Problem 1561

### Part (a)
1. Base Case:
For $n = 1$, the set $\{1\}$ contains only one number which is relatively prime to 1. Therefore, $S(1) = 1$. Hence,
$$2S(1) = 2 \times 1 = 2$$
Since 2 is not a perfect square, the statement holds for $n = 1$.

2. General Case:
For $n > 1$, consider the set $\{1, 2, \ldots, n\}$. If $a$ is relatively prime to $n$, then $n - a$ is also relatively prime to $n$. Therefore, the sum of all numbers in the set that are relatively prime to $n$ can be paired as $(a, n-a)$.

3. Sum Calculation:
The sum of each pair $(a, n-a)$ is $n$. If there are $\varphi(n)$ such pairs, where $\varphi(n)$ is Euler's totient function, then:
$$2S(n) = n \varphi(n)$$

4. Prime Factor Analysis:
Let $p$ be the largest prime that divides $n$ with exponent $k$. The exponent of $p$ in $n \varphi(n)$ is $k + (k-1) = 2k-1$, which is odd. Since a perfect square must have even exponents for all prime factors, $n \varphi(n)$ cannot be a perfect square.

5. Conclusion:
Therefore, $2S(n)$ is not a perfect square for any $n$.

$\blacksquare$

### Part (b)
1. Given:
We need to show that for given positive integers $m$ and $n$ with $n$ odd, the equation $2S(x) = y^n$ has at least one solution $(x, y)$ among positive integers such that $m \mid x$.

2. **Case $m = 2$:**
Let $n = 2k - 1$ (since $n$ is odd). Take $x = 2^k$. Then:
$$2S(2^k) = 2 \times \left( \sum_{1 \leq a \leq 2^k, \gcd(a, 2^k) = 1} a \right)$$
The numbers relatively prime to $2^k$ are the odd numbers from 1 to $2^k - 1$. The sum of the first $2^{k-1}$ odd numbers is $2^{k-1} \times 2^{k-1} = 2^{2k-2}$. Therefore:
$$2S(2^k) = 2 \times 2^{2k-2} = 2^{2k-1}$$
Since $2^{2k-1} = (2^{k-1})^n$ for $n = 2k - 1$, we have $y = 2^{k-1}$.

3. **General $m$:**
For any $m$, we can use a similar approach by considering $x$ as a multiple of $m$ and adjusting the powers accordingly. The key idea is to ensure that $x$ is chosen such that $2S(x)$ forms a perfect $n$-th power.

4. Conclusion:
Therefore, the equation $2S(x) = y^n$ has at least one solution $(x, y)$ among positive integers such that $m \mid x$.

$\blacksquare$