x+y+z=xyz?

Question

What are some non-trivial solutions to this equation?
Besides the obvious [6] 1+2+3=1x2x3 [6]

Answer 1

I would answer this but I'm really bad at math, right now. . . . . . . .

Answer 2

There are no other positive integral solutions to this equation, and it is easily proven.
Solve for x:

xyz - x = y+z
x(yz-1) = y+z
x= (y+z)/(yz-1)
Now clearly x can be an integer only if yz-1≤y+z , or:
yz-y ≤ 1+z ,
y(z-1) ≤ 1+z ,
or y ≤ (z+1)/(z-1)

So when z≥4, y≤5/3, or y=1
But then x = (z+1)/(z-1) and this cannot be an integer when z≥4 (it has a value of 5/3 at z=4 and gets continuously smaller, approaching 1 in the limit as z approaches infinity).

Answer 3

If any (x, y or z) is 0... and the other two are have the same absolute value, but diferent sign...

e.g.

x=0, y = -3, z = 3

0 -3 + 3 = (0)(-3)(3) as easy as that...or you can get...trigonometrially wild...


make x = sen^2 (o)
y = cos(o)
z=sen^2(o) + cos(o) / sen^2(o)cos(o) -1

it works XD...


Now really...it's not that hard... you can do reverse engeneering... meaning...

make... x = ANY NUMBER... y = ANY NUMBER... (both choosed arbitrairly by you...) then... z is a variable... and you just have to find z in the equation...

x + y + z = xyz

(x+y+z)/xy = z (god...where's LATEX when you need it)

(x+y)/xy = z-z/xy

(x+y)/xy = z(1- 1/xy)

(x+y)/xy = z((xy-1)/xy)

(x+y)/(xy-1) = z... (this is a magic formula that will give you the Z number required to make this equation true...with two...(x, y) given numbers...)

(it works for the 1,2,3....)

3/(1*2-1) = 3 XD< AWESOME!!

Answer 4

NIce work HDK.

When I saw the number 6 it reminded me of what are called 'perfect numbers'

A number that equals the sum of its proper divisors is perfect.

After 6, the next perfect number is 28 = 1+2+4+7+14 but that doesn't fit the problem.

Answer 5

Zero my Hero -- trivial I suppose but ver elegant all zeros

also the all negative version of 1,2,3

The question does not disallow decimals such as 0.6,2,13=15.6 there will be a huge family of these. 0.7,2,6.75=9.45

I suppose one could develop a rule for these families, dont have time to play with that now. I cant think how this could be true for other integers, but I'll check back for other answers myself.

Answer 6

The first link details a similar problem.

The second link explores a similar relationship for quadratic polynomials.

Edit:
The third link provides a general solution for integers.... there are no more such pairs.

search term in Google was: triples sum equal product

Answer 7

no, substitute: 2=x 3=y and 7=z
2+4+7=12
2*4*7=56

Answer 8

enable F=xyz+t(x^2+y^2+z^2-2) (lagrange) dF/dx=0 dF/dy=0 dF/dz=0 dF/dx= yz+2tx=0 dF/dy=xz+2ty=0 dF/dz=xy+2tz=0 t=-yz/2x=-xz/2y=-xy/2z yz/x=xz/y =xy/z you could merely locate x and y and z

Answer 9

any multiple (n) times by both sides

Nx+ny+nz=nxyz

ex: (5)1+5(2)+5(3)=5(1)(2)(3)

5+10+15 = 30
30=30

will work for anything, even fractions

Answer 10

How do you define trivial? Single digits?