Carmichael number divisible by a given positive integer 
Prove or disprove this conjecture : If $k>2$ is an integer with $\gcd(k,\varphi(k))=1$ , then there is a Carmichael number $N$ with $k\mid N$

The condition $\gcd(k,\varphi(k))=1$ is necessary. Otherwise, $k$ would not be squarefree and therefore also $N$ , which is impossible. Or there would be prime factors $p,q$ of $k$ and therefore $N$, with $p\mid q-1$. This is also impossible since this would imply $p\mid N-1$ because of $q-1\mid N-1$ contradicting $p\mid N$
But is it also sufficient ? And if yes, is there an efficient method to construct a Carmichael number divisible by $k$ ? A particular hard case seems to be $k=885=3\cdot 5\cdot 59$.

Is there a Carmichael number divisble by $885$ ?

 A: Your conjecture seems to be true. Here are some ideas i had (this was a bit too long for a comment)
Firts of all, by Bezout's lemma, as $\gcd(k,\varphi(k))=1$, then there exist $x,y$ such that $\forall z$
$$k\big(x+\varphi(k)\cdot z\big)=\varphi(k)\big(y+k\cdot z\big)+1$$
I will prove that there exists a $z$ such that $k\big(x+\varphi(k)\cdot z\big)$ is a Carmichael number. For this to happen, $\forall n$, $\gcd\big(n;k\big(x+\varphi(k)\cdot z\big)\big)=1$ we have
$$n^{k\big(x+\varphi(k)\cdot z\big)-1}=n^{\varphi(k)\big(y+k\cdot z\big)}\equiv 1\pmod{k\big(x+\varphi(k)\cdot z\big)}$$
Because $n$ and $k$ are coprime, $$n^{\varphi(k)\big(y+k\cdot z\big)}\equiv 1\pmod{k}$$.
So we must have $$n^{\varphi(k)\big(y+k\cdot z\big)}\equiv 1\pmod{x+\varphi(k)\cdot z}$$
From $$k\big(x+\varphi(k)\cdot z\big)=\varphi(k)\big(y+k\cdot z\big)+1$$  we can observe that $x$ and $\phi(k)$ are coprime and $kx\equiv1\pmod{\varphi(k)}$
From here I had some ideas of using Dirichlet's theorem, to make $x+\varphi(k)\cdot z$ a prime or to make $x+\varphi(k)\cdot z=k\cdot (x^2+l\cdot \varphi(k))$, but both approaches failed.
If my initial idea doesn't do the job, I think that some other application of Bezout's lemma finishes it.
A: The smallest Carmichael number that is a multiple of 885, is:
3399464645479365585

Below 2^64, there are only two more Carmichael numbers that are divisible by 885:
15165761619468172065
15885147377097813585

It's an open-problem whether every odd cyclic number has at least one Carmichael multiple.
See also:

*

*https://oeis.org/A253595

*https://www.numericana.com/data/crump.htm
Several methods for constructing Carmichael numbers (including source-code implementations), are listed at:

*

*https://trizenx.blogspot.com/2020/08/pseudoprimes-construction-methods-and.html
