Let n be a power of two: n = 2^m. Let us prove that first n - 1 jumps will end at different hassocks. Really, jump with number k ends at hassock . Suppose that jumps k₁ and k₂ ends at the same hassock. It means that . So we get that k₁² + k₁ - k₂² - k₂ = (k₁ - k₂)(k₁ + k₂ + 1) is divisible by 2^m + 1. But k₁ - k₂ and k₁ + k₂ + 1 have different parity and both less than 2^m + 1. A contradiction. So we derive that all hassocks will be visited after first n - 1 jumps.

If n isn't a power of two, then n is divisible by some odd prime p. Let's look at hassocks not mod n, but mod p. We will see that not all residues (mod p) will be visited. Indeed after p jumps flea returns to initial residue mod p, because is divisible by p. And after that jumps would be the same as in the beginning (mod p), because p is divisible by p. Moreover, even after p - 1 jumps flea returns to initial residue mod p. So there are p - 1 different residues as maximum. Thus not all hassocks will be visited.