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 $x &lt; p$ 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 &lt; 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.