k$ is divisible by 3 or not, we end up with an equation of the form $z^3=3x^2+y^2$. Multiply both sides by 27 and you get $(3z)^3=(9x)^2+3(3y)^2$. You get another triplet that satisfies the equation. Given one solution of this form we have infinitely many others. Not sure where to proceed from here.
Not sure if this is the smallest, but $a=17299, b=25469, c=9954$ will do the job, but I'm not sure how these numbers were arrived at.

COMMENT.-Apropos of Tomita's comment, I want to add that it is not necessary to go to the (more popularly known) Weierstrass form and that the curve $x^3+y^3=nz^3$ is also in their own right an elliptic curve in which the formulas of sum are given as follows:
If $A=(X_1,Y_1,Z_1)\ne B=(X_2,Y_2,Z_2)$ then $A+B=(X_3,Y_3,Z_3)$ where $$X_3=X_1Z_1Y_2^2-X_2Z_2Y_1^2 \\Y_3=Y_1Z_1X_2^2-Y_2Z_2X_1^2\\Z_3=X_1Y_1Z_2^2-X_2Y_2Z_1^2$$ and if $A=B=(X,Y,Z)$ (case of tangent, not a chord) $$X_3=-Y(2X^3+Y^3)\\Y_3=X(X^3+2Y^3)\\Z_3=Z(X^3-Y^3)$$

►I want to add that having deduced the point of the elliptic curve $x^3+y^3=22$ given by the O. P. and shown in the figure above (I want to believe that it has been calculated by him) is very meritorious and that it is clear that he does not have the tools to know if said point is minimum (i.e. it is a generator of the corresponding additive group). To this end, in the classic paper “The diophantine equation $ax^3+by^3+cz^3=0$, Acta Math. (Stockolm) 85 $(1951)\space p. 203-362$” written by E. S. Selmer, there is a very laborious table (calculated without a computer!) in which there are just over the first hundred values of cube-free integers, $n$, representable as sum of two cubes of rationals. This table also shows the range of the curves and then you can see if the point given by Turloc The Red is minimum or not (in Tomita's comment, it can also be deduced that the point in question is indeed a generator).
