Do there exist pairs of distinct real numbers whose arithmetic, geometric and harmonic means are all integers?
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 $y$.
$$\mathrm{GM}(x^2,y^2) = xy$$ $$\mathrm{AM}(x^2,y^2) = \frac{x^2+y^2}{2}$$ $$\mathrm{HM}(x^2,y^2) = \frac{2x^2y^2}{x^2+y^2}$$
Let $t = 2(x^2 + y^2)$ Let $a=tx^2$ Let $b=ty^2$. 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.
$$\mathrm{GM}(a,b) = txy$$ $$\mathrm{AM}(a,b) = t\frac{x^2+y^2}{2} = (x^2+y^2)^2$$ $$\mathrm{HM}(a,b) = t\frac{2x^2y^2}{x^2+y^2} = 4x^2y^2$$
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 $y=2$
$$t = 10$$ $$a = 10$$ $$b = 40$$ $$\mathrm{GM}(10,40) = 20$$ $$\mathrm{AM}(10,40) = 25$$ $$\mathrm{HM}(10,40) = 16$$
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 $x^{n!}$ so the GMs are all integers. Then work out the AMs and HMs and multiply through.
Take $1$ and any square $>1$. Multiply both of them with the smallest common denominator of their AM and HM.
For any $$ (a, b) = (km^2(m^2 + n^2), kn^2(m^2 + n^2)), $$ where $k, m, n \in \mathbb{N}_+$ and $2 \mid a - b$, there is$$ \frac{a + b}{2} \in \mathbb{Z}, \quad \sqrt{ab} = kmn(m^2 + n^2) \in \mathbb{Z}, \quad \frac{2ab}{a + b} = 2km^2 n^2 \in \mathbb{Z}. $$