Let $a$, $b$ and $c$ be real numbers such that $a^2+b^2=c^2$, solve the system:
\[ z^2=x^2+y^2 \]
\[ (z+c)^2=(x+a)^2+(y+b)^2 \]
in real numbers $x, y$ and $z$.
Given the system of equations:
\[ z^2 = x^2 + y^2 \]
\[ (z + c)^2 = (x + a)^2 + (y + b)^2 \]

We are also given that \( a^2 + b^2 = c^2 \).

1. Start by expanding the second equation:
\[
(z + c)^2 = (x + a)^2 + (y + b)^2
\]
\[
z^2 + 2cz + c^2 = x^2 + 2ax + a^2 + y^2 + 2by + b^2
\]

2. Substitute \( z^2 = x^2 + y^2 \) from the first equation into the expanded form:
\[
x^2 + y^2 + 2cz + c^2 = x^2 + 2ax + a^2 + y^2 + 2by + b^2
\]

3. Simplify by canceling out \( x^2 \) and \( y^2 \) on both sides:
\[
2cz + c^2 = 2ax + a^2 + 2by + b^2
\]

4. Use the given \( a^2 + b^2 = c^2 \) to replace \( c^2 \):
\[
2cz + c^2 = 2ax + 2by + c^2
\]

5. Subtract \( c^2 \) from both sides:
\[
2cz = 2ax + 2by
\]

6. Divide by 2:
\[
cz = ax + by
\]

7. Square both sides to use the Cauchy-Schwarz inequality:
\[
(cz)^2 = (ax + by)^2
\]
\[
c^2z^2 = (ax + by)^2
\]

8. By the Cauchy-Schwarz inequality:
\[
(a^2 + b^2)(x^2 + y^2) \ge (ax + by)^2
\]

9. Given \( a^2 + b^2 = c^2 \) and \( x^2 + y^2 = z^2 \):
\[
c^2z^2 \ge (ax + by)^2
\]

10. Since \( c^2z^2 = (ax + by)^2 \), equality holds in the Cauchy-Schwarz inequality, which implies that \( \frac{a}{c} = \frac{x}{z} \) and \( \frac{b}{c} = \frac{y}{z} \).

11. Therefore, we have:
\[
y = \frac{b}{a} x
\]

12. Substitute \( y = \frac{b}{a} x \) into \( z^2 = x^2 + y^2 \):
\[
z^2 = x^2 + \left( \frac{b}{a} x \right)^2
\]
\[
z^2 = x^2 + \frac{b^2}{a^2} x^2
\]
\[
z^2 = x^2 \left( 1 + \frac{b^2}{a^2} \right)
\]
\[
z^2 = x^2 \left( \frac{a^2 + b^2}{a^2} \right)
\]
\[
z^2 = x^2 \left( \frac{c^2}{a^2} \right)
\]
\[
z = \frac{c}{a} x
\]

Thus, the solutions are:
\[
y = \frac{b}{a} x \quad \text{and} \quad z = \frac{c}{a} x
\]

The final answer is \( \boxed{ y = \frac{b}{a} x } \) and \( z = \frac{c}{a} x \).