, 2015

First, it is easy to see that $f(x) \equiv 0$ is a solution to the equation, so assume $f(x)$ is not always $0$. Substituting $(x - 1, 1)$ into the original equation gives $f(2) = t$.

Let $f(1) = a$. Substituting $(x, 0)$ gives
$$(t + 1)a - f(x) = af(x + 1) \quad (1)$$
If $a = 0$, then by (1) we know $f(x) \equiv 0$, which is not allowed.
So
$$a \neq 0, \quad f(x + 1) = t + 1 - \frac{f(x)}{a} \quad (2)$$
Substituting $x - 1$ for $x$ in (1) and rearranging gives
$$f(x - 1) = (t + 1)a - af(x) \quad (3)$$
Substituting $1, 0, -1$ into (3) gives
$$f(0) = -a^2 + (t + 1)a,$$
$$f(-1) = a^3 - (t + 1)a^2 + (t + 1)a,$$
$$f(-2) = -a^4 + (t + 1)a^3 - (t + 1)a^2 + (t + 1)a.$$
Substituting into the original equation, $(t + 1)f(2) - f(-2) = f(0)^2$
So $(t + 1)t + a^4 - (t + 1)a^3 + (t + 1)a^2 - (t + 1)a = a^4 - 2(t + 1)a^3 + (t^2 + 2t + 1)a^2$
Rearranging gives $(a^2 - 1)(t + 1)(a - t) = 0$, and since $t \neq -1$, we know $a = \pm 1$ or $a = t$.

If $a = t \neq \pm 1$, then $f(0) = -a^2 + (t+1)a = t$
Substituting $(x, -1)$ into the original equation gives
$$(t+1)f(1-x) - f(x-1) = f(0)f(x+1) = tf(x+1) \quad (4)$$
Using (2) and (3) to convert (4) entirely into terms of $f(x)$ and $f(-x)$ and rearranging gives
$$tf(x) = f(-x) + t^2 - t$$
Therefore $t^2 f(x) = t(-x) + t^3 - t^2 = f(x) + t^3 - t^2 + t^2 - t^1 = f(x) + t^3 - t$,
so $(t^2 - 1)f(x) = t^3 - t$, and since $t \neq \pm 1$ we know $f(x) = t$.

If $a = 1$, then by (1) we know
$$f(x) + f(x+1) = t + 1. \quad (5)$$
So from $f(2) = t$ we get $f(3) = 1$.
Substituting $(2, \frac{1}{2})$ into the original equation gives $t^2 + t - f(\frac{5}{2}) - f(\frac{3}{2}) = 0$, but $f(\frac{5}{2}) + f(\frac{3}{2}) = t + 1$,
so $t^2 - 1 = 0$.
Since $t \neq -1$ we get $t = 1$. Substituting $(x, 2)$ gives $2f(2x+1)-f(x+2) = f(x+1)f(3)$.
By (5) we know $2f(2x+1) = f(x+1)+f(x+2) = 2$, so $f(2x+1) = 1$, that is to say $f(x) \equiv 1 = t$.

If $a = -1$, then (2) becomes
$$f(x+1) = f(x) + (t+1) \quad (6)$$
Let $g(x) = \frac{f(x)+(t+2)}{t+1}$ and substitute back into the original equation and rearrange to get
$$(t+1)g(xy) + g(x) + g(y) = (t+1)g(x)g(y) + g(x+y) \quad (7)$$
(6) becomes $g(x+1) = g(x) + 1$, and $g(2) = 2$, so $g(1) = 1$, $g(-1) = -1$.
Substituting $(x, -1)$ into (7) gives $g(-x) = -g(x)$, that is, $g$ is an odd function.

Substituting $(x, y)$ and $(-x, -y)$ into (7) and subtracting the two equations gives $g(x) + g(y) = g(x + y)$
Substituting back into (7) gives $g(xy) = g(x)g(y)$, and combining these two equations we get $g(x) = x$.
Therefore $f(x) = (t+1)g(x) - (t+2) = (t+1)x - (t+2)$.

Combining the above, we obtain three groups of solutions
$$f(x) \equiv 0, \quad f(x) \equiv t, \quad f(x) = (t+1)x - (t+2).$$