3
$\begingroup$

I created this problem looking at the hm-gm-am-qm inequality (hm = $\frac{2xy}{x + y}$, gm = $\sqrt{xy}$, am = $\frac{x + y}{2}$, qm = $\sqrt{\frac{x^2 + y^2}{2}}$ and hm $\le$ gm $\le$ am $\le$ qm).

If we set $x$ and $y$ without $x$ and $y$ being equal, can the hm, gm, am and qm of $x$ and $y$ all have integer values?

$\endgroup$

closed as off-topic by JonMark Perry, Glorfindel, Excited Raichu, rhsquared, ABcDexter Jan 7 at 16:54

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see "Are math-textbook-style problems on topic?" on meta." – JonMark Perry, Glorfindel, Excited Raichu, rhsquared, ABcDexter
If this question can be reworded to fit the rules in the help center, please edit the question.

4
$\begingroup$

It turns out that this is

possible.

As it happens, a closely related question -- requiring integer values for HM, GM, AM only -- was asked on math.stackexchange.com a while ago and I answered it: https://math.stackexchange.com/questions/2739592/do-there-exist-pairs-of-distinct-real-numbers-whose-arithmetic-geometric-and-ha/2741216#2741216

The solution turns out to be this (quoted from that answer with a change of variable names to match this question):

Choose positive integers $p,q,r$ with $r<q$. Write $n=pq^2$ and $k=p(q^2-r^2)$. Set $x,y=n\pm\sqrt{kn}$.

Now, when do we get an integer quadratic mean as well?

The quadratic mean is $\sqrt{\frac{x^2+y^2}2}=\sqrt{\frac{(n+\sqrt{kn})^2+(n-\sqrt{kn})^2}2}=\sqrt{n^2+kn}$ so what we need is for $n(n+k)$ to be a square. This equals $pq^2(2pq^2-pr^2)$, which is a square iff $2q^2-r^2$ is. So we need $2q^2=r^2+s^2$, say. Obviously and boringly this holds if $q=r=s$, but there are indeed other solutions; for instance, if we take $q=5$ then we can have $r=1,s=7$ yielding (with $p=1$) $n=25$, $k=24$, and $x,y=25\pm\sqrt{600}$. These have harmonic mean 1, geometric mean 5, arithmetic mean 25, and quadratic mean 35.

A related question we might ask, with the above known:

What do all the solutions look like? First rewrite that key equation $2q^2=r^2+s^2$ as $q^2=(r+s)^2+(r-s)^2$; we are looking for solutions to this that don't have $r\pm s=0$. It turns out that there are such solutions iff every prime number that divides $q$ an odd number of times is 1 (mod 4); the "right" way to think about this stuff involves working in the so-called Gaussian integers $a+ib$ where $a,b$ are ordinary integers and $i$ is a square root of -1. For details, consult an undergraduate or early-graduate textbook on number theory :-). So, anyway, this means there are plenty of solutions.

We might further ask

whether it can be done with $x,y$ both integers. The link above answers this question without the quadratic mean; I think the answer is that with the quadratic mean requirement it can't be done.

To begin with,

here is the solution from the math.stackexchange.com question: Choose positive integers $q,r,t$. If $q,r$ are of opposite parity, $t$ must be a multiple of 4; otherwise $t$ is unrestricted. Now write $p=t(q^2+r^2)/2$ and then $x=pq^2$ and $y=pr^2$. (This gives $x\neq y$ provided $q\neq r$.) Now, we need $\frac{x^2+y^2}2$ to be a square; that is, we need $\frac{q^4+r^4}2$ to be a square; that is, we need $q^4+r^4=2s^2$. Well, here is a link to (hidden among a bunch of other calculations) a proof that the only solution to this with $q,r,s$ coprime is $(1,1,2)$; any solution with $q,r,s$ not coprime can easily be transformed into one where they are coprime simply by dividing through by common factors; so there are no solutions with $q\neq r$.

$\endgroup$
0
$\begingroup$

Partial answer - three out of four

It's possible to make the HM, GM, and AM all integers: you just have to choose integers $x,y$ such that

  1. $x,y$ have the same parity (so that the AM is an integer);

  2. $xy$ is a square (so that the GM is an integer);

  3. $xy$ is a multiple of $x+y$ (so that the HM is an integer).

Given any $x,y$ satisfying the first two conditions, we can ensure the third condition too by

multiplying both $x$ and $y$ by the current value of $x+y$. This won't affect their equal parity, and will keep $xy$ as a square, but it will also ensure $xy$ is a multiple of $x+y$.

Thus, we start with the following simple pair which satisfy the first condition:

$x=2,y=4$.

Square them both to satisfy the second condition:

$x=4,y=16$.

Now $x+y=20$, so we can multiple both by $5$ to satisfy the third condition:

$x=20,y=80$.

Check:

HM is $\frac{20\times80}{20+80}=16$, GM is $\sqrt{20\times80}=40$, AM is $\frac{20+80}{2}=50$. All are integers.


In order to make the QM an integer as well, we just need to find $x$ and $y$ solutions to the Diophantine equation $x^2+y^2=2z^2$.

$\endgroup$

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