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Math Help - factorial equation

  1. #1
    Member disclaimer's Avatar
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    factorial equation

    Solve x!+y!=z! in the set of non-negative integers.

    It seems that 0!+1!=2! would work but I'm not totally sure. Is there any "official" solution or is guesswork the only way to go?
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  2. #2
    Super Member Bacterius's Avatar
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    I believe this should be in the Number Theory Section.

    You might want to consider the following expression which is equivalent :

    2 \times 3 \times \cdots \times x + 2 \times 3 \times \cdots \times y = 2 \times 3 \times \cdots \times z

    You can factorize this, then divide the RHS to the number-only LHS factor. Then try to see when it actually is an integer ()

    EDIT : this might not give all solutions. You can also study the equation x! - z! = y! which is conceptually simpler.
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  3. #3
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    Assume without loss of generality that x\le{y}, and divide by x!:

    1+\frac{y!}{x!}=\frac{z!}{x!}

    Now, what is the relationship between y! and x!?

    You should get that there are 3 solutions besides the one you gave.
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  4. #4
    Moo
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    Hello,

    If x=y, then we'd have 2*x!=z!
    But unless x=1, it's not possible to have a solution.
    So (1,1,2) is a solution.

    If x<y (wlog), we can divide each side by x! and we'd have 1+(x+1)\cdots (y-1)y=(x+1)\cdots (z-1)z

    It's obvious that y<z. So the RHS can be written (x+1)\cdots(y-1)y(y+1)\cdot (z-1)z

    Hence 1=(x+1)\cdots(y-1)y\cdot[z(z-1)\cdots (y+1)-1]

    Since all of these are integers, we must have (x+1)\cdots(y-1)y=1 and z(z-1)\cdots (y+1)-1=1

    Again, since all the factors are integers, we must have x+1=y=1 \Rightarrow x=0,y=1

    Thus there are only 2 solutions : (0,1,2) and (1,1,2)
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  5. #5
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    I was thinking of (0,0,2) and (1,0,2) in addition to your solutions (0,1,2) and (1,1,2).
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