Originally Posted by
PaulRS Since the RHS is a power of 2, we can't have $\displaystyle a,b,c\geq{3}$ at the same time, because the LHS would be a multiple of 3.
Without loss of generality -becuase of the symmetry on a,b,c- suppose $\displaystyle a<3$ , then $\displaystyle a!=1$ or $\displaystyle a!=2$
- Case $\displaystyle a!=1$ then $\displaystyle a=0$ or $\displaystyle a=1$ and $\displaystyle b!+c! = 2^{d!}-1$.
Note that 2 doesn't divide the RHS, and so $\displaystyle b$ and $\displaystyle c$ can't be greater than 1 at the same time.
WLOG consider $\displaystyle b=1$ or $\displaystyle b=0$ then $\displaystyle c!=2(2^{d!-1}-1)$ since the RHS is not divisible by 4, c is at most 3., and no less than 2 -since the RHS is even-.
For $\displaystyle c=2$ we get $\displaystyle d=2$ and for $\displaystyle c=3$ we have no solution since $\displaystyle d!$ should be equal to 3.
Thus we have $\displaystyle (a,b,c,d)=(0,1,2,2)$; $\displaystyle (a,b,c,d)=(0,0,2,2)$; $\displaystyle (a,b,c,d)=(1,1,2,2)$ and all the permutations of the values of a,b,c there.
-Case $\displaystyle a!=2$ then $\displaystyle a=2$
Then $\displaystyle b!+c!=2(2^{d!-1}-1)$ , again, since 4 doesn't divide the RHS, b and c cannot be both greater than 3.
WLOG assume $\displaystyle b\leq 3$. If $\displaystyle b=0,1$ -0 or 1- then $\displaystyle c=0,1$ and $\displaystyle d=2$, so assume $\displaystyle 3\geq b\geq 2$, then -by looking back at the equation- $\displaystyle 2|c!$ and so $\displaystyle c\geq{2}$
If $\displaystyle b=2$ then $\displaystyle c!=4(2^{d!-2}-1)$ (1), so actually 4 divides $\displaystyle c!$, and then we are forced to have$\displaystyle c\geq 4$ , but that would imply that 8 divides $\displaystyle c!$, which doesn't happen in (1) !!
If $\displaystyle b=3$ then $\displaystyle c!=2(2^{d!-1}-3)$ (2), so actually 2 divides $\displaystyle c!$, but 8 doesn't, so $\displaystyle 3\geq c\geq{2}$ , here note that $\displaystyle c=3$ is impossible since this would imply $\displaystyle c!+6=2^{d!-1}$ which is a power of 2! -and can't be divisible by 3 -, so $\displaystyle c=2$ and we must have - in (2)-, $\displaystyle 2=2(2^{d!-1}-3)$ and so $\displaystyle 2^{d!-1}-3=1$ i.e. $\displaystyle d!-1=2$ which is impossible!.
So we get $\displaystyle (a,b,c,d)=(2,0,1,2)$, $\displaystyle (a,b,c,d)=(2,0,0,2)$ , $\displaystyle (a,b,c,d)=(2,1,1,2)$ and basically we find that these are the same solutions we found on the other part! -considering the permutations-