Block cipher block and key sizes Is there any point in making block size less than key size in block cipher?
For example, Russian GOST symmetric cipher uses 256-bit keys to encrypt 64-bit blocks. That means there are many keys which give the same result, so why making keys so large?
GOST symmetric cipher uses 256-bit keys to encrypt 64-bit blocks. That means there are many keys which give the same result


This is actually incorrect. A block cipher, when given a key, maps $n$-bit plaintext blocks to $n$-bit ciphertext blocks (and vice versa for decryption). Here's a toy example for $n=2$:


  $E_k(00) = 01 \\
E_k(01) = 10 \\
E_k(10) = 00 \\
E_k(11) = 11$


Although there are $2^n$ possible plaintext blocks and $2^n$ possible ciphertext blocks, there are actually $2^n!$ different possible ways to map plaintext blocks to ciphertext blocks. In this toy example, there are actually $2^2! = 4! = 1\times2\times3\times4 = 24$ different ways of mapping each two-bit input to a unique two-bit output.

Back to your question, GOST uses 256-bit keys, meaning that there are $2^{256}$ possible keys. However, with a 64-bit block size, there are $2^{64}!$ different possible permutations, and $2^{64}!$ is much larger than $2^{256}$.