Let $n$ be a positive integer. A German set in an $n \times n$ square grid is a set of $n$ cells which contains exactly one cell in each row and column. Given a labelling of thecells with the integers from $1$ to $n^2$ using each integer exactly once, we say that an integer is a German product if it is the product of the labels of the cells in a German set.

(a) Let $n=8$. Determine whether there exists a labelling of an $8 \times 8$ grid such that the following condition is fulfilled: The difference of any two German products is alwaysdivisible by $65$.

(b) Let $n=10$. Determine whether there exists a labelling of a $10 \times 10$ grid such that the following condition is fulfilled: The difference of any two German products is always divisible by $101$. 
### Part (a)

1. **Understanding the Problem:**
   We need to determine if there exists a labelling of an $8 \times 8$ grid such that the difference of any two German products is always divisible by $65$.

2. **Prime Factorization:**
   \[
   65 = 5 \times 13
   \]
   We need to consider the divisibility by both $5$ and $13$.

3. **Counting Multiples:**
   In the set $\{1, 2, \ldots, 64\}$:
   - There are $\left\lfloor \frac{64}{5} \right\rfloor = 12$ integers divisible by $5$.
   - There are $\left\lfloor \frac{64}{13} \right\rfloor = 4$ integers divisible by $13$.

4. **Existence of German Set:**
   Since there are only $4$ integers divisible by $13$, it is impossible to form a German set (which requires $8$ integers) where all integers are divisible by $13$. Therefore, there must be at least one German set whose product is not divisible by $13$.

5. **Conclusion:**
   Since we cannot ensure that all German products are divisible by $65$, the condition cannot be fulfilled.

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

### Part (b)

1. **Understanding the Problem:**
   We need to determine if there exists a labelling of a $10 \times 10$ grid such that the difference of any two German products is always divisible by $101$.

2. **Prime Number Property:**
   Since $101$ is a prime number, there exists a primitive root $g \pmod{101}$.

3. **Labelling the Grid:**
   We can label the grid using powers of the primitive root $g$. Specifically, we label the cell in the $a$-th row and $b$-th column with $g^{10a + b} \pmod{101}$ for $0 \leq a, b \leq 9$.

4. **Covering All Residue Classes:**
   Since $g$ is a primitive root modulo $101$, the powers $g^0, g^1, \ldots, g^{99}$ cover all $100$ residue classes in $(\mathbb{Z}/101\mathbb{Z})^*$.

5. **Product of German Set:**
   The product of any German set will be:
   \[
   g^{(0+1+\ldots+9)(10a+b)} = g^{45(10a+b)}
   \]
   Since $45 \equiv 0 \pmod{100}$, the product of any German set is congruent to $1 \pmod{101}$.

6. **Difference of German Products:**
   The difference of any two German products will be:
   \[
   g^{45(10a+b)} - g^{45(10c+d)} \equiv 1 - 1 \equiv 0 \pmod{101}
   \]
   Thus, the difference is always divisible by $101$.

\[
\boxed{\text{True}}
\]