Problem 1725

1. Initial Assumptions and Simplifications:
- We are given that $n$ is a positive integer with exactly 36 different prime divisors.
- We need to prove that $\sum_{1 \le i < j \le 5} (c_i - c_j)^2 \geq 2^{36}$.

2. **Understanding $c_n$:**
- $c_n$ is the number of integers that are coprime to $n$ in the interval $[ \frac{(k-1)n}{5}, \frac{kn}{5} ]$ for $k = 1, 2, 3, 4, 5$.

3. Using the Möbius Function:
- The number of integers coprime to $n$ and less than $\frac{an}{5}$ is given by:
$$\sum_{d|n} \mu(d) \left\lfloor \frac{an}{5d} \right\rfloor = \sum_{d|n} \mu(d) \frac{an}{5d} - \sum_{d|n} \mu(d) \left\{ \frac{an}{5d} \right\}$$
- Here, $\mu(d)$ is the Möbius function, and $\left\{ x \right\}$ denotes the fractional part of $x$.

4. **Expressing Differences $c_2 - c_1$ and $c_3 - c_1$:**
- We have:
$$c_2 - c_1 = \sum_{d|n} \mu(d) \left( 2 \left\{ \frac{n}{5d} \right\} - \left\{ \frac{2n}{5d} \right\} \right)$$
$$(c_2 - c_1) + (c_3 - c_1) = \sum_{d|n} \mu(d) \left( 3 \left\{ \frac{n}{5d} \right\} - \left\{ \frac{3n}{5d} \right\} \right)$$

5. **Introducing $v_i(n)$:**
- Define:
$$v_i(n) = \sum_{d|n, n/d \equiv i \mod 5} \mu(d)$$
- It follows that:
$$\sum_{i=0}^4 v_i(n) = 0 \quad \text{for any } n > 1$$

6. **Expressing $c_i - c_j$ in terms of $v_i(n)$:**
- We have:
$$c_2 - c_1 = v_3(n) + v_4(n)$$
$$(c_2 - c_1) + (c_3 - c_1) = v_2(n) + v_3(n) + 2v_4(n)$$
$$c_1 - c_2 = v_1(n) + v_2(n)$$
$$c_1 - c_3 = v_1(n) + v_3(n)$$
$$c_2 - c_3 = v_2(n) - v_3(n)$$

7. **Calculating $S(n)$:**
- We need to compute:
$$S(n) = \sum_{1 \le i < j \le 5} (c_i - c_j)^2$$
- Substituting the expressions for $c_i - c_j$:
$$S(n) = 4(v_1(n) + v_2(n))^2 + 2(v_1(n) + v_3(n))^2 + 2(v_2(n) - v_3(n))^2$$

8. **Considering the Case $5 \not| n$:**
- If $5 \not| n$, we analyze the remainders of the reciprocals of the prime divisors of $n$ modulo 5.
- Let $n = \prod_{i=1}^s p_i^{\alpha_i}$, and $r_i$ be the remainder of $p_i^{-1} \mod 5$.

9. Analyzing Different Cases:
- Case 1: $r_j = 1$ for some $j$:
- $v_i(n) = 0$ for all $i$, hence $S(n) = 0$.
- Case 2: $r_j = 2$ and $r_k = 3$ for some $j, k$:
- $S(n) = 4S(n')$.
- Case 3: $r_j = r_k = 4$ for some $j \neq k$:
- $S(n) = 16S(n')$.
- Case 4: $r_j = r_k = r_l = r_m = 2$ for some distinct $j, k, l, m$:
- $S(n) = 16S(n')$.
- Case 5: $r_j = r_k = r_l = r_m = 3$ for some distinct $j, k, l, m$:
- $S(n) = 16S(n')$.

10. Induction and Conclusion:
- Using induction on the number of prime divisors $s$:
- Base case: $s = 0$, $n = 1$, $S(n) = 6 \geq 2^1$.
- Inductive step: If $S(n') \geq 2^{s+1}$, then $S(n) \geq 2^{s+2}$.
- For $n$ with 36 prime divisors, $S(n) \geq 2^{36}$.

$\blacksquare$