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

  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. 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. 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. 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. 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. 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. morphism

    morphism 2,020
    Science Advisor
    Homework Helper

    It's an example of a (non-linear) Diophantine equation (in 4 variables).
     
  9. 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. 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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