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 :)

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

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naturalnumbers", 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. – Calvin Lin Staff · 4 years, 7 months agoLog in to reply

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

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– Calvin Lin Staff · 4 years, 7 months ago

@Saurabh Be careful, Esteban is only interested in integer solutions. For example, the hyperbola \( y = \frac {1}{x}\) only has 2 integer solutions.Log in to reply

@Esteban. I think you mean "(in other words,

No answers)". It clearly either has infinite answers, or a finite number of answers. – Calvin Lin Staff · 4 years, 7 months agoLog in to reply

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

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– Esteban Gomezllata · 4 years, 7 months ago

@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)Log in to reply

– Calvin Lin Staff · 4 years, 7 months ago

@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.Log in to reply

– Lawrence Sun · 4 years, 7 months ago

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.Log in to reply