I know that Bob can calculate the shared DH key without knowing the private key. If he sends to Alice a public key = 1, then the the DH key would be 1.
Can i achieve something like this in ECDH? where Bob can calculate the shared key without knowing the private key?
Similarly, just send the identity element $\mathcal{O}$ as the public key.
Alice must be a fool to not validate the public key to fall into this trap. The first rule is to check that $P \neq \mathcal{O}$
Update per user's comment
How can i find the identity element of curve 25519?
The identity of some curves like Curve25519 is the point of infinity ( represented as $\mathcal{O} \text { or } \infty$) and this can't be represented in the affine coordinates. We need to use projective coordinate systems.
We took the Curve25519 equation $$y^2 = x^3 + 486662 x^2 + x$$ and convert it into homogeneous equation $$Y^2 Z - X^3 - 486662 X^2 Z - X Z^2$$
Now the points are triple $(X:Y:Z)$ where actually a point now represent a line $(\lambda X:\lambda Y: \lambda Z)$
The point at infinity is $(0:1:0)$ and this cannot be treated as an affine point.
Any affine point $(x,y)$ can be turned into homogenous by $(x:y:1)$ and any homogenous point other than the infinity $(X:Y:Z)$ can be turned back with $(X/Z,Y/Z)$.
