What is the precise description of such positive integers $n$: for any integer $a, a^{n+1} \equiv a \pmod n$?

Question

Here's what I've managed to get till now(of course, we only have to consider $a$ from $0$ to $n-1$):

If $n \ne 2, n$ isn't prime. Otherwise we'll have $2^n \equiv 1 \pmod n$, a contradiction.
Do prime factorization of $n$. Then the power of each prime factor of $n$ is $1$. Otherwise, we can consider $a=p_1p_2...p_k$, where $p_i$ are the prime factors of $n$. Then we would definitely have $a \mid n$, another contradiction.
Besides, $n$ has $2$ as its factor. Otherwise, we would have $(-1)^{n+1} \equiv 1 \pmod n$.

All above arguments has excluded the case $n=2$, where $n$ satisfies the statement.

That's all i've got. How can I arrange further progress? The Euler $\phi(n) = (p_1-1)(p_2-1)...(p_k-1)$, where $p_i$ are $n$'s prime factors. How can I link it with $n$ itself, and use some results like Fermat's little Theorem?

Update: Found on this page http://www-groups.mcs.st-andrews.ac.uk/~john/Zagier/Problems.html.

Please use MathJax in questions to make them more easily readable! That said... I'm not sure what your question is here. Are you looking for the kind/class of numbers that can be $n$? Are you looking for proof that such numbers exist? It's really unclear. The exponent of $n+1$ makes for a much different question from FLT; I'm not sure if they can be related (or if this class of numbers exists!) — Eric Snyder, Sep 21 '22 at 10:12
@EricSnyder Thank for editing it for me! I would learn to give a cleaner question (in both sense) next time XD. What i'm looking for is a clearer description of such kind of numbers. Clearly 2 satisfies the statement: 0^3 = 0 and 1^3 =1. Another example is 6=2*3, which can be verified. Actually I guess that all number satisfying 2. above is one such number. So maybe I just need a proof of such thought; maybe it's wrong. But I'm sure there's a brief description of them, since this is left as an exercise afterclass. Here FLT may be irrelated, but the euler generalization of it should help. — HStc, Sep 21 '22 at 10:33
See also Classifying all numbers $n$ with the property that if $p$ is a prime, then $p \mid n \iff p-1 \mid n$ and Compute a natural number $n\geq 2$ s.t. $p\mid n \Longrightarrow p^2\nmid n$ AND $p-1\mid n \Longleftrightarrow p\mid n$ for all prime divisor p of n. and Sequence given by $a_0=2$, $a_{n+1} = a_n \cdot(a_n+1)$ not prime for all $n$ — Bill Dubuque, Sep 21 '22 at 11:47

What is the precise description of such positive integers $n$: for any integer $a, a^{n+1} \equiv a \pmod n$?

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