# Find the possible values of k: x+1/y=y+1/z=z+1/x=k

• MHB
• lfdahl
In summary, the possible values of $k$ are $\pm1$, where $x, y,$ and $z$ are real numbers (not all equal) that follow the condition $x + \frac{1}{y} = y + \frac{1}{z} = z + \frac{1}{x} = k$. These values can be obtained through the equations $x^2 - kx + 1 \ne 0$ and $(k^2-1)(x^2 - kx + 1) = 0$. However, they can also be achieved through specific values of $x, y,$ and $z$, such as $(2,-1,\frac12)$ for $k = lfdahl Gold Member MHB$x, y$and$z$are real numbers (not all equal), that obey $x + \frac{1}{y} = y + \frac{1}{z} = z + \frac{1}{x} = k$ Find the possible values of$k$. Source: Nordic Math. Contest lfdahl said:$x, y$and$z$are real numbers (not all equal), that obey $x + \frac{1}{y} = y + \frac{1}{z} = z + \frac{1}{x} = k$ Find the possible values of$k$. Source: Nordic Math. Contest [sp]Let$f(x) = k - \frac1x$, and define$f^{(2)}(x) = f(f(x))$and$f^{(3)}(x) = f(f^{(2)}(x))$. We want to find$x$such that$f^{(3)}(x) = x$but$f(x) \ne x$. The condition for$f(x) \ne x$is$k - \frac1x \ne x$, or$x^2 - kx + 1 \ne 0$. Next, $$\displaystyle f^{(2)}(x) = k - \frac1{k - \frac1x} = k - \frac x{kx-1} = \frac{k^2x - x - k}{kx-1},$$ and $$\displaystyle f^{(3)}(x) = k - \frac{kx-1}{k^2x - x - k} = \frac{k^3x - kx - k^2 - kx + 1}{k^2x - x - k}.$$ The condition$f^{(3)}(x) = x$then becomes$k^3x - kx - k^2 - kx + 1 = x(k^2x - x - k)$, or$(k^2-1)x^2 - (k^3-k)x + k^2 - 1 = 0.$So$(k^2-1)(x^2 - kx + 1) = 0$. But we already know that$x^2 - kx + 1 \ne 0$, and so$k^2 -1 = 0$. So the only possible values of$k$are$\pm1$. But if$(x,y,z) = (2,-1,\frac12)$then$x + \frac{1}{y} = y + \frac{1}{z} = z + \frac{1}{x} = 1$, and if$(x,y,z) = (-2,1,-\frac12)$then$x + \frac{1}{y} = y + \frac{1}{z} = z + \frac{1}{x} = -1$. So the values$k = \pm1$can both occur, and these are the only possible values of$k\$.

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Thankyou, Opalg for your participation and another exemplary solution!

## 1. What is the equation x+1/y=y+1/z=z+1/x=k asking for?

The equation is asking for the possible values of k that satisfy the given equation.

## 2. How many solutions can this equation have?

This equation can have an infinite number of solutions, as there are an infinite number of possible values for k.

## 3. Can this equation have negative values for k?

Yes, this equation can have negative values for k as long as the given equation is satisfied.

## 4. How can I find the possible values of k?

To find the possible values of k, you can manipulate the equation by solving for one variable in terms of the others and then substituting that into the other equations. This will give you a quadratic equation in terms of k, which you can then solve for the possible values of k.

## 5. Can this equation have imaginary solutions for k?

Yes, this equation can have imaginary solutions for k if the values of x, y, and z are complex numbers.

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