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 :)

#Algebra #NumberTheory #HelpMe! #Proofs #Math

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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

Harshit Kapur - 8 years, 5 months ago

@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.

Calvin Lin Staff - 8 years, 5 months ago

@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 - 8 years, 5 months ago

Its represents a hyperbolic curve hence will have infinite solutions

Saurabh Dubey - 8 years, 5 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.

Calvin Lin Staff - 8 years, 5 months ago

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

Lawrence Sun - 8 years, 5 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)

Esteban Gomezllata - 8 years, 5 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.

Calvin Lin Staff - 8 years, 5 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.

Lawrence Sun - 8 years, 5 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$