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added a top-level tag; https://meta.mathoverflow.net/questions/1457/why-are-mo-tags-formatted-as-they-are
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Martin Sleziak
Martin Sleziak
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I was interested in create and solve a diophantineDiophantine 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 diophantineDiophantine 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).

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).

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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user142929
user142929
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Solutions of $y^2=\binom{x}{0}+2\binom{x}{1}+4\binom{x}{2}+8\binom{x}{3}$ for positive integers $x$ and $y$

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).