[i][b]a)[/b][/i] Determine if there are matrices $A,B,C\in\mathrm{SL}_{2}(\mathbb{Z})$ such that $A^2+B^2=C^2$.

[b][i]b)[/i][/b] Determine if there are matrices $A,B,C\in\mathrm{SL}_{2}(\mathbb{Z})$ such that $A^4+B^4=C^4$.

[b]Note[/b]: The notation $A\in  \mathrm{SL}_{2}(\mathbb{Z})$ means that $A$ is a $2\times 2$ matrix with integer entries and $\det A=1$.
### Part (a)
Determine if there are matrices \( A, B, C \in \mathrm{SL}_2(\mathbb{Z}) \) such that \( A^2 + B^2 = C^2 \).

1. **Understanding the Problem:**
   - We need to find matrices \( A, B, C \) in \( \mathrm{SL}_2(\mathbb{Z}) \) such that \( A^2 + B^2 = C^2 \).
   - \( \mathrm{SL}_2(\mathbb{Z}) \) denotes the group of \( 2 \times 2 \) matrices with integer entries and determinant equal to 1.

2. **Properties of Matrices in \( \mathrm{SL}_2(\mathbb{Z}) \):**
   - If \( A \in \mathrm{SL}_2(\mathbb{Z}) \), then \( \det(A) = 1 \).
   - The determinant of the square of a matrix in \( \mathrm{SL}_2(\mathbb{Z}) \) is also 1, i.e., \( \det(A^2) = (\det(A))^2 = 1 \).

3. **Exploring the Equation \( A^2 + B^2 = C^2 \):**
   - Consider the determinants on both sides of the equation:
     \[
     \det(A^2 + B^2) = \det(C^2)
     \]
   - Since \( \det(C^2) = (\det(C))^2 = 1 \), we have:
     \[
     \det(A^2 + B^2) = 1
     \]
   - However, the determinant of the sum of two matrices is not generally equal to the sum of their determinants. Therefore, we need to explore further.

4. **Counterexample Approach:**
   - Suppose \( A = I \) (the identity matrix), \( B = I \), and \( C = \sqrt{2}I \). Clearly, \( A, B, C \in \mathrm{SL}_2(\mathbb{Z}) \) and:
     \[
     A^2 + B^2 = I + I = 2I \quad \text{and} \quad C^2 = (\sqrt{2}I)^2 = 2I
     \]
   - This satisfies \( A^2 + B^2 = C^2 \), but \( \sqrt{2}I \notin \mathrm{SL}_2(\mathbb{Z}) \) because \( \sqrt{2} \) is not an integer.

5. **Conclusion:**
   - There are no matrices \( A, B, C \in \mathrm{SL}_2(\mathbb{Z}) \) such that \( A^2 + B^2 = C^2 \).

\[
\boxed{\text{False}}
\]

### Part (b)
Determine if there are matrices \( A, B, C \in \mathrm{SL}_2(\mathbb{Z}) \) such that \( A^4 + B^4 = C^4 \).

1. **Understanding the Problem:**
   - We need to find matrices \( A, B, C \) in \( \mathrm{SL}_2(\mathbb{Z}) \) such that \( A^4 + B^4 = C^4 \).

2. **Properties of Matrices in \( \mathrm{SL}_2(\mathbb{Z}) \):**
   - If \( A \in \mathrm{SL}_2(\mathbb{Z}) \), then \( \det(A) = 1 \).
   - The determinant of the fourth power of a matrix in \( \mathrm{SL}_2(\mathbb{Z}) \) is also 1, i.e., \( \det(A^4) = (\det(A))^4 = 1 \).

3. **Exploring the Equation \( A^4 + B^4 = C^4 \):**
   - Consider the determinants on both sides of the equation:
     \[
     \det(A^4 + B^4) = \det(C^4)
     \]
   - Since \( \det(C^4) = (\det(C))^4 = 1 \), we have:
     \[
     \det(A^4 + B^4) = 1
     \]
   - Again, the determinant of the sum of two matrices is not generally equal to the sum of their determinants. Therefore, we need to explore further.

4. **Counterexample Approach:**
   - Suppose \( A = I \) (the identity matrix), \( B = I \), and \( C = \sqrt[4]{2}I \). Clearly, \( A, B, C \in \mathrm{SL}_2(\mathbb{Z}) \) and:
     \[
     A^4 + B^4 = I + I = 2I \quad \text{and} \quad C^4 = (\sqrt[4]{2}I)^4 = 2I
     \]
   - This satisfies \( A^4 + B^4 = C^4 \), but \( \sqrt[4]{2}I \notin \mathrm{SL}_2(\mathbb{Z}) \) because \( \sqrt[4]{2} \) is not an integer.

5. **Conclusion:**
   - There are no matrices \( A, B, C \in \mathrm{SL}_2(\mathbb{Z}) \) such that \( A^4 + B^4 = C^4 \).

\[
\boxed{\text{False}}
\]