<s>[INST] Find all non-negative integers $a, b, c ,d$ such that $a! + b! + c! = 2^d$ 
Find all non-negative integers $a, b, c ,d$ such that $a! + b! + c! = 2^d$.

By trial I found  $a= 2 , b= 3 , c= 5$ and $d= 7$ which is one solution. How to find all the solutions of it ?
 [/INST] Without loss of generality, we can assume $a\leq b\leq c$. Let's do some remarks. I will suppose $d> 3$, the first cases are easy to check by hands and they will coincides with the solutions in $a=1$.
If $a,b,c\geq 3$ then $3$ divides $a!+b!+c!$, but $3$ does not divide $2^d$.  Hence $a=1,2$. (the case $a=0$ is equal to case $a=1$; in the solutions you can replace $0$ with $1$)
Case $a=2$
We have $2! + b! + c! = 2^d$ that is $1+\frac{b!}{2}+\frac{c!}{2}=2^{d-1}$.
If $b,c\geq 4$ then the LHS is odd and the RHS is even. Then $b$ has to be $2$ or $3$.
If $b=2$ we have $c!=2^{d}-4 = 4(2^{d-2}-1)$. So we need $c\geq 4$ in order to have a factor $2^2$ in the LHS. But now the LHS have a factor $2^3$ in its factorization, and the RHS don't, a contradiction.
If $b=3$ we have $c!=2^d-8=8(2^{d-3}-1)$. As above we need $c\geq 4$ in order to have a factor $2^3$ in the LHS, but if $c\geq 6$ we have a factor $2^4$ in the LHS factorization and the RHS don't. So $c$ can be only $4$ or $5$.
With these considerations, the solutions are:
$$(a,b,c,d) = (2,3,4,5), (2,3,5,7)$$

Case $a=1$
We have $1+b!+c! = 2^d$ that is $b!+c! = 2^d-1$. The RHS is odd, so $b!+c!$ has to be odd. For, we need $b!$ odd and $c!$ even (because $b\leq c$). Hence, the unique case is $b=1$.
So now we have $c! = 2^{d}-2 = 2(2^{d-1}-1)$, and using the same argument used in the case above, we will find that $c$ can be only $2$ or $3$.
With these considerations the solutions are:
$$
(a,b,c,d) = (1,1,2,2), (1,1,3,3)
$$
Edit: Thanks to Gareth Ma for his remark (case $a=1$).
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