Problem 1625

1. Initial Setup and Simplification:
Given the functional equation:
$$f(x^2 - g(y)) = g(x)^2 - y \quad \forall x, y \in \mathbb{R}$$
We introduce a new notation for convenience:
$$P(a, b, c, d): a^2 - g(b) = c^2 - g(d) \implies g(a)^2 - b = g(c)^2 - d$$

2. **Claim 1: $g$ is strictly increasing**:
- Assume $g(b) \leq g(d)$ for some $b, d \in \mathbb{R}$.
- For any $a \in \mathbb{R}$, there exists $c \in \mathbb{R}$ such that $c^2 = a^2 - g(b) + g(d)$.
- Using $P(a, b, c, d)$:
$$g(a)^2 - b + d = g(c)^2 \geq m^2$$
where $m = \inf_{x \in \mathbb{R}} |g(x)|$.
- This implies:
$$d - b \geq m^2 - g(a)^2$$
- Since this holds for all $a$, we can take $g(a)^2$ as close to $m^2$ as possible:
$$d - b \geq 0 \implies b \leq d$$
- Taking the contrapositive, $b > d \implies g(b) > g(d)$, proving $g$ is strictly increasing. $\blacksquare$

3. **Claim 2: $g$ is pseudo-odd**:
- Using $P(x, 0, -x, 0)$:
$$g(x)^2 = g(-x)^2 \implies g(-x) = -g(x) \quad \text{(by injectivity since \( x \neq -x \))}$$
- Thus, $g$ is pseudo-odd. $\blacksquare$

4. **Behavior of $g$ at zero and positive values**:
- For $x > 0$:
$$x > -x \implies g(x) > g(-x) = -g(x) \implies g(x) > 0 \implies g(x) \geq m$$
- For $x < 0$:
$$g(x) \leq -m$$
- If $|g(0)| > m$, there exists $x$ with $|g(0)| > |g(x)| \geq m$. We can take $x > 0$ due to pseudo-oddness.
- This leads to contradictions, so $g(0) = \pm m$.

5. **Claim 3: Range of $g$**:
- If $|k| > m$, then $k \in R$ (range of $g$).
- By Claim 1, $g(k^2 - m^2) > g(0)$.
- There exists $c$ such that:
$$c^2 - g(k^2 - m^2) = 0^2 - g(0) \implies g(c)^2 = k^2 - m^2 + g(0)^2 = k^2$$
- Since $c \neq 0$, $k = g(c)$ or $k = g(-c)$ by pseudo-oddness. $\blacksquare$

6. **Determining the form of $g$**:
- Assume $g(0) = m$ (similar method works for $g(0) = -m$).
- By Claim 3, $[m, +\infty) \subseteq R$.
- For $x, y > 0$:
$$g(x) + g(y) - m > m \implies g(x) + g(y) = g(z) + m = g(z) + g(0)$$
- Choose $a, c \in \mathbb{R}$ such that:
$$a^2 - c^2 = g(x) + g(y) = g(z) + g(0)$$
- Using pseudo-oddness and comparing $P(a, x, c, -y)$ and $P(a, 0, c, -z)$:
$$x + y = g(a)^2 - g(c)^2 = z$$
- Therefore:
$$g(x) + g(y) = g(x + y) + m$$
- The function $g(x) - m$ satisfies Cauchy's functional equation on $\mathbb{R}^+$ and is bounded below by 0:
$$g(x) = m + px \quad \text{for some constant } p$$

7. **Final form of $g$ and $f$**:
- Transforming $P(a, b, c, d)$:
$$a^2 - m - pb = c^2 - m - pd \implies (m + pa)^2 - b = (m + pc)^2 - d$$
- Comparing coefficients:
$$p = 1 \quad \text{and} \quad m = 0 \implies g(x) = x \quad \forall x \in \mathbb{R}$$
- Bringing back $f$:
$$f(x^2 - y) = x^2 - y \implies f(x) = x \quad \forall x \in \mathbb{R}$$

The final answer is $\boxed{f(x) = g(x) = x \ \ \forall x \in \mathbb{R}}$