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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.

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In base $2$ the solutions are 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.

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 
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