Let $ABC$ be an equilateral triangle and let $P_0$ be a point outside this triangle, such that $\triangle{AP_0C}$ is an isoscele triangle with a right angle at $P_0$. A grasshopper starts from $P_0$ and turns around the triangle as follows. From $P_0$ the grasshopper jumps to $P_1$, which is the symmetric point of $P_0$ with respect to $A$. From $P_1$, the grasshopper jumps to $P_2$, which is the symmetric point of $P_1$ with respect to $B$. Then the grasshopper jumps to $P_3$ which is the symmetric point of $P_2$ with respect to $C$, and so on. Compare the distance $P_0P_1$ and $P_0P_n$. $n \in N$.
1. **Initial Setup and Symmetry:**
   - Let \( \triangle ABC \) be an equilateral triangle with side length \( a \).
   - Let \( P_0 \) be a point outside \( \triangle ABC \) such that \( \triangle AP_0C \) is an isosceles right triangle with a right angle at \( P_0 \).
   - The grasshopper starts at \( P_0 \) and jumps to \( P_1 \), the symmetric point of \( P_0 \) with respect to \( A \). This means \( AP_0 = AP_1 \).

2. **Translation and Symmetry:**
   - From \( P_1 \), the grasshopper jumps to \( P_2 \), the symmetric point of \( P_1 \) with respect to \( B \). This implies \( BP_1 = BP_2 \).
   - From \( P_2 \), the grasshopper jumps to \( P_3 \), the symmetric point of \( P_2 \) with respect to \( C \). This implies \( CP_2 = CP_3 \).

3. **Equilateral Triangle Formation:**
   - The point \( P_2 \) is obtained from \( P_0 \) by a translation in the direction \( \vec{AB} \) by the distance \( 2a \).
   - Similarly, \( P_4 \) is obtained from \( P_2 \) by a translation in the direction \( \vec{CA} \) by the distance \( 2a \), and \( P_6 \) from \( P_4 \) by a translation in the direction \( \vec{BC} \) by the distance \( 2a \).
   - It follows that the points \( P_0, P_2, P_4 \) form an equilateral triangle with side lengths \( 2a \) and the sides \( P_0P_2 \parallel AB, P_2P_4 \parallel CA, P_4P_0 \parallel BC \). The point \( P_6 \equiv P_0 \) are identical.

4. **Further Translations:**
   - For similar reasons, \( P_3 \) is obtained from \( P_1 \) by a translation in the direction \( \vec{BC} \) by the distance \( 2a \), \( P_5 \) is obtained from \( P_3 \) by a translation in the direction \( \vec{AB} \) by the distance \( 2a \), and \( P_7 \) is obtained from \( P_5 \) by a translation in the direction \( \vec{CA} \) by the distance \( 2a \).
   - It follows that the points \( P_1, P_3, P_5 \) also form an equilateral triangle with side lengths \( 2a \) and the sides \( P_1P_3 \parallel BC, P_3P_5 \parallel AB, P_5P_1 \parallel CA \). The point \( P_7 \equiv P_1 \) are identical.

5. **Congruence and Parallelism:**
   - Thus, the equilateral triangles \( \triangle P_0P_2P_4 \cong \triangle P_3P_5P_1 \) are congruent and have parallel sides. The second is obtained from the first by a translation in the direction \( \vec{P_0P_3} \) by the distance \( P_0P_3 \).

6. **Using the Equilateral and Isosceles Properties:**
   - Since \( P_6 \equiv P_0 \) are identical and \( P_0A = P_0C, AP_0 = AP_1, CP_0 = CP_5 \), the triangle \( \triangle P_1P_0P_5 \) is also isosceles.
   - This means that the point \( P_0 \) lies on the altitude of the equilateral triangle \( \triangle P_3P_5P_1 \) from the vertex \( P_3 \). Since in addition, the isosceles triangle \( \triangle P_1P_0P_5 \) is right, we have:
     \[
     P_0P_3 = P_1P_5 \cdot \frac{\sqrt{3}}{2} - \frac{P_1P_5}{5} = a (\sqrt{3} - 1)
     \]

7. **Distance Calculations:**
   - Obviously, \( P_0P_5 = P_0P_1 = 2AP_0 = 2CA \cdot \frac{\sqrt{2}}{2} = a \sqrt{2} \) and \( P_0P_2 = P_0P_4 = 2a \).
   - As a result:
     \[
     P_0P_2 = P_0P_4 = P_0P_1 \sqrt{2}
     \]
     \[
     P_0P_3 = P_0P_1 \cdot \frac{\sqrt{3} - 1}{\sqrt{2}} = P_0P_1 \cdot \frac{\sqrt{6} - \sqrt{2}}{2}
     \]
     \[
     P_0P_5 = P_0P_1
     \]
     \[
     P_0P_6 = 0
     \]
     \[
     P_0P_n = P_0P_{(n \mod 6)}
     \]

The final answer is \( \boxed{ P_0P_n = P_0P_{(n \mod 6)} } \).