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

Question

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.

Shouldn't it be $$4(u^4-v^4)=w^2$$? Quadratic mean is RMS $\sqrt{\frac{1}{2}(a^2+b^2)}$ right? — N8tron, Jun 20 '18 at 11:11
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. — Phira, Jun 20 '18 at 11:14

score -1 · Answer 1 · answered Jun 26 '18 at 21:51

-1

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.

answered Jun 26 '18 at 21:51

Narasimham

42,260

2

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

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

1 Answers1