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Math Help - Factoriel : 2

  1. #1
    Super Member dhiab's Avatar
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    Factoriel : 2

    Solve in N : a! + b! + c! = 2^{d!} <br />
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  2. #2
    Super Member PaulRS's Avatar
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    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-
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  3. #3
    Super Member dhiab's Avatar
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    Quote Originally Posted by PaulRS View Post
    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
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  4. #4
    Super Member PaulRS's Avatar
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    Quote Originally Posted by dhiab View Post
    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
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