Let the sides be $a$, $ar$, and $ar^2$.
$\begin{aligned}
\therefore\enspace a^2 + (ar)^2 &= (ar^2)^2 &&\{\text{Pythagoras}\} \\ \\
\therefore\enspace 1+r^2&=r^4 \\ \\
\therefore\enspace r^4-r^2-1&=0 \\ \\
\therefore\enspace r^2 &= \,\!\rlap{\dfrac{1\pm\sqrt{1-4(1)(-1)}}{2}} \\ \\
\therefore\enspace r^2 &= \dfrac{1+\sqrt{5}}{2} &&\{r^2>0\}
\end{aligned}$
But if $a$ is an integer, then $ar^2$ cannot be an integer.
Thus the integer side lengths of a right angled triangle cannot form a geometric progression.
