Geometric proof of number theory result: If the geometric and quadratic means of integers $a$ and $b$ are themselves integers, then $a=b$ We start with the following problem:


  Let $a$ and $b$ be positive integers such that their geometric and quadratic means are integers. Show that $a=b$.


One possible approach is to write down the corresponding diophantine equations and to do infinite descent on $u^4-v^4=w^2$ where $w=a^2-b^2$ which then has to be zero.

However, the means also have simple geometric representations. So my question is if there is a way to do the infinite descent geometrically.
Even if the two numbers are non-integer positive real numbers

$$ \sqrt{\frac{a^2+b^2}{2}} = \sqrt{ab} $$

requires that ( by squaring and simplfying )

$$  a^2+b^2-2ab =0 \rightarrow (a=b) $$

The RMS of $(a,b)$  is in general greater than GM. They are equal only if the numbers $(a,b)$ are themselves equal.
Even if the two numbers are non-integer positive real numbers

$$ \sqrt{\frac{a^2+b^2}{2}} = \sqrt{ab} $$

requires that ( by squaring and simplfying )

$$  a^2+b^2-2ab =0 \rightarrow (a=b) $$

The RMS of $(a,b)$  is in general greater than GM. They are equal only if the numbers $(a,b)$ are themselves equal.