, 2024

We claim that the only integers that work are prime powers $p^k$ in which $p \equiv 5$ or $7$ modulo $8$. Let define $\omega, \Omega$ as function from the set of odd integers to $\{0; 1\}$ by
$$\omega(n) = \begin{cases} 0 & \text{if } n \equiv 1,3 \pmod 8 \\ 1 & \text{if } n \equiv 5,7 \pmod 8 \end{cases}$$
and
$$\Omega(n) \equiv |\{k\mid 0 < k < n/4, \gcd(k, n) = 1\}| \pmod 2.$$
Then we want to find $n$ such that $\Omega(n) = 1$. It is easy to verify
$$\omega(ab) \equiv \omega(a) + \omega(b) \pmod 2 \qquad (\dagger)$$
Let $n = p_1^{a_1} p_2^{a_2} \dots p_k^{a_k}$ which $p_1, p_2, \dots, p_k$ are distinct primes and $a_1, a_2, \dots, a_k$ are integers. Note that $n = 1$ does not work so consider $k \ge 1$.

Lemma: $\Omega(n) \equiv \omega(n) + \sum_{1 \le i \le k} \omega\left(\frac{n}{p_i}\right) + \sum_{1 \le i < j \le k} \omega\left(\frac{n}{p_i p_j}\right) + \dots + \omega\left(\frac{n}{p_1 p_2 \dots p_k}\right) \pmod 2$.

Proof: the idea to proof this lemma is similar to prove the Euler totient function. It is easy to check that $\omega(n)$ is indeed the parity of $|\{k\mid 0 < k < n/4\}| = \lfloor \frac{n}{4} \rfloor$. To remove numbers divisible by some $p$, one can subtract from above the value
$$|\{pk\mid 0 < k < n/(4p)\}| = |\{k\mid 0 < k < (n/p)/4\}| = \omega\left(\frac{n}{p}\right).$$
But that removes numbers divisible by $pq$ (for $p \ne q$) twice so we will add $\omega(pq)$ then continue as the principle of inclusion and exclusion, we add until $\omega\left(\frac{n}{p_1 p_2 \dots p_k}\right)$. Note that the formula should be added and subtracted alternatively but since we consider modulo $2$ so all of the signs can be consider as plus. Back to the original problem, by applying the lemma and $(\dagger)$, one can get
$$\begin{aligned} & \Omega(n) + \sum_{1 \le i \le k} \omega(p_i) + \sum_{1 \le i < j \le k} \omega(p_i p_j) + \dots + \omega(p_1 p_2 \dots p_k) \\ & \equiv \omega(n) + \left( \sum_{1 \le i \le k} \omega\left(\frac{n}{p_i}\right) + \sum_{1 \le i \le k} \omega(p_i) \right) + \dots + \left( \omega\left(\frac{n}{p_1 p_2 \dots p_k}\right) + \omega(p_1 p_2 \dots p_k) \right) \\ & \equiv \binom{k}{0} \omega(n) + \binom{k}{1} \omega(n) + \binom{k}{2} \omega(n) + \dots + \binom{k}{k} \omega(n) \\ & \equiv 2^k \omega(n) \equiv 0 \pmod 2. \end{aligned}$$
From this, one can conclude that
$$\begin{aligned} \Omega(n) &\equiv \omega(1) + \sum_{1 \le i \le k} \omega(p_i) + \sum_{1 \le i < j \le k} \omega(p_i p_j) + \dots + \omega(p_1 p_2 \dots p_k) \\ &= \sum_{i=1}^{k} \left( \binom{k-1}{0} + \binom{k-1}{1} + \dots + \binom{k-1}{k-2} \right) \omega(p_i) \\ &= 2^{k-1} \sum_{i=1}^{n} \omega(p_i) \\ &= 2^{k-1} \omega(p_1 p_2 \dots p_k) \pmod 2. \end{aligned}$$
Thus if $k \ge 2$ then $\Omega(n) = 0$ and if $k = 1$, then $n = p^k$, in this case we can check that $\Omega(n) = \omega(p) = 1$ if and only if $p \equiv 5$ or $7$ modulo $8$. $\square$