Hint : if k comprime with n we have n-k also And that kill the problem

Given that if \((x, n) = 1\), then \((x, n - x) = 1\) as well. Let's list the numbers relatively prime to \(n\) in such pairs. Note that their sum is always \(n\), and there are \(\frac{\phi(n)}{2}\) such pairs, so \(\beta(n) = \frac{\frac{n \cdot \phi(n)}{2}}{\phi(n)} = \frac{n}{2}\). Now, observe that \(\alpha(n) = \frac{\sigma(n)}{d(n)}\). List the divisors of \(n\) in pairs of \(\left(k, \frac{n}{k}\right)\). Notice that as \(k\) increases, \(k + \frac{n}{k}\) gets smaller. If \(n\) is prime, we have \(\alpha(n) = \frac{n + 1}{2}\), \(\beta(n) = \frac{n}{2}\), which is clearly impossible. When \(n = p^2\), where \(p\) is prime, we have \(\beta(n) = \frac{p^2}{2}\), \(\alpha(n) = \frac{1 + p + p^2}{3}\). If \(\alpha(n) = \beta(n)\), \(2p^2 - 2p + 2 = 0\) which yields no integer solutions. So \(n\) has at least 4 divisors. \(\sigma(n) \leq \frac{d(n)}{4} \cdot \left(n + 1 + 2 + \frac{n}{2}\right) = \frac{d(n)}{4} \cdot \left(\frac{3}{2}n + 3\right)\) (here we take the cases when \(n = 1, 2\) which gives the largest value). \(\implies \frac{\sigma(n)}{d(n)} \leq \frac{3}{8}n + \frac{3}{4}\) Since \(\frac{\sigma(n)}{d(n)} = \frac{n}{2}\), \(\frac{3}{8}n + \frac{3}{4} \geq \frac{n}{2} \implies n \leq 6\). Testing \(n = 1, 2, 3, 4, 5, 6\) gives \(n = 6\) and $n = 1$ as the only solutions.

Doesn't $n=1$ work as well?