ATGY wrote: If $n$ is prime, we have $n - 1 = 1 \implies n = 2$, which works. So assume $n$ is not prime. Observe that $n = d_k = 1 + (1 + ... + 2k - 3) = (k - 1)^2 = k^2 - 2k$. We prove that $v_2(n) = 1$ by proving that all the even divisors of $n$ have a $v_2$ of $1$. Notice that the divisors with even indices (aka $d_{2m}$) form up the even divisors while the divisors with odd indices form up the odd divisors. Notice that $d_2 = 2$, so say even divisors with indices $\{2, \dots, 2m\}$ have a $v_2$ of $1$. We have $d_{2m + 2} = d_{2m} + (2x + 1) + (2x + 3) = d_{2m} + (4x + 4)$. Notice that $v_2(4x + 4) = 2$, and $v_2(d_{2m} + (4x + 4)) = \min(v_2(d_{2m}), v_2(4x + 4)) = 1$ by our inductive hypothesis, which means all even divisors $n$ have a $v_2$ of 1. Since $2 \mid k^2 - 2k$, $2 \mid k^2$, which means $k$ is even and $4 \mid k^2$, $4 \mid 2k$, which means $4 \mid k^2 - 2k = n$. However, this means $v_2(n) \geq 2$, which is a contradiction. So our only solution is $\boxed{n = 2}$. You missed "in some order" Also, $1+(1+\ldots+2k-3)=(k-1)^2+1=k^2-2k+2$

nAalniaOMliO wrote: Let $1=d_1<d_2<\ldots<d_k=n$ be all divisors of $n$. It turned out that numbers $d_2-d_1,\ldots,d_k-d_{k-1}$ are $1,3,\ldots,2k-3$ in some order. Find all possible values of $n$ Nice! As above, sum $d_t-d_{t-1}$ to get $n=1+(k-1)^2$. Note that $n$ is even, since for some $i$ difference $d_{i+1}-d_{i}=1$, so $d_id_{i+1}$ is even and so $n$. By this reason, $d_2=2$. Now we can see that difference $n/2=n-n/2=d_k-d_{k-1}$ is maximal of all differences $d_{i}-d_{i-1}$, because $d_i-d_{i-1}\leqslant n/2-d_{i-1}<n/2$ for $i<k$. Thus we must have $2k-3=d_k-d_{k-1}=n/2=(1+(k-1) ^2) /2$ and so $k=2, 4$ and $n=2, 10$. They both work. Answer: $n=2$ and $n=10$.