For example if they are two private keys from a public derivation wallet and you know the chain code you can recover the keys.
Also, the next paragraph about linear equations is quite important: as I described many times, each signature is just a relation between some public key, and signature R-value, that also is in fact some ECDSA point. That means you just have two points, that are connected in a linear way, like in "y=ax+b" function. You have one key under "x", and you can reach another key under "y" by applying "a" and "b" constants, that are publicly known. This equation is true for both cases, for private and public keys. Both sides should be always equal modulo "n", for each and every signature.
The same way of thinking can lead you to another important facts: deterministic wallets are widely used, even in the famous puzzle from 2015, you have some HD wallet.
A few words about the puzzle. There is no pattern. It is just consecutive keys from a deterministic wallet (masked with leading 000...0001 to set difficulty). It is simply a crude measuring instrument, of the cracking strength of the community.
That means, there is some single key, that was used to generate all 256 addresses. Also, it means bruteforcing next keys is not the only way to unlock them. Another method is to pick some HD wallet, go through all steps, and apply more restrictions to the initial key, based on values derived from it. For example, the first key is of course equal to one. But the second key can be two or three. In this puzzle, it is three, that means, around half of the seeds can be rejected.
However, that is not the end of the road. Computers are deterministic, if you know all inputs, you can always reach the same outputs. That means, even if you have some keys from non-HD wallet, then still, you can find some value, that will allow you to deteministically generate all keys. It is all about the size of your inputs, and it is related to exhausting the whole space of possibilities, and applying the proper compression to that. You can see it clearly in neural networks, where you can begin with some data, and you will always reach some model, even if it will be overfitted.
The same can be applied here: even if the author of the original puzzle used just some single key to derive all addresses, then still, it is technically possible to find another seed, and even another algorithm, that will lead to the same solution. However, practically speaking, I think if we will ever get there, it will be long after breaking all hash functions and ECDSA, where all of those things will be obsolete, and replaced with something else through another soft-fork. But mathematically speaking, I think there exist such solutions.
