Solutions of $y^2=\binom{x}{0}+2\binom{x}{1}+4\binom{x}{2}+8\binom{x}{3}$ for positive integers $x$ and $y$

Question

I was interested in create and solve a Diophantine equation similar than was proposed in the section D3 of [1]. I would like to know what theorems or techniques can be applied to prove or refute that the Diophantine equation of the title has a finite number of solutions, I don't have the intuition to know it. Our equation is given as $$y^2=2^0\binom{x}{0}+2^1\binom{x}{1}+2^2\binom{x}{2}+2^3\binom{x}{3},$$ thus using the definition of binomial coefficients we are interested in to solve this equation for positive integers $x\geq 0$ and $y\geq0$ $$3y^2=4x^3-6x^2+8x+3.$$

Computational fact. I got up to $10^4$ that the only solutions $(x,y)$ for positive integers $x,y\geq 0$ are $(x,y)=(0,1)$,$(2,3)$, $(62,557)$ and $(144,1985)$. For example our third solution is $$3\cdot 557^2=930747=4\cdot(62)^3-6\cdot (62)^2+8\cdot(62)+3.$$

Question. Does the equation $$y^2=\binom{x}{0}+2\binom{x}{1}+4\binom{x}{2}+8\binom{x}{3}$$ have a finite number of solutions for positive integers $x,y\geq0$ ? If it is very difficult to solve, what work can be done? Many thanks.

References:

[1] Richard K. Guy, Unsolved Problems in Number Theory, Problem Books in Mathematics, Unsolved Problems in Intuitive Mathematics Volume I, Springer-Verlag (1994).

Answer 1

Start with $$ 3𝑦^2=4𝑥^3−6𝑥^2+8𝑥+3. $$ The change of variables $x=X/12$ and $y=Y/36$ gives the equation $$ Y^2 = X^3 - 18X^2 + 288X + 1296. $$ Entering this into the LMFDB leads to the page http://www.lmfdb.org/EllipticCurve/Q/315936/g/1 . So your elliptic curve, after another change of variables to get rid of the $X^2$ term (replace $X$ by $X+6$) gives the curve $$ Y^2 = X^3 + 180X + 2592. \quad(*) $$ According to the LMFDB, the curve $(*)$ has Mordell-Weil rank 2, generated by $ \left(-6, 36\right) $ and $ \left(18, 108\right) $. And in the form $(*)$, there are the following integral points, where only the point with positive $Y$ is listed: $$ (-6, 36) ,\; (-3, 45) ,\; (18, 108) ,\; (24, 144) ,\; (28, 172) ,\; (738, 20052) ,\; (1722, 71460) ,\;(6189, 486891) .$$ This leads to the points $$ (0, 1),\; (1/4, 5/4),\; (2, 3),\; (5/2, 4),\; (17/6, 43/9),\; (62, 557),\; (144, 1985),\; (2065/4, 54099/4) $$ on your original curve. giving the 4 integral points that you found.

Answer 2

The curve $y^2 = 1+\frac{8}{3} x - 2 x^2 + \frac{4}{3} x^3$ is elliptic. Siegel's theorem says it has only finitely many integral points.