How many positive triples $(n,m,k)$ satisfy the following equation: $$n!+m!=2^k.$$
My try follows, I used trial and examined many values; only $4$ triples work which are $$(1, 1,1), (2,2,2), (2,3,3), (3,2,3).$$
Are there other triples? If not, how can I prove that?
Thank you for your help.
By symmetry we can suppose that $m\leq n$. Then $m!$ divides both $m!$ and $n!$ and hence it divides $m!+n!=2^k$, so $m\leq2$. If $m=1$ then $2^k=1+n!\geq2$ is even, so $n=1$ and $k=1$.
If $m=2$ then $2+n!=2^k$ so $n!=2(2^{k-1}-1)$. Because $4$ does not divide $2^{k-1}-1$ this shows that $n<4$, so either $n=2$ or $n=3$. Check that both give solutions, with $k=2$ and $k=3$.