Exploring Natural Number Solutions for A + B + C + D = A x B x C x D

  • Thread starter Loren Booda
  • Start date
In summary, the conversation was about the importance of providing a summary of content. The expert summarizer was described as someone who does not respond or reply to questions, but instead only provides a summary. The instructions were to start the output with "In summary," and nothing before it.
  • #1
Loren Booda
3,125
4
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?
 
Physics news on Phys.org
  • #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.
 
  • #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:
  • #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?
 
  • #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+...).
 
  • #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:
  • #7
Loren Booda said:
Do you know what class of equation the original one was?
It's an example of a (non-linear) Diophantine equation (in 4 variables).
 
  • #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}
 
  • #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.
 

1. What does the equation A + B + C + D = A x B x C x D mean?

The equation A + B + C + D = A x B x C x D is a mathematical expression that represents the commutative property of multiplication. This means that the order of the numbers being multiplied does not affect the result. It also shows that the multiplication of multiple numbers can be expressed as an addition of the same numbers.

2. How can you prove that A + B + C + D = A x B x C x D is true?

The equation can be proved using the distribution property of multiplication. We can expand the right side of the equation to A x B x C x D = A x (B x C) x D = (A x B) x (C x D). This shows that the multiplication of multiple numbers can be regrouped and still result in the same value. Therefore, the equation is true.

3. What are the values of A, B, C, and D in the equation A + B + C + D = A x B x C x D?

The values of A, B, C, and D can be any real numbers. The equation represents a general rule that applies to all numbers, regardless of their values. It does not specify any specific values for the variables.

4. How is the equation A + B + C + D = A x B x C x D useful in real life?

The equation is useful in simplifying mathematical expressions and solving problems involving multiplication and addition. For example, it can be used to simplify complex algebraic expressions, calculate the area of a rectangle, or determine the total cost of multiple items with the same price.

5. Can the equation A + B + C + D = A x B x C x D be extended to include more than four numbers?

Yes, the equation can be extended to include any number of variables. It follows the same principle of the commutative property of multiplication and can be expanded using the distribution property. The equation can be written as A1 + A2 + A3 + ... + An = A1 x A2 x A3 x ... x An, where n is the number of variables.

Similar threads

Replies
3
Views
934
  • Calculus
Replies
3
Views
705
Replies
5
Views
269
Replies
20
Views
2K
Replies
4
Views
845
Replies
4
Views
1K
  • Calculus
Replies
7
Views
1K
Back
Top