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).