I'm working in the following Wilson's theorem excercise:
Show that each number of $100!+1,100!+2,100!+3,…,100!+100$ is composite number.
I'm starting from:
$$100! \equiv -1 \pmod {101}$$
My thoughts about this is that I may be able to add the required number each time to that congruence, so for the first case I would have:
$$100! +1\equiv (-1)(+1) \pmod {101}$$
$$100! +1\equiv 0 \pmod {101}$$
Showing that $101k=100!+1$ so $100!+1$ is a composite number and I would basically do the same for each number.
Is that correct? Am I missing something? Any hint, help or correction will be really appreciated.
Best Answer
$101$ is prime so by Wilson's theorem $100!\equiv -1\pmod {101}$ so $100!+1\equiv 0\pmod {101} $ so $101|100!+1$ .
For $1 <k\le 100$ we simply note the $k|100! $ (by definition) and $k|k $ so $k|100!+k $.
Note $N! +1$ might be prime for some $N $ but not if $N+1$ is prime. $3!+1=7$ is prime (but $3+1$ isnt).