Find all integers $n=2k+1>1$ so that there exists a permutation $a_0, a_1,\ldots,a_{k}$ of $0, 1, \ldots, k$ such that 
\[a_1^2-a_0^2\equiv a_2^2-a_1^2\equiv \cdots\equiv a_{k}^2-a_{k-1}^2\pmod n.\]

[i]Proposed by usjl[/i]
1. **Identify the problem and given conditions:**
   We need to find all odd integers \( n = 2k + 1 > 1 \) such that there exists a permutation \( a_0, a_1, \ldots, a_k \) of \( 0, 1, \ldots, k \) satisfying:
   \[
   a_1^2 - a_0^2 \equiv a_2^2 - a_1^2 \equiv \cdots \equiv a_k^2 - a_{k-1}^2 \pmod{n}.
   \]
   Let \( d \) be the common difference, i.e., \( a_{i+1}^2 - a_i^2 \equiv d \pmod{n} \) for all \( i \).

2. **Check for small values of \( n \):**
   - For \( n = 3 \):
     \[
     k = \frac{3-1}{2} = 1.
     \]
     The permutation \( a_0, a_1 \) of \( 0, 1 \) can be \( 0, 1 \). We have:
     \[
     a_1^2 - a_0^2 = 1^2 - 0^2 = 1 \equiv 1 \pmod{3}.
     \]
     Thus, \( n = 3 \) works.

   - For \( n = 5 \):
     \[
     k = \frac{5-1}{2} = 2.
     \]
     The permutation \( a_0, a_1, a_2 \) of \( 0, 1, 2 \) can be \( 0, 1, 2 \). We have:
     \[
     a_1^2 - a_0^2 = 1^2 - 0^2 = 1,
     \]
     \[
     a_2^2 - a_1^2 = 2^2 - 1^2 = 3 \equiv 3 \pmod{5}.
     \]
     Thus, \( n = 5 \) works.

3. **Prove that \( n \neq 3, 5 \) does not work:**
   - Assume \( n \) is a composite number. Let \( p \) be a prime divisor of \( n \). The sequence \( a_i^2 \) modulo \( p \) is periodic with period \( \frac{p+1}{2} \). Since \( k < \frac{p+1}{2} \), the sequence must repeat, implying \( d \equiv 0 \pmod{p} \), which contradicts the periodicity.

4. **Consider \( n = p \) where \( p \) is a prime:**
   - We count the quantity \( 2\sum_{i=0}^{k-1} \left( \frac{a_i^2 + d}{p} \right) \) using the Legendre symbol \( \left( \frac{\bullet}{p} \right) \):
     \[
     2\sum_{i=0}^{k-1} \left( \frac{a_i^2 + d}{p} \right) = 2\sum_{i=1}^{k} \left( \frac{a_i^2}{p} \right) = \begin{cases} 
     2k & \text{if } a_0 = 0, \\
     2k - 2 & \text{if } a_0 \neq 0 
     \end{cases} \equiv -1, -3 \pmod{p}.
     \]
   - On the other hand:
     \[
     2\sum_{i=0}^{k-1} \left( \frac{a_i^2 + d}{p} \right) = 2\sum_{i=0}^{k} \left( \frac{a_i^2 + d}{p} \right) - 2\left( \frac{a_k^2 + d}{p} \right).
     \]
     Using properties of the Legendre symbol and sums of powers, we get:
     \[
     \sum_{i=0}^{p-1} \left( \frac{i^2 + d}{p} \right) + \left( \frac{d}{p} \right) - 2\left( \frac{a_k^2 + d}{p} \right) \equiv -4, -2, 0, 2 \pmod{p}.
     \]
   - This implies \( p = 3 \) or \( p = 5 \).

Thus, the only possible values for \( n \) are \( 3 \) and \( 5 \).

The final answer is \( \boxed{ n = 3, 5 } \).