Erdős Problem #403

Does the equation \[2^m=a_1!+\cdots+a_k!\] with \(a_1<a_2<\cdots <a_k\) have only finitely many solutions? Asked by Burr and Erdős. Frankl and Lin [Li76] independently showed that the answer is yes, and the largest solution is \[2^7=2!+3!+5!.\] In fact Lin showed that the largest power of \(2\) which can divide a sum of distinct factorials containing \(2\) is \(2^{254}\), and that there are only 5 solutions to \(3^m=a_1!+\cdots+a_k!\) (when \(m=0,1,2,3,6\)). See also [404]. A solution is encoded below as a pair \((m, s)\) where \(s\) is the finite set \(\{a_1 < a_2 < \cdots < a_k\}\) of positive integers, so the distinctness of the \(a_i\) is given by set membership. The empty set contributes no solutions since \(2^m \geq 1 > 0\). The linked proof gives more than finiteness: it classifies the solutions outright, as \((0,\{1\})\), \((1,\{2\})\), \((3,\{2,3\})\), \((5,\{2,3,4\})\) and \((7,\{2,3,5\})\), so the set below has exactly five elements.