<s>[INST] Find all positive integers $(a,b,c,n)$ such that $2^n=a!+b!+c!$ I have found the solutions by a little calculation $(2,3,5,7)$ and $(2,3,4,5)$. But I don't know if there's any other solutions or not?
 [/INST] Let $a \le b \le c$.
If $m \ge 2$ m! is even so if $a = 1$ (odd), $b = 1$ and $c! = 2^n - 2 = 2(2^{n-1} - 1)$ so $c = 2; n=1$ or $c = 3; n = 2$.  Two answers so far. (1,1,2,1)(1,1,3,2)
If $a > 1$ then $a! + b! + c! = a!(1 + b!/a! + c!/a!) = 2^n$ If $a \ge 3$ $3|a!$ and so $3|2^n$ which is impossible so $a = 2$.
So $1 + b!/2 + c!/2 = 2^{n-1}$  $n$ can't be $1$. so $b!/2$ must be odd so $b= 3$.  So $c!/2 = 2^{n -1} - 4= 4(2^{n -3} - 1)$ so $c$ = 4 and $n = 5$. Or 3*5...c = $2^{n-3}-1).$  Possibly, $c=5$ and $15 = $ $2^{n-3} - 1$; $n = 7$.  So two more answers $(2,3,4,5)$ and $(2,3,5,7)$.
Any more will require $3*5...c = 2^{n-3}-1)$ which is odd so $c<6$.  So those $4$ are the only solutions.
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