Pythagorean Triples in a GP

Let our three integers in a Pythagorean triple be x, y, and z.Assume that there is a Pythagorean triple like this in a geometric series: $\frac{y}{x}$ = $\frac{z}{y}$ . Cross-multiplying, $y^{2} = xz$ . Plugging this into the Pythagorean theorem, $x^{2} + xz = z^{2}$ . Also, $x^{2} + xz - z^{2} = 0$ . Solve this as a quadratic in terms of x to find the discriminant is 5, which is not a perfect square, meaning that any solutions to this quadratic will be irrational. But, for this to be a Pythagorean triple, x, y, and z must be integers. So, the answer is No.

Let our three integers in a Pythagorean triple be called $a$ , $b$ , and $c$ with $a<b<c$ . If there is an integer solution to this problem, we can rewrite these using the formula for a term in a geometric progression with $a$ as the starting term and $r$ as the common ratio. This gives $b=ar$ and $c=ar^2$ . Plugging these into the Pythagorean theorem, we get $a^2+(ar)^2=(ar^2)^2$ . Simplifying and gathering like terms gives $a^2r^4-a^2r^2-a^2=0$ . As $a$ must be a positive integer, we can simplify by dividing by $a^2$ , giving $r^4-r^2-1=0$ . Now set $x=r^2$ giving $x^2-x-1=0$ . Using the quadratic formula, we get $x=\frac{1+\sqrt5}{2}$ meaning $r=\sqrt\frac{1+\sqrt5}{2}$ . As $r$ is irrational, no integer value of $a$ can result in a geometric progression of integers that is also a Pythagorean triple. Therefore, the answer is no.