Let's consider hashing of odd 256-bit number $L$ with 256-bit odd key $K$: $z = (2^{256}-L) cdot K bmod 2^{256}$ $z$ is the hashed number. If $K$ is selected randomly is it secure to hashing numbers this way? It is as far as I know just linear congruential generator, they are not secure. But this is slightly different (we have exponentiation in modulus and we are not using it to generate some sequence of numbers, we have no seed, just numbers to hash). Why exactly it can't be secure? Is it secure if $L$ and $K$ are selected randomly, to hash $L$ this way?

Modulo hashing scheme

Cryptography Asked on October 24, 2021

Let’s consider hashing of odd 256-bit number $L$ with 256-bit odd key $K$:

$z = (2^{256}-L) cdot K bmod 2^{256}$

$z$ is the hashed number.

If $K$ is selected randomly is it secure to hashing numbers this way? It is as far as I know just linear congruential generator, they are not secure. But this is slightly different (we have exponentiation in modulus and we are not using it to generate some sequence of numbers, we have no seed, just numbers to hash). Why exactly it can’t be secure?
Is it secure if $L$ and $K$ are selected randomly, to hash $L$ this way?

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