# Does any $x < p$ satisfy the curve equation of X25519?

I've been reading about the famous X25519, a montgomery curve from wikipedia and in that article they say that we do not have to check for point validity. Is it because that any $$x < p$$ satisfy the curve equation ? Is this possible for X25519 because it is a Montgomery curve or because it was specifically designed to do so (for the sake of efficiency, and more possibilities for public keys) by its discoverer Dan Bernstein ? Why every $$x < p$$ does not satisfy the curve equation for other curves like SECP256k1, where the equations are of the Weierstrass form, $$y^2 = x^3 + ax + b \mod p$$ ? I would really like to know the mathematical reasons behind it. Thankyou everyone in advance!

If you have an elliptic curve given by the equation $$y^2 = f(x) \bmod p$$, then for each $$x$$, either $$f(x)$$ is a square modulo $$p$$, and there exists a square root $$y$$ such that $$(x,y)$$ and $$(x,-y)$$ satisfy the curve equation. If $$f(x)$$ is not a square modulo $$p$$, then this value $$x$$ does not correspond to a point on the curve, but to a point on the quadratic twist of the curve.
Therefore knowing only $$x$$, we know it corresponds to a point on the curve or its quadratic twist. The thing is that many standardized curves have a weak quadratic twist (see this page on SafeCurves so it is mandatory to check the point is on the curve, and if we use only the $$x$$-coordinate, we still need to check it is not on the quadratic twist to avoid invalid curve attacks.
Dan Bernstein designed X25519 as a Diffie-Hellman function using Curve25519 in its Montgomery form. On this curve, the Montgomery ladder scalar multiplication is efficient and uses only the $$x$$-coordinate of the points. Then, manipulated points lie on the curve or its quadratic twist which is as secure as the original curve. Then any $$x$$ is valid, but does not necessarily correspond to a point on the curve.