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# An algebra conjecture: y^2 = ax^2 + b

I once had this idea that the equation $$y^2 = ax^2 + b$$, where a, b, x and y are natural numbers, would have either infinite answers $$(x,y)$$ or no answer (in other words, it has not a finite positive number of answers). I think that this is false when "a" is a squared number, and I found a proof for a = 2 and a = 3, but I have not found the proof for the rest of numbers ( a = 5 for example ).

Has anyone an idea of how we could prove this (or prove this false if it is false)? I leave you the proof of a = 2,3 as a challenge :)

Note by Esteban Gomezllata
4 years, 7 months ago

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It will always have infinite solutions. Look carefully, its a hyperbola. I think it holds good for all natural numbers as you say, even squares · 4 years, 7 months ago

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@Harshit Note the extra condition "$$x$$ and $$y$$ are natural numbers", which makes this problem more difficult that talking about the graph, since we do not know when it will pass through a lattice point. Once again $$\frac {3}{2} = \sqrt{\frac {3}{2} } ^2$$ is not considered a square. Staff · 4 years, 7 months ago

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Its represents a hyperbolic curve hence will have infinite solutions · 4 years, 7 months ago

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@Saurabh Be careful, Esteban is only interested in integer solutions. For example, the hyperbola $$y = \frac {1}{x}$$ only has 2 integer solutions. Staff · 4 years, 7 months ago

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@Esteban. I think you mean "(in other words, No answers)". It clearly either has infinite answers, or a finite number of answers. Staff · 4 years, 7 months ago

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This is trivial, just use properties of Pell-Equations. · 4 years, 7 months ago

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@Lawrence Thank you for the information, but I think to have understood that Pell-Equations require "b" to be 1. For some other values of "b", even with "a" being a non-square number, it had no solutions (below 2000 or 20000; I made a simulation to see patterns) · 4 years, 7 months ago

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@Esteban If you understood how to deal with Pell Equation for $$b=1$$ through the fundamental solution, think about how this idea can be extended to $$b\neq 1$$. In particular, the case $$b=-1$$ is very often used. Note that it is often hard to determine if a fundamental solution exists in these other cases. Staff · 4 years, 7 months ago

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Clearly you don't understand them well enough if you believe you need $$b=1$$... using very basic ideas the equation $$x^2 - dy^2 = a$$ for $$d$$ not a perfect square has infinitely many solutions iff it has at least one solution. · 4 years, 7 months ago

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