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It’s as easy as a, b, c

Posted by Chris on October 7, 2014 – 12:30 pm

Find all solutions in positive integers a, b, c to the equation
a! b! = a! + b! + c!


This post is under “MathsChallenge” and has 6 respond so far.
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6 Responds so far- Add one»

  1. 1. Wizard of Oz Said:

    Divide throughout by a! to get:

    b! = 1 + b!/a! + c!/a!

    Since there are no odd factorials then b! is even so that either b = a + 1 (with b odd) or c = a + 1 (with c odd).

    By inspection, even for small a we find that c >> a+1. So b = a+1 and our equation now becomes:

    (a+1)! = 1 + (a+1) + c!/a!
    c! = a!(a+1)! – a!(a+2)

    This is where it gets beyond me. Even if I generate solutions to this equation for successive (even) values of a, how do I test for the result being a factorial number?

  2. 2. Catherine Said:

    Divide throughout by a! to get:
    b! = 1 + b!/a! + c!/a!

    Divide throughout by b! to get:
    a! = 1 + a!/b! + c!/b!

    Since factorials are integers, so a! = b!.

    Now we can convert the equation into

    a!a! = 2a! + c!

    let a! = x
    x^2 – 2X – c! = 0

    Determinant must be a square number,
    square root of 4 + 4c! is 2(1+c!)^1/2;
    so c! + 1 must be a square number.

    1! = 1, 1 + 1 = 2, invalid
    2! = 2, 2 + 1 = 3, invalid
    3! = 6, 6 + 1 = 7, invalid
    4! = 24, 24 + 1 + 25, valid
    5! = 120, 121 valid
    ….

    If c = 4, then a = b = 3
    If c = 5 , a and b cannot be found.

    (This is the only solution I can find, and I don’t know if c could be a large number)

  3. 3. kaliprasad Said:

    continuing from above

    we have another choice that is b = a and c= a + 1 and c is even

    we can divide by a! to get c = (a+1) or b = a+ 1

    combining the 2 we get b = a and c = a+ 1

    so (b!)^2 = 2b! + (b+1) ! = b!(b+3)
    so b! = b + 3

    we get b = 3 or solution set = a = b = 3 and c = 4

    check 3! * 3! = 36 = 6+6+ 24 = 3! + 3! + 4!

  4. 4. kaliprasad Said:

    above is not 100% correct as if b = a the c need not be a +

    so b! = 2 + c!/b!
    b = 3 , c = 4 is a solution
    b = 4 LHS = 24 c = 5=> RHS = 7
    c = 6 => RHS = 30
    no solution
    I cannot work out

  5. 5. Zorglub Said:

    Here is a complete solution.

    Without any loss of generality, assume that b ≥ a

    First, observe that if a = 0 or a = 1 then the equation becomes b! = 1 + b! + c!, which is impossible.
    Also if a = 2 we have 2b! = 2 + b! + c! which implies that b! = 2 + c! which is also impossible.
    Conclusion: b ≥ a ≥ 3

    Second, if c ≤ b then a!b! ≤ a! + 2b! ≤ 3b! which implies that a! ≤ 3 which is impossible since a ≥ 3.
    In summary : c > b ≥ a ≥ 3

    As Catherine said:
    Divide throughout by a! to get: b! = 1 + b!/a! + c!/a!
    Divide throughout by b! to get: a! = 1 + a!/b! + c!/b!
    implies that a = b and the equation simplifies to

    a!a! = 2a! + c!

    The trick is to replace c = a + d with d ≥ 1 and to divide throughout by a!. We get

    a! = 2 + (a+1)(a+2)…(a+d).

    Next, divide both sides by 3. a!/3 is an integer. 2/3 is not an integer. Therefore (a+1)(a+2)…(a+d)/3 is not an integer, which implies that d ≤ 2 . There are only 2 cases to consider

    Case d=1: c = a+1
    —————
    a! = 2 + (a+1) = a+3

    (a-1)! = 1 + 3/a and the only integer solution is a = 3 (and b=3, c=4)

    Case d=2: c = a+2
    —————
    a! = 2 + (a+1)(a+2) = a^2 + 3a + 4
    (a-1)! = a + 3 + 4/a and the only integer solution would be with a=4, but
    3! differs from 4 + 3 + 1

    CONCLUSION: There is a unique solution (a,b,c) = (3,3,4)

  6. 6. Chris Said:

    Hi Zorglub. Thanks for that. I didn’t have an official solution. Yours look spot on.

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