
For every integer $n	\ge 2$ let $B_n$ denote the set of all binary $n$-nuples of zeroes and ones, and split $B_n$ into equivalence classes by letting two $n$-nuples be equivalent if one is obtained from the another by a cyclic permutation.(for example 110, 011 and 101 are equivalent). Determine the integers $n 	\ge 2$ for which $B_n$ splits into an odd number of equivalence classes.
1. **Define the problem and notation:**
   For every integer \( n \ge 2 \), let \( B_n \) denote the set of all binary \( n \)-tuples of zeroes and ones. We split \( B_n \) into equivalence classes by letting two \( n \)-tuples be equivalent if one is obtained from the other by a cyclic permutation. We need to determine the integers \( n \ge 2 \) for which \( B_n \) splits into an odd number of equivalence classes.

2. **Apply Burnside's Lemma:**
   By Burnside's Lemma, the number of orbits (equivalence classes) of \( B_n \) under the action of cyclic permutations is given by:
   \[
   \frac{1}{n} \sum_{i=0}^{n-1} 2^{\gcd(i, n)}
   \]
   This can be rewritten using the properties of the Euler's totient function \( \varphi \):
   \[
   \frac{1}{n} \sum_{d \mid n} \varphi\left(\frac{n}{d}\right) 2^d
   \]
   Let \( S = \sum_{d \mid n} \varphi\left(\frac{n}{d}\right) 2^d \). Then the number of orbits is \( \frac{S}{n} \).

3. **Analyze the valuation of \( S \):**
   We need to determine when \( \frac{S}{n} \) is an odd number. This requires \( S \) to be divisible by \( n \) but not by \( 2n \). We claim that:
   \[
   v_2(S) \geq v_2(\varphi(n)) + 1
   \]
   where \( v_2(x) \) denotes the 2-adic valuation of \( x \), i.e., the highest power of 2 dividing \( x \).

4. **Prove the claim:**
   It suffices to check that:
   \[
   v_2\left(\varphi\left(\frac{n}{d}\right)\right) + d \geq v_2(\varphi(n)) + 1
   \]
   for all \( d > 1 \). This is because \( d \) should increase more than \( v_2\left(\varphi\left(\frac{n}{d}\right)\right) \) decreases. Consider the bound:
   \[
   v_2\left(\varphi\left(\frac{n}{d}\right)\right) + v_2(\varphi(d)) = v_2\left(\varphi\left(\frac{n}{d}\right) \varphi(d)\right) \geq v_2(\varphi(n)) - 1
   \]
   This follows by observing individual prime powers dividing \( \frac{n}{d} \) and \( d \). Then we have:
   \[
   v_2\left(\varphi\left(\frac{n}{d}\right)\right) + d \geq v_2(\varphi(n)) + 1 + d - v_2(\varphi(d)) - 2 \geq v_2(\varphi(n)) + 1
   \]
   because \( d - v_2(\varphi(d)) - 2 = 0 \) for \( d = 2, 3 \) and \( d \geq \log_2(d) + 2 \geq v_2(\varphi(d)) + 2 \) for \( d \geq 4 \).

5. **Implication of the claim:**
   This claim implies:
   \[
   v_2(S) \geq v_2(\varphi(n)) + 1 \geq v_2(n)
   \]
   This is because \( \varphi \) can only knock off one 2 from \( n \), but every odd prime dividing \( n \) contributes to \( v_2(\varphi(n)) \). In particular, if the inequality is strict, then the number of orbits \( \frac{S}{n} \) is even. Thus, we must have equality in both inequalities, i.e., \( n \) is a power of 2 and \( n \in \{2, 3, 4\} \).

6. **Manual verification:**
   Manually checking, \( B_2 \) has 3 orbits, and \( n = 2 \) is the only \( n \) that works.

The final answer is \( \boxed{ n = 2 } \).