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user6976
user6976

For your equation $x^2+y^2=z^2$, $A$ can have $N/2$ numbers. For example, take all odd numbers $\le N$. This set does not have any solutions of your (Pythagorean) equation. For the new (Pell) equation, the set of even numbers will do the trick.

For your equation $x^2+y^2=z^2$, $A$ can have $N/2$ numbers. For example, take all odd numbers $\le N$. This set does not have any solutions of your (Pythagorean) equation.

For your equation $x^2+y^2=z^2$, $A$ can have $N/2$ numbers. For example, take all odd numbers $\le N$. This set does not have any solutions of your (Pythagorean) equation. For the new (Pell) equation, the set of even numbers will do the trick.

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Source Link
user6976
user6976

For your equation $x^2+y^2=z^2$, $A$ can have $N/2$ numbers. For example, take all odd numbers $\le N$. This set does not have any solutions of your (Pythagorean) equation.