Hash collisions in sha512 hash of secp256k1 public keys For a sha512 hash of a secp256k1 public key, how many other public keys could generate that hash? I would assume zero since the key size is equal to the hash length (secp256k1 public keys are 64 bytes, 512-bit), and there are only $2^{256}$ public keys as the private keys are 256-bit.
The output of the SHA512 is 512-bit, so there are at most $2^{512}$ different outputs. 

For random inputs, the expected hash collision with 50% probability is $\sqrt(2^{512}) = 2^{256}$ by Birthday Paradox ($\sqrt(n)$). 

The public keys of secp256k1 have 64-byte = 512-bit. To calculate SHA512  hash the input values are padded so that the input size is always multiple of 1024. After $2^{256}$ SHA512 hash of the public keys, we expect to find collisions with 50% probability.


SHA512: For finding a collision with birthday attack for SHA512 is almost impossible. You will need immense time, storage, and power for $2^{256}$ input values.
SHA256: For finding a collision with birthday attack for SHA256 is almost the same as breaking AES-128 by brute force, need for $2^{128}$ hash values. If we assume that the BitCoin miners top peak performance is $\approx 2^{91}$ hashes in a year, you will need $2^{37}$ years with the same computing platform and power consumption.