A Poulet-number is a Fermat pseudoprime to base $2$ that is it is a composite number $N$ such that $$2^{N-1}\equiv 1\mod N$$ holds.
Let $m$ be a positive integer and consider $f(m)=(2m^2+1)(8m^2+1)$
If $f(m)$ is a Poulet-number , must at least one of $2m^2+1$ and $8m^2+1$ be prime ?
Numerical analysis :
- For $1\le m\le 1.9\cdot 10^9$ , a Poulet-number is only possible , if at least one of the factors is prime.
- If $30\mid m$ , this is even true upto $40\cdot 10^9$.
Cases with composite $2m^2+1$ :
gp > forstep(m=3,2*10^9,3,a=2*m^2+1;b=8*m^2+1;if(Mod(2,a)^(a*b-1)==1,if(ispseudoprime(a)==0,if(Mod(2,b)^(a*b-1)==1,print(m)))))
7332
315171
3092400
27489840
517041000
gp >
Cases with composite $8m^2+1$
gp > forstep(m=3,2*10^9,3,a=2*m^2+1;b=8*m^2+1;if(Mod(2,b)^(a*b-1)==1,if(ispseudoprime(b)==0,if(Mod(2,a)^(a*b-1)==1,print(m)))))
6264
2498430
1907691930
gp >
We must have $3\mid m$ for a Poulet-number , otherwise both factors are divisible by $3$ making a Poulet-number impossible. $f(m)$ is a Poulet-number iff $f(m)$ is squarefree and for every prime $p\mid f(m)$ , the order of $p$ with respect to the base $2$ divides $(2m^2+1)(8m^2+1)-1=2m^2(8m^2+5)$
I could not find further useful conditions.
Any idea or reference ?
