1. ## Factoriel : 2

Solve in N : $a! + b! + c! = 2^{d!}
$

2. Since the RHS is a power of 2, we can't have $a,b,c\geq{3}$ 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 $a<3$ , then $a!=1$ or $a!=2$

- Case $a!=1$ then $a=0$ or $a=1$ and $b!+c! = 2^{d!}-1$.
Note that 2 doesn't divide the RHS, and so $b$ and $c$ can't be greater than 1 at the same time.

WLOG consider $b=1$ or $b=0$ then $c!=2(2^{d!-1}-1)$ since the RHS is not divisible by 4, c is at most 3., and no less than 2 -since the RHS is even-.

For $c=2$ we get $d=2$ and for $c=3$ we have no solution since $d!$ should be equal to 3.

Thus we have $(a,b,c,d)=(0,1,2,2)$; $(a,b,c,d)=(0,0,2,2)$; $(a,b,c,d)=(1,1,2,2)$ and all the permutations of the values of a,b,c there.

-Case $a!=2$ then $a=2$

Then $b!+c!=2(2^{d!-1}-1)$ , again, since 4 doesn't divide the RHS, b and c cannot be both greater than 3.

WLOG assume $b\leq 3$. If $b=0,1$ -0 or 1- then $c=0,1$ and $d=2$, so assume $3\geq b\geq 2$, then -by looking back at the equation- $2|c!$ and so $c\geq{2}$

If $b=2$ then $c!=4(2^{d!-2}-1)$ (1), so actually 4 divides $c!$, and then we are forced to have $c\geq 4$ , but that would imply that 8 divides $c!$, which doesn't happen in (1) !!

If $b=3$ then $c!=2(2^{d!-1}-3)$ (2), so actually 2 divides $c!$, but 8 doesn't, so $3\geq c\geq{2}$ , here note that $c=3$ is impossible since this would imply $c!+6=2^{d!-1}$ which is a power of 2! -and can't be divisible by 3 -, so $c=2$ and we must have - in (2)-, $2=2(2^{d!-1}-3)$ and so $2^{d!-1}-3=1$ i.e. $d!-1=2$ which is impossible!.

So we get $(a,b,c,d)=(2,0,1,2)$, $(a,b,c,d)=(2,0,0,2)$ , $(a,b,c,d)=(2,1,1,2)$ and basically we find that these are the same solutions we found on the other part! -considering the permutations-

3. Originally Posted by PaulRS
Since the RHS is a power of 2, we can't have $a,b,c\geq{3}$ 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 $a<3$ , then $a!=1$ or $a!=2$

- Case $a!=1$ then $a=0$ or $a=1$ and $b!+c! = 2^{d!}-1$.
Note that 2 doesn't divide the RHS, and so $b$ and $c$ can't be greater than 1 at the same time.

WLOG consider $b=1$ or $b=0$ then $c!=2(2^{d!-1}-1)$ since the RHS is not divisible by 4, c is at most 3., and no less than 2 -since the RHS is even-.

For $c=2$ we get $d=2$ and for $c=3$ we have no solution since $d!$ should be equal to 3.

Thus we have $(a,b,c,d)=(0,1,2,2)$; $(a,b,c,d)=(0,0,2,2)$; $(a,b,c,d)=(1,1,2,2)$ and all the permutations of the values of a,b,c there.

-Case $a!=2$ then $a=2$

Then $b!+c!=2(2^{d!-1}-1)$ , again, since 4 doesn't divide the RHS, b and c cannot be both greater than 3.

WLOG assume $b\leq 3$. If $b=0,1$ -0 or 1- then $c=0,1$ and $d=2$, so assume $3\geq b\geq 2$, then -by looking back at the equation- $2|c!$ and so $c\geq{2}$

If $b=2$ then $c!=4(2^{d!-2}-1)$ (1), so actually 4 divides $c!$, and then we are forced to have $c\geq 4$ , but that would imply that 8 divides $c!$, which doesn't happen in (1) !!

If $b=3$ then $c!=2(2^{d!-1}-3)$ (2), so actually 2 divides $c!$, but 8 doesn't, so $3\geq c\geq{2}$ , here note that $c=3$ is impossible since this would imply $c!+6=2^{d!-1}$ which is a power of 2! -and can't be divisible by 3 -, so $c=2$ and we must have - in (2)-, $2=2(2^{d!-1}-3)$ and so $2^{d!-1}-3=1$ i.e. $d!-1=2$ which is impossible!.

So we get $(a,b,c,d)=(2,0,1,2)$, $(a,b,c,d)=(2,0,0,2)$ , $(a,b,c,d)=(2,1,1,2)$ and basically we find that these are the same solutions we found on the other part! -considering the permutations-
Hello , Thank you :
if to explain me the sense of words: RHS, LHS, WLOG

4. Originally Posted by dhiab
Hello , Thank you :
if to explain me the sense of words: RHS, LHS, WLOG
RHS = Right Hand Side
LHS = Left Hand Side
WLOG = Without Loss of Generality