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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).

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  • $\begingroup$ Hi @GerryMyerson many thanks for your comments and attention on my posts, I read these and I appreciate all the comments. If this question result a good post I would like to dedicate this to you. $\endgroup$ – user142929 Jul 26 '19 at 22:20
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    $\begingroup$ Using @ only alerts users who have already engaged with the question. $\endgroup$ – Gerry Myerson Jul 26 '19 at 23:53
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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.

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  • $\begingroup$ Many thanks also for your great answer. I wondered that maybe an interesting problem can be, if it has a good mathematical content, to study the solutions of $P(x)={n\brace m}$, where $P\in\mathbb{Z}[X]$ and ${n\brace m}$ are the Stirling numbers of the second kind. If do you like this problem feel free to study it as a present for you. $\endgroup$ – user142929 Aug 28 '19 at 8:37
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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.

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  • $\begingroup$ Many thanks I'am going to accept the other answer but your is also perfect. I provide you other equations if you want to study these. The first is $y^2=(x)_0+(x)_1+(x)_2+(x)_3$, where $(x)_n$ denotes the Pochhammer symbol. It is the equation $y^2=(x+2)^3-2(x+2)^2+1$, and the second is $y^2={x\brace 0}+{x\brace 1}+{x\brace 2}+{x\brace 3}$, where ${n\brace m}$ denotes Stirling numbers of second kind. Using the MathWolrd article Stirling Number of the Second Kind it should be the equation $2y^2=2^x+3^{x-1}+1$. $\endgroup$ – user142929 Jul 28 '19 at 18:49
  • $\begingroup$ Few minutes ago I wondered about a different problem as a present for you: the relationship of Iannucci's equation ([1]) and the sequence A277172 (see also the author of this sequence) from the OEIS. Exponentiating Iannucci's equation and using Fermat's little theorem one has that the solutions $n$ satisfy the congruence. I think that maybe you're interested in this because you've added comments in this article from OEIS. See Theorem 2 from Iannucci's article. References: [1] Iannucci, On the Equation $\sigma(n)=n+\phi(n)$, Journal of Integer Sequences, Vol. 20 Article 17.6.2 (2017). $\endgroup$ – user142929 Sep 9 '19 at 21:01

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