<s>[INST] Find two least composite numbers $n$ such that $n\mid2^n-2$ and $n\mid3^n-3$. 
$n\mid2^n-2$
  $n\mid3^n-3$. 

It comes down to 

$2^n \equiv 2 \mod n$
    $3^n \equiv 3 \mod n$

But how can I solve this?
 [/INST] As suggested in the comments, one clear path for the solution is to consider Carmichael numbers. The first two work: $561= 3 \cdot 11 \cdot 17$ and $1105= 5 \cdot 13 \cdot 17$.
The solutions $n \le 10000$ are
$$
\begin{align}
561&= 3 \cdot 11 \cdot 17 \\
1105&= 5 \cdot 13 \cdot 17 \\
1729&= 7 \cdot 13 \cdot 19 \\
2465&= 5 \cdot 17 \cdot 29 \\
2701&= 37 \cdot 73 \\
2821&= 7 \cdot 13 \cdot 31 \\
6601&= 7 \cdot 23 \cdot 41 \\
8911&= 7 \cdot 19 \cdot 67 \\
\end{align}
$$
Of these, $2701$ is not a Carmichael number.
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