Task 1. The sequence of positive integers $a_{0}, a_{1}, a_{2}, \ldots$ is defined by $a_{0}=3$ and

$$
a_{n+1}-a_{n}=n\left(a_{n}-1\right)
$$

for all $n \geq 0$. Determine all integers $m \geq 2$ for which $\operatorname{gcd}\left(m, a_{n}\right)=1$ for all $n \geq 0$.
Solution. A direct formula for the sequence is given by $a_{n}=2 \cdot n !+1$ for $n \geq 0$. (We use the common definition $0 !=1$, which satisfies $1 !=1 \cdot 0$ !, and for larger $n$ it holds that $n !=n \cdot(n-1) !$.$) We prove the direct formula by induction. It holds that$ $a_{0}=3$ and this is equal to $2 \cdot 0 !+1$. Suppose now that for some $k \geq 0$ it holds that $a_{k}=2 \cdot k !+1$, then

$$
a_{k+1}=a_{k}+k\left(a_{k}-1\right)=2 \cdot k !+1+k \cdot 2 \cdot k !=2 \cdot k ! \cdot(1+k)+1=2 \cdot(k+1) !+1 .
$$

This completes the induction.

We see that $a_{n}$ is always odd, so $\operatorname{gcd}\left(2, a_{n}\right)=1$ for all $n$. It follows that $\operatorname{gcd}\left(2^{i}, a_{n}\right)=1$ for all $i \geq 1$. Therefore, $m=2^{i}$ with $i \geq 1$ satisfies. Now take an $m \geq 2$ that is not a power of two. Then $m$ has an odd prime divisor, say $p$. We show that $p$ is a divisor of $a_{p-3}$. By Wilson's theorem, $(p-1) ! \equiv-1 \bmod p$. Thus,

$2 \cdot(p-3) ! \equiv 2 \cdot(p-1) ! \cdot((p-2)(p-1))^{-1} \equiv 2 \cdot-1 \cdot(-2 \cdot-1)^{-1} \equiv 2 \cdot-1 \cdot 2^{-1} \equiv-1 \quad \bmod p$.

So indeed, $a_{p-3}=2 \cdot(p-3) !+1 \equiv 0 \bmod p$. We conclude that $m$ does not satisfy. Therefore, the only values of $m$ that satisfy are the powers of two.