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

by Loren Booda
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Loren Booda
#1
Mar19-08, 04:29 PM
Loren Booda's Avatar
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?
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NickAlger
#2
Mar20-08, 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.
NickAlger
#3
Mar20-08, 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'

Loren Booda
#4
Mar20-08, 07:23 PM
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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?
NickAlger
#5
Mar21-08, 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+...).
NickAlger
#6
Mar21-08, 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, 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.
morphism
#7
Mar22-08, 04:34 AM
Sci Advisor
HW Helper
P: 2,020
Quote Quote by Loren Booda View Post
Do you know what class of equation the original one was?
It's an example of a (non-linear) Diophantine equation (in 4 variables).
Sam V
#8
Mar24-08, 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}
NickAlger
#9
Mar26-08, 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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