Cute. Suppose not. Let $n = 4k$ for convenience and consider the largest even $d$ for which $d$ and $d+2$ both divide $n$. (This exists as $d = 2$ works.) Then $d+1$ also divides $n$. If $d \equiv 0 \pmod{4}$ then $2d + 2$ and $2d + 4$ both divide $n$. If $d \equiv 2 \pmod{4}$ then $2d$ and $2d + 2$ both divide $n$. In both cases, we find a larger pair of consecutive evens dividing $n$, contradiction.

What is IOM 2018?

pavel kozlov wrote: What is IOM 2018? International Olympiad of Metropolises: http://megapolis.educom.ru/en

a really cute one suppose the contrary Consider the following sequence of integer $a_1=3$ and $a_{n+1}=2a_n -1$ Claim: $a_n | 4k \forall n >0$ Proof: $2|4k ,4|4k \implies 3|4k$ So $6|4k ,4|4k \implies 5|4k$ now I'll use induction on $n$ suppose $a_n,a_{n+1} $ divide $4k$ so $4a_n$ and $2a_{n+1}=4a_n+2$ both divide $4k$ so $4a_n+1=2a_{n+1}-1=a_{n+2}$ divides $4k$ $\blacksquare$ since $a_{n\ge 1}$ is strictly increasing there's some $j$ such that $a_j >4k$ but $a_j |4k$ contradiction and we win

Cant we just take 4 and 2?

Mikeglicker wrote: Cant we just take 4 and 2? what if $3|k$

Solved with Jeffrey Chen, Max Lu, and Raymond Feng int a = 2; while (4*k % (a+1) == 0) { if (a%4 == 2) a = 2*a; else a = 2*a+2; } cout << a << " " << a+2 << endl;

@Ali3085 Then take 3 and 1

@above $3$ and $1$ are not consecutive divisors. $4k$ must be divisible by $2$.