<s>[INST] Solving $a! + b! = 2^n$ 
$a! + b! = 2^n$, find $a,b,n$

My obseravtions thus far:


*

*If $a>4$, $b<4$, otherwise $a! + b!  = 0 \mod 10$

*If $a=1$, $b=1$


If we try all combination with the numbers 2 to 4, we get following solutions: 
(1,1,1), (2,2,2), (2,3,3),(3,2,3)
Are there more solutions and how to find them?
 [/INST] There are no other solutions in $\mathbb N = \{ 1, 2, \dots \}$. There are a few more if you allow $0 \in \mathbb N$.
Indeed, we may assume that $a\le b$. Then $2^n = a! + b! = (a!)(1+c)$ and so $a!$ divides $2^n$. Therefore, $a!=1$ or $a!=2$, and so $a=1$ or $a=2$.
If $a=1$, then $b! = 2^n-1$ is odd and so $b!=1$, that is $b=1$. Then $n=1$.
If $a=2$, then $b! = 2^n-2 = 2(2^{n-1}-1)$ is even but not a multiple of $4$. Therefore, $b\le 3$.
If $a=2$ and $b=1$, then $a!+b!=3$, not a power of $2$.
If $a=2$ and $b=2$, then $a!+b!=4=2^2$, and so $n=2$.
If $a=2$ and $b=3$, then $a!+b!=8=2^3$, and so $n=3$.
If you allow $0 \in \mathbb N$, then you also get $(a,b,c)=(0,0,1), (0,1,1), (1,0,1)$.
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