Solving a Diophantine equation $3a^3+3b^3=a^3+c^3$

Question

A friend of mine gave me the following problem

Find all integers solutions to $$3a^3+3b^3=a^3+c^3$$

Of course $(0,0,0)$ is a solution, and I think that there are no others, but I can’t prove it. I tried factorising (since they are cubic) or taking it modulo 3, but it took me nowhere...

Edit: $a,b,c\ge 0$

Any hints

Answer 1

The point of this answer is not to give a full list of solutions. Below I simply describe how the chord-tangent method gives us new solutions starting from the known one(s).

The key is to observe that the equation describes a non-singular curve in the projective plane. We can do the usual dehomogenization trick of writing $x=a/c$, $y=b/c$, when the equation becomes $$ 2x^3+3y^3=1,\qquad(*) $$ an elliptic curve $E$. If we find any rational solutions $x=p/q,y=r/s$ to $(*)$ we can clear the denominators and produce integer solution $[a:b:c]=[ps:rq:qs]$.

The chord-tangent method of producing new rational points on $E$ starting from known one(s) is based on the observations that

• If $P=(x_0,y_0)\in E$ is a rational point, then the slope $m$ of the tangent of $E$ at $P$ is also rational (implicit differentiation makes this obvious). The equation of the tangent is thus $y=m(x-x_0)+y_0$ and has rational coefficients. Plugging that into $(*)$ gives then the equation $$2x^3+3(m(x-x_0)+y_0)^3=1.$$ We know by tangency that $x=x_0$ is a double zero of that cubic. Therefore the third root $x=x_1$ is also rational (because the Vieta relations imply that the sum of the three zeros is rational), and gives us a new rational point $(x_1,y_1)\in E$ with $y_1=m(x_1-x_0)+y_0$.
• Similarly, if $P=(x_1,y_1)$ and $Q=(x_2,y_2)$ are two rational points on $E$, then the chord $$ y=\frac{y_2-y_1}{x_2-x_1}(x-x_1)+y_1 $$ has rational coefficients, intersects $E$ at the points $P$ and $Q$, and hence its third point of intersection $R=(x_3,y_3)$ with $E$ must also have rational coefficients.
• There are a few exceptional situations. For some cubics it may happen that the substitution $y=mx+b$ into the cubic cancels all the cubic terms. Geometrically this means that the third point of intersection lies on the line at infity of the projective plane. With the present cubic this cannot happen, because then we would have $2+3m^3=0$ contradicting the assumption that $m$ is rational. But, if we try to apply the chord-tangent process to the Fermat cubic, $x^3+y^3=1$, and to the points $P=(1,0)$ and $Q=(0,1)$ that mechanism becomes a killjoy - the resulting third (projective) point escapes to a point at infinity with homogeneous coordinates $[x:y:z]=[1:-1:0]$.

Applying this to our starting point $P_1=(x,y)=(-1,1)\in E$ gives the following other points:

• The tangent (red line in the figure) of $E$ at $P_1$ has slope $m=-2/3$, and intersects $E$ at $P_2=(4/5,-1/5)$. This gives us solutions $[a:b:c]=[4:-1:5]$.
• The tangent (green line in the figure) of $E$ at $P_2$ has slope $m=-32/3$, and intersects $E$ at $P_3=(488/655,253/655)$, giving solutions $[a:b:c]=[488:253:655]$.
• The chord joining (orange line in the figure) $P_1$ and $P_3$ intersects $E$ at $P_4=(-3449/26309,18269/26309)$ giving us more points.

The curve $E$ itself is the blue curve. $P_1$ is the intersection of red and orange lines, $P_2$ the intersection of red and green lines, $P_3$ the intersection of green and orange lines, and $P_4$ the extra intersection of $E$ and the orange line.

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In a comment under the question Álvaro Lozano-Robledo confirmed that the group of rational points of this elliptic curve is infinite and cyclic (i.e. rank one and torsion-free). As user670344 explains in a comment, this implies that the first quadrant $x\ge0,y\ge0$ will also contain infinitely many rational points.