proving $a^2_{n+1}-a^2_{n}=a_{n+1} \cdot a_{n} $ is a Geometric progression

Question

How can i prove that: $a^2_{n+1}-a^2_{n}=a_{n+1} \cdot a_{n} $ is a Geometric progression?

I think i can solve Quadratic equation but maybe there is a simpler way?

Thanks

Answer 1

Divide both sides by $a_n^2$ and define $b=\frac{a_{n+1}}{a_n}$; you should find that the resulting equation is particularly simple. (And in particular, that $b$ is a constant independent of $n$, as the notation suggests.)

(ETA: As Jack D'Aurizio notes in a comment, this isn't quite a complete solution - in fact, the original claim is false without the restriction that we take $a_n\geq 0$ for all $n$, as otherwise one could choose 'alternating' solutions of the quadratic. With that restriction, this proof goes through smoothly.)