One can easily show that if $p$ is a prime that does not divide $a$, then $a^{p(p-1)}\equiv 1 \pmod{p^2}$.


However, my question is: If instead of $p$ being a prime, it were a pseudoprime to the base $a$, would the result still be valid? Moreover, in general, for what (if any) composite numbers $p$ would the above result be valid?


My interest in this question stems from a study of the extended Lucas sequences (with sequences of the form $a^{n} - 1$ being special cases of them.)


Thank you.



On composite divisors of certain terms in the extended Lucas sequences
In base $2$ the solutions are [A306259](https://oeis.org/A306259):


Composite numbers k such that $2^{(k(k-1))} \equiv 1 \pmod{ k^2}$


From a comment: It contains all Fermat pseudoprimes to base 2,
[A001567](https://oeis.org/A001567).


The sequence starts:



```
21, 105, 165, 205, 231, 273, 301, 341, 385, 465

```

There are many solutions in other bases, the following
pari/gp program computes them:



```
c=0;a=2;for(p=1,10^4,if(isprime(p),next);b=Mod(a,p^2)^((p-1)*p);if(b==1,print1(p,",");c++ ));c 

```

