I assume that you mean the following: Let $ABCD$ rectangle. We consider the points $E\in CA,F\in AB,G\in BC$ such that $DE\perp CA,EF\perp AB$ and $EG\perp BC$. Solve in the set of rational numbers the equation $AC^x=EF^x+EG^x$. Proof: Let $AB = a, BC =b, AC = c$ $AE\cdot AC = AD^2\Rightarrow AE = \frac{b^2}{c}$ $\frac{EF}{BC} = \frac{AE}{AC} \Rightarrow EF = AE\cdot \frac{b}{c} = \frac{b^3}{c^2}$ Similarly $EG = \frac{a^3}{c^2}$ $AC^x = EF^x+EG^x \Leftrightarrow c^x = \left(\frac{a^3}{c^2}\right)^x + \left(\frac{b^3}{c^2}\right)^x \Leftrightarrow 1 = \frac{a^{3x}}{c^{3x}}+\frac{b^{3x}}{c^{3x}}$ $\Leftrightarrow \left(\frac{a^2}{a^2+b^2}\right)^{\frac{3x}2}+\left(\frac{b^2}{a^2+b^2}\right)^{\frac{3x}2} = 1$ $\Leftrightarrow p^y + (1-p)^y = 1$ where $p=\frac{a^2}{a^2+b^2}\in (0, 1)$ and $y = \frac{3x}{2}$ Now consider the function $f(y) = p^y + (1-p)^y$ with $p\in (0,1)$ $f'(y) = p^y\ln p+(1-p)^y\ln (1-p) < 0$ since $p^y, (1-p)^y > 0$ and $\ln p, \ln (1-p) < 0$ Then $f(y)$ is strictly decreasing. We have that $f(1) = 1$, then the only solution to $f(y) = 1$ is $y=1$ Then the solution is $x=\frac 23$