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Is it possible for $aG \equiv bG$, with $a, b$ are scalars and $G$ is a point on the curve?

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Yes, it is possible, but only if $a, b$ differ by a multiple of the order of the point $G$.

When we perform elliptic curve cryptography, we select a curve and a point $G$ where the order is a large prime (typically, circa the field size). When we select a random multiplier, we most often select a random one uniformly between 1 and $\operatorname{order}(G)-1$; if both $a, b$ were selected that way, then the only way for $aG = bG$ would be if $a=b$

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