Geometric Pythagorean Triples

Suppose there does exist a pythagorean triple satisfying the property. Then let $k$ be the common ratio.

$a^2+k^2a^2=k^4a^2 \implies k^2+1=k^4$ so $k^2$ is the golden ratio ( $\phi$ ), and irrational which implies that $k$ is irrational (You can prove this by contrapositive). A similar approach can be taken if $k<1$ in which $k^2$ must equal $\phi-1$

The $m, n$ formula can come in handy here.

If we take two non-negative integers i.e. $m \quad n$

Then the Pythagorean triplets are ${m^2-n^2}, {m^2+n^2} and {2mn}$

And can be related to each other as $({m^2-n^2})^2 + ({2mn})^2 = ({m^2+n^2})^2$

Considering $a=({m^2-n^2}), \quad b=({2mn}) and \quad c=({m^2+n^2})$ Condition for geometric progression $b^2=ac$ , which will not be followed by the above mentioned value of $a$ , $b$ and $c$ . Thus, the given statement is not possible.