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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## Comments

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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

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@Harshit Note the extra condition "\(x\) and \(y\) are

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.Log in to reply

Its represents a hyperbolic curve hence will have infinite solutions

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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.

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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.Log in to reply

This is trivial, just use properties of Pell-Equations.

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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)

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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.

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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.

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