2019 IMO Problems/Problem 4

Problem

Find all pairs $(k,n)$ of positive integers such that

$k!=(2^n-1)(2^n-2)(2^n-4)\dots(2^n-2^{n-1}).$

$LHS$ $k$ ! = 1(when $k$ = 1), 2 (when $k$ = 2), 6(when $k$ = 3)

$RHS = 1$ (when $n$ = 1), 6 (when $n$ = 2)

Hence, (1,1), (3,2) satisfy

{\Large aa large}{\large For $k$ = 2: large} RHS is strictly increasing, and will never satisfy $k$ = 2 for integer n since RHS = 6 when $n$ = 2.

For $k$ > 3, $n$ > 2:

LHS: Minimum two odd terms other than 1.

RHS: 1st term odd. No other term will be odd. By parity, LHS not equal to RHS.

Hence, (1,1), (3,2) are the only two pairs that satisfy.

~flamewavelight and phoenixfire