Let x, y, and z be distinct nonzero real numbers and
x + (1/y) = y + (1/z) = z + (1/x)
Find all possible values of xyz.
I don't even know where to start!
Any help is appreciated, thank you :)
Just by inspection, it should be clear that
(x,y,z) = ±(1,1,1), ±(-1, 1/2, 2) are solutions
Then, set up the three equations
x + 1/y = y + 1/z
y + 1/z = z + 1/x
x + 1/y = z + 1/x
Some manipulation yields
y = -1/(x+1)
z = -(x+1)/x
so, pick any other values for x that you want.
Thanks for the response!
I still have a question though - what do you mean by picking any other values of x? Is there an equation that I'm supposed to solve?
I ran a computer simulation and the only solutions were of the type
(k,k,k), where k is any real number, k ≠ 0
but the values of (x,y,z) were to be distinct, so other than the above, there is no solution
e.g. oobleck has
y = -1/(x+1)
z = -(x+1)/x
but (2,2,2), which satisfies the original set of equations, does not satisfy this simplification.
I suspect an error in the manipulation
Thank you for the response.
Yes, I do think there might be a miscalculation. From running this through a calculator, I got y=1/(1-x).
Another thing: I don't get how one could get ±(-1, 1/2, 2) as a solution through just inspection. Is there math behind this?
Also, it appears that ±(-1, 1/2, 2) the only solution to the equation.
Oh, I've figured it out! Thanks for the help, but I've found another method.
To solve this problem, we need to manipulate the given equation to find a relationship between x, y, and z. Let's start by expanding the equation:
x + (1/y) = y + (1/z) = z + (1/x)
Multiplying the first term by y, the second term by z, and the third term by x, we get:
xy + 1 = yz + 1 = zx + 1
From this equation, we can see that xy + 1, yz + 1, and zx + 1 are all equal. Since we are given that x, y, and z are distinct nonzero real numbers, we can assume that none of them equal zero.
Let's set those equal to a constant k:
xy + 1 = k
yz + 1 = k
zx + 1 = k
Now, we can solve these equations to find the values of x, y, and z. We can start by subtracting 1 from each equation:
xy = k - 1
yz = k - 1
zx = k - 1
Now, let's take the product of all three equations:
xy * yz * zx = (k - 1) * (k - 1) * (k - 1)
Since xyz = xy * yz * zx, we can substitute and solve for xyz:
xyz = (k - 1) * (k - 1) * (k - 1)
So, xyz equals the cube of the quantity (k - 1).
To find all possible values of xyz, we need to consider all possible values of k. Since k is a constant, it can take on any real value. This means that xyz can take on any real value as well.
In conclusion, there are infinitely many possible values of xyz.