7
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Shouldn't it be $$4(u^4-v^4)=w^2$$? Quadratic mean is RMS $\sqrt{\frac{1}{2}(a^2+b^2)}$ right? $\endgroup$ – N8tron Jun 20 '18 at 11:11
  • 1
    $\begingroup$ There are several ways to reduce the question to the stated diophantine equation. That is why I did not write it down explicitly to no prejudice a particular approach. And yes, in one approach, you would get $w= (a^2-b^2)/2$, but in another one, the value I gave in my question. $\endgroup$ – Phira Jun 20 '18 at 11:14
-2
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ The geometric and arithmetic means are allowed to be different in the original problem. $\endgroup$ – wnoise Jun 26 '18 at 21:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.