I self-realized an interesting property today that all numbers (a,b) belonging to the infinite set {(a,b):a=(2l+1)2,b=(2k+1)2; l,k∈N; l,k≥1} have their AM and GM

bothintegers.Now I wonder if there exist distinct

realnumbers (a,b) such that their arithmetic mean, geometric mean and harmonic mean (AM, GM, HM)all threeare integers. Also, I wonder if a stronger result for (a,b) both beingintegersexists.I tried proving it, but I did not find it easy. For the AM, it is easy to assume a real a and an AM m1 such that the second real b equals 2m1−a. For the GM, we get a condition that m2=√(2m1−a)a. If m2 is an integer, then… what? I am not sure

exactlyhow we can restrict the possible values of a and m1 in this manner.

**Answer**

Expanding on Christian Blatter’s answer.

There are a few key points.

- The arithmetic mean of two rational numbers is always rational.
- The harmonic mean of two non-zero rational numbers is always rational.
- The geometric mean of two squared positive integers is always an integer.
- For all three types of mean if we multiply every input by a positive real value we also multiply the result by that same value.

These key points lead to a strategy for finding numbers whose am, gm and hm are all integers.

- pick a pair of integers whose GM is an integer.
- calculate the AM and HM
- multiply through by the denominators of the AM and HM.

Now to work this through, pick any two distinct positive integers x and y.

GM(x2,y2)=xy

AM(x2,y2)=x2+y22

HM(x2,y2)=2x2y2x2+y2

Let t=2(x2+y2) Let a=tx2 Let b=ty2. Since only addition, multiplication and squaring of positive integers is involved it is clear that t, a and b are all positive integers. It is also clear that a and b are distinct.

GM(a,b)=txy

AM(a,b)=tx2+y22=(x2+y2)2

HM(a,b)=t2x2y2x2+y2=4x2y2

Again since all these values can be calculated merely by adding, multiplying and squaring positive integers they are all positive integers.

Lets plug in some numbers, for example x=1 and y=2

t=10

a=10

b=40

GM(10,40)=20

AM(10,40)=25

HM(10,40)=16

Indeed we can extend this techiquie to find an arbitary size list of integers the AM, GM, and HM of any subset of which are integers. Just start with integers of the form xn! so the GMs are all integers. Then work out the AMs and HMs and multiply through.

**Attribution***Source : Link , Question Author : Gaurang Tandon , Answer Author : Peter Green*