Cassarrim1

2022-02-01

If k is even number then for what value of x and y,
${x}^{2k}-{\left(xy\right)}^{k}+{y}^{2k}=0$

nebajcioz

Expert

Suppose y is zero, then the only solution to . Thus, $x=y=0$ is one of the possible solutions.
Now assume y is not zero. Then dividing everything by ${y}^{2k}$ you get
${\left(\frac{x}{y}\right)}^{2k}-{\left(\frac{x}{y}\right)}^{k}+1=0$
or, after substituting $z={\left(\frac{x}{y}\right)}^{k}$,
${z}^{2}-z+1=0$
This quadratic equation has no real solutions.
This means $x=y=0$ is the only possible solution, and even that for $k>0$ because when $k=0$ the expression ${0}^{0}$ is not well-defined.

Do you have a similar question?