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This is not true. $f(x)$ and $f(-x)$ will not be in the same equivalence class in general, yet their images are the same.

This is not true. $f(x)$ and $f(2x)$ will not be in the same equivalence class in general, yet their images will agree on infinitely many points. Hence there is no fixed value $m$ such that any $m$ points determine a single equivalence class

This is not true. $f(x)$ and $f(2 x)$ will not be in the same equivalence class in general, yet the images of $f(2 x)$ and $f(x)$ share infinitely many values.

Note: This does not show that no finite set of values of a polynomial can determine it; rather, it shows that there is no fixed $m$ such that all finite sets of size $m$ determine the polynomial.

This is not true. $f(x)$ and $f(-x)$ will not be in the same equivalence class in general, yet their images are the same.

This is not true. $f(x)$ and $f(2 x)$ will not be in the same equivalence class in general, yet the images of $f(2 x)$ and $f(x)$ share infinitely many values.

This is not true. $f(x)$ and $f(2 x)$ will not be in the same equivalence class in general, yet the images of $f(2 x)$ and $f(x)$ share infinitely many values.

Note: This does not show that no finite set of values of a polynomial can determine it; rather, it shows that there is no fixed $m$ such that all finite sets of size $m$ determine the polynomial.