A + B + C + D = A x B x C x Dby Loren Booda Tags: None 

#1
Mar1908, 04:29 PM

P: 3,408

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
Mar2008, 05:59 PM

P: 15

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. 



#3
Mar2008, 06:05 PM

P: 15

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' 



#4
Mar2008, 07:23 PM

P: 3,408

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



#5
Mar2108, 01:03 AM

P: 15

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+...). 



#6
Mar2108, 11:59 PM

P: 15

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, n2} 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. 



#7
Mar2208, 04:34 AM

Sci Advisor
HW Helper
P: 2,020





#8
Mar2408, 06:17 PM

P: 1

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} 



#9
Mar2608, 03:47 AM

P: 15

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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