Whats to prevent me from finding 'almost all' zeros by trying enough (say 10 000 000) versions of enc(v) - enc(v). That, combined with comparison based on inte

Whats to prevent me from finding 'almost all' zeros by trying enough (say 10 000 000) versions of enc(v) - enc(v). That, combined with comparison based on integer division and underflow would still make decryption quite easy.