IMO Team Selection Test 1

The sequence is given by the formula $a_n=2\cdot n!+1$ for $n\ge 0$. (We use the usual definition $0!=1$, which satisfies $1!=1\cdot 0!$, in the same way we have $n!=n\cdot (n-1)!$ for other positive integers $n$.) We will prove the equality by induction. We have $a_0=3$, which equals $2\cdot 0!+1$. Now suppose for certain $k\ge 0$ that $a_k=2\cdot k!+1$, then $$a_{k+1}=a_k+k(a_k-1)=2\cdot k!+1+k\cdot 2\cdot k!=2\cdot k!\cdot (1+k)+1=2\cdot (k+1)!+1.$$ This finishes the induction.

We see that $a_n$ is always odd, hence $\gcd (2,a_n)=1$ for all $n$. It follows also that $\gcd (2^i,a_n)=1$ for all $i\ge 1$. Hence, $m=2^i$ with $i\ge 1$ satisfies the condition. Now consider an $m\ge 2$ which is not a power of two. Then $m$ has an odd prime divisor, say $p$. We will show that $p$ is a divisor of $a_{p-3}$. By Wilson's theorem, we have $(p-1)!\equiv -1(\mathrm{mod}p)$. Hence, $$\begin{array}{rl}2\cdot (p-3)!& \equiv 2\cdot (p-1)!\cdot ((p-2)(p-1){)}^{-1}\\ & \equiv 2\cdot -1\cdot (-2\cdot -1{)}^{-1}\equiv 2\cdot -1\cdot 2^{-1}\equiv -1(\mathrm{mod}p).\end{array}$$ So indeed we have $a_{p-3}=2\cdot (p-3)!+1\equiv 0(\mathrm{mod}p)$. We conclude that $m$ does not satisfy the condition. Hence, the only values of $m$ satisfying the condition are powers of two. $\square$