This problem raised in my mind while solving an elementary number theory problem.
Problem statement: For $n\in \mathbb{N}$ let $\tau(n)$ notes the number of positive divisors of $n$ including $1$ and $n$. Find all $n\in \mathbb{N}$ such that, $n\equiv1 \pmod{\tau(n)}$.
Partial progress: We know by Fermat's little theorem, $a^{p-1}\equiv1\pmod{p}$, where $a$ is a natural number, $p$ is a prime and $\mathrm{gcd}(a,p)=1$. If we take $n=p^{q-1}$ for two different primes $p,q$, then $\tau(n)=\tau(p^{q-1})=(q-1+1)=q$. So by Fermat's little theorem we get that $n\equiv1\pmod{\tau(n)}$. In fact let $n=p_1^{(q_1-1)}p_2^{(q_2-1)}\ldots p_k^{(q_k-1)}$, where $p_i$'s and $q_i$'s are distinct primes and $\mathrm{gcd}(p_i,q_i)=1$ $\forall\;i\in\{1,2,\ldots,k\}$. Then $p_i^{(q_i-1)}\equiv\pmod{q_i}$ as before for all $i\in\{1,2,\ldots,k\}$. Now $\tau(n)=(q_1-1+1)(q_2-1+1)\ldots(q_k-1+1)=q_1q_2\ldots q_k$. Since $\mathrm{gcd}(q_i,q_j)=1$ $\forall\;1\leq i\neq j\leq k$, therefore, $n=p_1^{(q_1-1)}p_2^{(q_2-1)}\ldots p_k^{(q_k-1)}\equiv1\pmod{q_1q_2\ldots q_k}$ or $n\equiv1\pmod{\tau(n)}$.
Main question: Are these numbers are the only solutions? If not, what is the set of solutions to the congruence $n\equiv1\pmod{\tau(n)}$?. Can we find the whole set parametrically?
Thanks in anticipation!