Is it possible to compute the y-coordinate of a point on SECP256K1, given only the x-coordinate Given an x-coordiante of a point on the SECP256K1 curve, is it possible to calculate the corresponding y-coorindate? (Assuming the point is a verifying public key that complies with the Bitcoin standards.)
I am new to the cryptographic realm so please forgive me if the question is naive. From what I know, the public key is a point, or a pair of integers. The SECP256K1 curve is a curve where any point (x, y) on it satisfies
(y ** 2) mod p == (x ** 3 + 7) mod p

where p = 2**256 - 2**32 - 977.
Now let's confine the discussion within the Bitcoin scope. Assume we have a private key that complies with the Bitcoin standards, and from it we can derive the public key, which can be represented as a point (x, y) on the SECP256K1 curve.
Now given only such a x, is it possible to calculate the y?
As a real example, given only x as
0x6778ec0abf66f1ba4d93aa45cad77dc26c593f520448f6fff5b70357270154ba

is it possible get the y as
0x6a5e8cd7276f80ee2f7c081702eff3e14134b006acd0afc8467be94a0a3a0558
Given an $x$-coordinate of a point on the SECP256K1 curve, is it possible to calculate the corresponding $y$-coordinate?

Yes, if there exists such $y$ for the given $x$. And, absent other indication, such $y$ can only be found within sign (or equivalently, parity). That limitation is because if $y^2\equiv x^3+7\pmod p$ with $p=2^{256}−2^{32}−2^{10}+2^6-2^4−1$ as in secp256k1 has a solution $y_0$ in $[0,p)$, then $y_1=p-y_0$ also is a solution.
Note: in some cases including secp256k1 as used in Bitcoin, a public key with $x$ and without $y$ (that is, in compressed form) comes with a prefix of 02 if $y$ is even, 03 if $y$ is odd, and that allows to fully recover $y$.
By Euler's criterion, $x^3+7$ has a square root modulo $p$ if and only if $(x^3+7)^{(p-1)/2}\bmod p=1$. That holds for the question's $x$, thus there are solutions.
In the general case, the Tonelli–Shanks algorithm can be used to find modular square roots. Since $p\equiv3\pmod4$, that algorithm reduces to computing $y_0\gets (x^3+7)^{(p+1)/4}\bmod p$ and $y_1\gets p-y_0$. The question's $y$ happens to be $y_1$.
Justification: when we have checked $(x^3+7)^{(p-1)/2}\bmod p=1$, and computed $y_0$ as $(x^3+7)^{(p+1)/4}\bmod p$, the later is such that $$\begin{array}{}
{y_0}^2 &amp;\equiv&amp;\left((x^3+7)^{(p+1)/4}\right)^2 &amp;\pmod p \\
&amp; \equiv&amp;(x^3+7)^{(p+1)/2} &amp;\pmod p \\
&amp;\equiv&amp;(x^3+7)^{(p-1)/2}\,(x^3+7)&amp;\pmod p\\
&amp;\equiv&amp;x^3+7 &amp;\pmod p &amp; \text{since}\;(x^3+7)^{(p-1)/2}\bmod p=1\end{array}$$
thus $y_0$ is a solution to $y^2\equiv x^3+7\pmod p$.

Definitions:
$$\begin{array}{l}
u\equiv v\pmod p&amp;\underset{\text{def}}\iff v-u\;\text{ is a multiple of }\;p\\
u=v\bmod p&amp;\underset{\text{def}}\iff v-u\;\text{ is a multiple of }\;p\;\text{ and }0\le u&lt;p\;
\end{array}$$