Peter wrote: A natural number $n$ is said to have the property $P$, if whenever $n$ divides $a^{n}-1$ for some integer $a$, $n^{2}$ also necessarily divides $a^{n}-1$. Show that every prime number $n$ has the property $P$. Show that there are infinitely many composite numbers $n$ that possess the property $P$. (a) We have that $n=p$ is a prime number. From Fermat's theorem $p|a^{p-1}-1$ and $p|a^{p}-1$, then $p|a^{\gcd(p;p-1)}-1$ so $p|a-1$. We got that $a=pk+1$ for some $k$ then \[a^{n}-1=(pk+1)^{n}-1 \equiv C_{n}^{p-1}(pk)=p^{2}k\equiv 0(mod p^{2})\] (b) Let us prove that the number $n=pq$ possess the property $P$, for every different primes $p$ and $q$. Indeed, we have that if $pq|a^{pq}-1$ then $p|(a^{q})^{p}-1$ thus from (a) follows that $p^{2}|a^{pq}-1$ and similarly we can prove that $q^{2}|a^{pq}-1$. Therefore ,whenever $pq|a^{pq}-1$ then $(pq)^{2}|a^{pq}-1$.

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