MSTang wrote:

Claim 1: $-1$ is in the range of $f$ .

Proof: Taking $y=f(x)$ gives $f(x-f(f(x)) = -1$ for all $x$ . $\spadesuit$

Claim 2: Over all $x$ , the quantity $f(f(x)) - f(x)$ is constant.

Proof. Let $a, b$ be integers. Taking $x=f(a)+f(b)$ and $y=a$ gives $f(f(b)) = f(f(f(a)+f(b))) - f(a) - 1.$ Taking $x=f(a)+f(b)$ and $y=b$ gives $f(f(a)) = f(f(f(a)+f(b))) - f(b) - 1.$ Comparing the two equations gives $f(f(a)) - f(a) = f(f(b)) - f(b)$ , as claimed. $\spadesuit$

Finisher. Let $f(f(x)) - f(x) = k$ for some constant $k$ . Then the given equation becomes $f(x-f(y)) = f(x) - f(y) + (k-1)$ for all $x, y$ . Using Claim 1, choose $y$ in the above equation such that $f(y) = -1$ , giving $f(x+1) = f(x) + k$ for all $x$ . Since $f$ has domain $\mathbb{Z}$ , we conclude that $f$ is of the form $f(x) = kx + c$ for some constants $k$ and $c$ . In the original equation, we need $k(x-ky-c)+c = k(kx+c) + c - ky - c - 1$ or $kx - k^2y - ck + c = k^2x + ck - ky - 1.$ Comparing $x$ coefficients gives $k = k^2$ , so $k \in \{0, 1\}$ . If $k=0$ , then $c = -1$ , giving the solution $f(x) = -1$ for all $x$ . If $k=1$ , then $c = 1$ , giving the solution $f(x) = x + 1$ for all $x$ . It is easy to check that both these functions work. $\square$

It is somewhat similar to the solution I had for this problem.