A + B + C + D = A x B x C x D

  1. Mar 19, 2008 #1
    Are there any natural number solutions for

    A + B + C + D = A x B x C x D

    besides

    {1, 1, 2, 4}?

    What class of equation does that above belong to?
     
  2. jcsd
  3. Mar 20, 2008 #2
    Suppose there is another solution set A', B', C', D' with larger numbers.
    A' = A + n (=1+n)
    B' = B + m (=1+m)
    C' = C + o (=2+o)
    D' = D + p (=4+p)
    where n,m,o,p are some nonnegative integers that you add to the original numbers to get the new solution numbers.

    Then
    A' x B' x C' x D' = (A + n)(B + m)(C + o)(D + p)
    = ABCD + ABCp + ABoD + ABop + AmCD + AmCp + AmoD + Amop + nBCD + nBCp + nBoD + nBop + nmCD + nmCp + nmoD + nmop

    If none of n,m,o, or p are zero, then you can just match terms to find that
    ABCD + ABCp + ABoD + ABop + AmCD + AmCp + AmoD + Amop + nBCD + nBCp + nBoD + nBop + nmCD + nmCp + nmoD + nmop
    > A + n + B + m + C + o + D + p
    = A' + B' + C' + D'
    So the multiplication is too big for another solution to work if all the additive terms are positive.

    Now it gets a little tricky when some of the additive terms are terms (n,m,o,p) are zero. If all 4 are zero (n=m=o=p=0), then we just have the origional solution, so we can ignore that possibility.

    If 3 are zero, then
    A + B + C + D = A x B x C x D and
    A' + B + C + D = A' x B x C x D so subtracting the two, we have
    A - A' = (A - A') x B x C x D
    which can only happen if B,C,D are all 1, so we can ignore that case too.

    This leaves the 2 cases left: either 1 or 2 of the additive numbers (n,m,o,p) is zero. I will just choose particular variables to set to zero, but you could switch them around.

    If n=0, but no others, then
    A' x B' x C' x D'
    = ABCD + ABCp + ABoD + ABop + AmCD + AmCp + AmoD + Amop
    = (A + 0)(BCD) + (B + m)(ACp) + (C + o)(AmD) + (D + p)(ABo) + Amop
    = A' x (BCD) + B' x (ACp) + C' x (AmD) + D' x (ABo) + Amop
    > A' + B' + C' + D'

    If n and m are both zero (n=m=0) but no others, then
    A' x B' x C' x D'
    = ABCD + ABCp + ABoD + ABop
    = A + B + C + D + ABCp + ABoD + ABop, remember A+B+C+D = ABCD!
    > A + B + C + o + D + p, by matching terms
    = A' + B' + C' + D'

    Soooo..... there are no larger solutions, and smaller possibilities ({1,1,1,1}, {1,1,1,2},...{1,1,2,3}) can be checked by hand.

    Looks like the solution you got is the only one.
     
  4. Mar 20, 2008 #3
    lol, this proof can be made much simpler:

    A' x B' x C' x D' = (A + n)(B + m)(C + o)(D + p)

    = ABCD + ABCp + ABoD + AmCD + nBCD + other nonnegative terms

    = A + B + C + D + ABCp + ABoD + AmCD + nBCD + other (remember ABCD=A+B+C+D still)

    > A + n + B + m + C + o + D + p

    = A' + B' + C' + D'
     
    Last edited: Mar 20, 2008
  5. Mar 20, 2008 #4
    Thanks, Nick. The simplest proofs are the best. Occam dices, slices, chops and minces.

    Do you know what class of equation the original one was?
     
  6. Mar 21, 2008 #5
    No clue. Perhaps someone else will know?

    I can say that the above proof strategy will work for any number of variables in the same form (ABCDEFG... = A+B+C+D+E+F+G+...).
     
  7. Mar 21, 2008 #6
    Another idea about this,

    For n variables,
    X1 x X2 x X3 x ... x Xn = X1 + X2 + X3 + ... + Xn

    if n is even then {1, 1, 1, ..., 2, n-2} is the solution. eg:
    1,1,2,4
    1,1,1,1,2,6
    1,1,1,1,1,1,2,8

    why? Because
    1 + 1 + ... + 2 + n = (n - 2) + 2 + n = 2*n
    You could also do this by induction on n.

    I'm not sure about odd numbers though.
    Edit: duh, it works for odd numbers too.
     
    Last edited: Mar 22, 2008
  8. Mar 22, 2008 #7

    morphism

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    It's an example of a (non-linear) Diophantine equation (in 4 variables).
     
  9. Mar 24, 2008 #8
    Odds

    For odd numbers of variables there seems to be more than one solution.

    for
    [tex]
    A + B + C + D + E = A * B * C * D * E
    [/tex]
    {1,1,2,2,2} works, as does
    {1,1,1,3,3}
     
  10. Mar 26, 2008 #9
    Ahh yes, interesting. Technically for the proof I posted, you need to check all combinations of sets involving only the numbers 1,2,3, and 4. I just (stupidly) assumed there were no other small solutions.

    I wonder how many other solutions there are for big odd N, as for N variables the proof does not rule out solutions that are combinations of 1,2,3,...,N.
     
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