We know by Fermat’s little theorem that if $m$ is a prime then for any $n \in \mathbb{Z}$, then $$ n^m \equiv n (mod \ m). $$ I was wondering if there was a composite $m$ that satisfies this condition? (Maybe this condition is equivalent to being a Carmichael number, but I wasn’t sure…) Thank […]

Let $n$ be an odd integer, we say that $n$ is $a$-pseudoprime if $gcd(a,n)=1$ and : $$\begin{pmatrix}\frac{a}{n}\end{pmatrix}=a^{\frac{n-1}{2}}\text{ mod } n $$ Euler’s criterion states that if $n$ is not prime, then at most half of the elements $a$ in $\frac{\mathbb{Z}}{n\mathbb{Z}}^*$ are such that $n$ is $a$-pseudoprime. Now I would like to know if we can […]

Can a Mersenne number ever be a Carmichael number? More specifically, can a composite number $m$ of the form $2^n-1$ ever pass the test: $a^{m-1} \equiv 1 \mod m$ for all intergers $a >1$ (Fermat’s Test)? Cases potentially proved so far: (That are never Carmichael numbers) where $n$ is odd where $n$ is prime Work […]

When a number $n$ does not have primitive roots modulo n, $Pr(n)$, it is possible to generate the set $M$ of those numbers $m$ whose order $ord_n(m)$ is the maximum multiplicative order of $k$ in $\Bbb Z/n \Bbb Z$, also named as the maximum possible order mod n depending on the reference you use. The […]

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