Problem 1829

Let $f: \mathbb{R} \to \mathbb{R}$ be a function such that for any $x, y \in \mathbb{R}$, the multiset $\{ f(xf(y) + 1), f(yf(x) - 1) \}$ is identical to the multiset $\{ xf(f(y)) + 1, yf(f(x)) - 1 \}$.

We aim to find all such functions $f$.

Let $P(x, y)$ denote the assertion that $\{ f(xf(y) + 1), f(yf(x) - 1) \} = \{ xf(f(y)) + 1, yf(f(x)) - 1 \}$.

First, consider $P(0, 0)$:
$$\{ f(1), f(-1) \} = \{ 1, -1 \}.$$
Thus, $f(1) = 1$ and $f(-1) = -1$ or $f(1) = -1$ and $f(-1) = 1$.

### Claim 1: $f$ is surjective.
Consider $P(x-1, 1)$:
$$\{ f((x-1)f(1) + 1), f(f(x-1) - 1) \} = \{ x, f(f(x-1)) - 1 \}.$$
This implies there exists $t$ such that $f(t) = x$ for all $x \in \mathbb{R}$.

### Claim 2: $f(0) = 0$.
Suppose $f(0) \neq 0$. Let $f(a) = 0$ for some $a \neq 0$. Then, consider $P(a, a)$:
$$\{ f(af(a) + 1), f(af(a) - 1) \} = \{ af(f(a)) + 1, af(f(a)) - 1 \}.$$
We get $\{ 1, -1 \} = \{ af(f(a)) + 1, af(f(a)) - 1 \}$, so $af(f(a)) = 0$, which implies $f(0) = 0$.

### Case 1: $f(1) = 1$.
We claim $f(x) \equiv x$.

Assume for contradiction $f(x) \neq x$ for some $x \neq 0$.

Consider $P(x-1, 1)$:
$$\{ f(x), f(f(x-1) - 1) \} = \{ x, f(f(x-1)) - 1 \}.$$
Since $f(x) \neq x$, it follows that $f(f(x-1) - 1) = x$ and $f(x) = f(f(x-1)) - 1$.

Consider $P(1, 1 + x)$:
$$\{ f(f(x+1) + 1), f(x) \} = \{ f(f(x+1)) + 1, x \}.$$
Since $f(x) \neq x$, it follows that $f(f(x+1) + 1) = x$ and $f(x) = f(f(x+1)) + 1$.

### Claim 3: If $f(a) = 0$ for some $a \neq 0$, then $f$ is injective.
Consider $P(a, y)$:
$$\{ f(af(y) + 1), f(yf(a) - 1) \} = \{ af(f(y)) + 1, yf(f(a)) - 1 \}.$$
Since $f(0) = 0$, we have:
$$\{ f(af(y) + 1), f(-1) \} = \{ af(f(y)) + 1, -1 \}.$$
It follows that $f(af(y) + 1) = af(f(y)) + 1$ for all $y$.

Similarly, $P(y, a)$ gives $f(ay - 1) = af(y) - 1$ for all $y$. Therefore, $f(y + 1) - f(y - 1) = 2$ for all $y$.

### Claim 4: $f$ is injective.
Assume for contradiction $f(u) = f(v)$ for $u \neq v$.

Consider $P(u, y)$ and $P(v, y)$:
$$\{ f(uf(y) + 1), f(yf(u) - 1) \} = \{ uf(f(y)) + 1, yf(f(u)) - 1 \},$$
$$\{ f(vf(y) + 1), f(yf(v) - 1) \} = \{ vf(f(y)) + 1, yf(f(v)) - 1 \}.$$
Since $f(u) = f(v)$, it follows that $f(yf(u) - 1) = f(yf(v) - 1)$ and $yf(f(u)) - 1 = yf(f(v)) - 1$.

Assume for contradiction $f(yf(u) - 1) \neq yf(f(u)) - 1$ for some $y \neq 0$. We have $f(yf(u) - 1) = uf(f(y)) + 1$ and $f(yf(v) - 1) = vf(f(y)) + 1$, so $f(f(y)) = 0$, contradicting our lemma.

Therefore, $f(yf(u) - 1) = yf(f(u)) - 1$ for all $y$. Similarly, $f(yf(u) + 1) = yf(f(u)) + 1$ for all $y$.

### Finish:
Now, consider $f(x+1) + 1 = f(x-1) - 1$. If $f$ were not a fixed point, we would have:
$$x = f(f(x+1) + 1) = f(f(x-1) - 1),$$
so $f(x+1) + 1 = f(x-1) - 1$.

We also know $f(x) = f(f(x-1)) - 1 = f(f(x+1)) + 1$.

Let $m = f(x-1) - 1 = f(x+1) + 1$. If $f(m) \neq m$, we have $f(m+1) + 1 = f(m-1) - 1$. Therefore, $f(f(x-1)) + 1 = f(f(x+1)) - 1$, but this contradicts our earlier equations.

Therefore, $f(m) = m$. We also know $f(f(x+1) + 1) = x$, so $m = f(m) = x$, contradicting our assumption $f(x) \neq x$.

Hence, the only solutions are:
$$f(x) \equiv x \quad \text{or} \quad f(x) \equiv -x.$$

The answer is: \boxed{f(x) \equiv x \text{ or } f(x) \equiv -x}.