Find all rational solutions to: \begin{eqnarray*} t^2 - w^2 + z^2 &=& 2xy \\ t^2 - y^2 + w^2 &=& 2xz \\ t^2 - w^2 + x^2 &=& 2yz . \end{eqnarray*}
To find all rational solutions to the system of equations:

\[
\begin{cases}
t^2 - w^2 + z^2 = 2xy \\
t^2 - y^2 + w^2 = 2xz \\
t^2 - w^2 + x^2 = 2yz
\end{cases}
\]

we will proceed step-by-step.

1. **Rewrite the equations for clarity:**

\[
\begin{cases}
t^2 - w^2 + z^2 = 2xy \quad \text{(1)} \\
t^2 - y^2 + w^2 = 2xz \quad \text{(2)} \\
t^2 - w^2 + x^2 = 2yz \quad \text{(3)}
\end{cases}
\]

2. **Subtract equation (1) from equation (2):**

\[
(t^2 - y^2 + w^2) - (t^2 - w^2 + z^2) = 2xz - 2xy
\]

Simplify:

\[
w^2 - y^2 - w^2 + z^2 = 2xz - 2xy
\]

\[
z^2 - y^2 = 2xz - 2xy
\]

Factor both sides:

\[
(z - y)(z + y) = 2x(z - y)
\]

If \(z \neq y\), we can divide both sides by \(z - y\):

\[
z + y = 2x
\]

Thus, we have:

\[
x = \frac{z + y}{2} \quad \text{(4)}
\]

3. **Subtract equation (1) from equation (3):**

\[
(t^2 - w^2 + x^2) - (t^2 - w^2 + z^2) = 2yz - 2xy
\]

Simplify:

\[
x^2 - z^2 = 2yz - 2xy
\]

Factor both sides:

\[
(x - z)(x + z) = 2y(z - x)
\]

If \(x \neq z\), we can divide both sides by \(x - z\):

\[
x + z = -2y
\]

Thus, we have:

\[
y = -\frac{x + z}{2} \quad \text{(5)}
\]

4. **Subtract equation (2) from equation (3):**

\[
(t^2 - w^2 + x^2) - (t^2 - y^2 + w^2) = 2yz - 2xz
\]

Simplify:

\[
x^2 - w^2 - y^2 + w^2 = 2yz - 2xz
\]

\[
x^2 - y^2 = 2yz - 2xz
\]

Factor both sides:

\[
(x - y)(x + y) = 2z(y - x)
\]

If \(x \neq y\), we can divide both sides by \(x - y\):

\[
x + y = -2z
\]

Thus, we have:

\[
z = -\frac{x + y}{2} \quad \text{(6)}
\]

5. **Combine equations (4), (5), and (6):**

From equation (4):

\[
x = \frac{z + y}{2}
\]

From equation (5):

\[
y = -\frac{x + z}{2}
\]

From equation (6):

\[
z = -\frac{x + y}{2}
\]

6. **Substitute \(x\), \(y\), and \(z\) into each other:**

Substitute \(y = -\frac{x + z}{2}\) into \(z = -\frac{x + y}{2}\):

\[
z = -\frac{x - \frac{x + z}{2}}{2}
\]

Simplify:

\[
z = -\frac{2x - (x + z)}{4}
\]

\[
z = -\frac{x - z}{4}
\]

Multiply both sides by 4:

\[
4z = -x + z
\]

\[
3z = -x
\]

\[
x = -3z
\]

Substitute \(x = -3z\) into \(y = -\frac{x + z}{2}\):

\[
y = -\frac{-3z + z}{2}
\]

\[
y = -\frac{-2z}{2}
\]

\[
y = z
\]

Thus, we have:

\[
x = -3z, \quad y = z
\]

7. **Verify the solution:**

Substitute \(x = -3z\) and \(y = z\) into the original equations:

\[
t^2 - w^2 + z^2 = 2(-3z)z
\]

\[
t^2 - w^2 + z^2 = -6z^2
\]

\[
t^2 - w^2 = -7z^2 \quad \text{(7)}
\]

\[
t^2 - y^2 + w^2 = 2(-3z)z
\]

\[
t^2 - z^2 + w^2 = -6z^2
\]

\[
t^2 + w^2 = -5z^2 \quad \text{(8)}
\]

\[
t^2 - w^2 + x^2 = 2yz
\]

\[
t^2 - w^2 + 9z^2 = 2z^2
\]

\[
t^2 - w^2 + 9z^2 = 2z^2
\]

\[
t^2 - w^2 = -7z^2 \quad \text{(9)}
\]

Equations (7) and (9) are consistent, but equation (8) is not consistent with the others unless \(z = 0\).

Thus, the only rational solution is when \(z = 0\), which implies \(x = 0\) and \(y = 0\).

The final answer is \( \boxed{ (x, y, z, t, w) = (0, 0, 0, t, w) } \) for any rational \(t\) and \(w\).