Baltic Way 2023 Shortlist

Answer: $1$, $2$.

Let $n=p_1^{{\alpha }_1}\cdots p_k^{{\alpha }_k}$ be the canonical representation of $n$, where ${\alpha }_1,\dots ,{\alpha }_k$ are positive integers. It is known that $$\delta (n)=({\alpha }_1+1)\dots ({\alpha }_k+1)$$ and $$\varphi (n)=(p_1-1)\dots (p_k-1)\cdot p_1^{{\alpha }_1-1}\dots p_k^{{\alpha }_k-1}.$$

Hence, the given equation reduces to $$({\alpha }_1+1)\dots ({\alpha }_k+1)=\frac{p_1\dots p_k}{(p_1-1)\dots (p_k-1)}.(\ast )$$ Note that ${\alpha }_i+1\ge 2\ge 1+\frac{1}{p_i-1}=\frac{p_i}{p_i-1}$, for every $i=1,\dots ,k$, whereby equalities hold if and only if ${\alpha }_i=1$ and $p_i=2$. Multiplying the inequalities ${\alpha }_i+1\ge \frac{p_i}{p_{i-1}}$, for $i=1,\dots ,k$, leads to the inequality obtained from $(\ast )$ by replacing equality with inequality. Hence equality $(\ast )$ holds if and only if every inequality that was multiplied holds as an equality. I.e., if ${\alpha }_i=1$ and $p_i=2$, for every $i=1,\dots ,k$. As primes in the canonical representation are distinct, this holds if and only if $k=0$ or $k=1$ and $p_1=2$, i.e., if $n=1$ or $n=2$.