Pythagorean triple means Just out of curiosity, do there exist two positive integers whose arithmetic mean ($A $), geometric mean ($G $) and harmonic mean ($H $) constitute a Pythagorean triple?
That is,  $A $,  $G $ and  $H $ are positive integers, and $H^2 + G^2 = A^2$.

Let $a$ and $b$ be positive integers with $a \ge b$; then you're asking whether we can ever have
$$\left(\frac{2}{\frac 1 a + \frac 1 b}\right)^2 + \Big(\sqrt{ab}\Big)^2 = \left(\frac{a + b}{2}\right)^2$$
or upon some simplification,
$$\frac{4a^2 b^2}{(a+b)^2} + ab = \frac{(a + b)^2}{4}$$
Rearranging,
$$\frac{4a^2 b^2}{(a + b)^2} = \frac{(a - b)^2}{4}$$
Taking square roots,
$$\frac{2ab}{a + b} = \frac{a - b}{2}$$
$$4ab = a^2 - b^2$$
We now show there are no positive integer solutions to this. Suppose there was; since this is homogeneous, we can cancel any common factors in $a$ and $b$, so they're relatively prime. Clearly $a$ and $b$ must have the same parity, so they're both odd.
Add $2b^2$ to both sides, leading to
$$4ab + 2b^2 = a^2 + b^2$$
$$2ab + 2b^2 = (a - b)^2$$
$$2b (a + b) = (a - b)^2$$
Here we have the desired contradiction. We have that $b | (a - b)^2$, and upon expanding, this leads to $b | a^2$. As $b$ and $a$ are relatively prime, this forces $b = 1$, so that $4a = a^2 - 1$. It's easy to verify that this has no solutions in the integers, and we're done.

Actually, it's simpler to just add $b^2$ to both sides, leading to $$a^2 = b(4ab + b)$$
hence $b | a^2$ and so on.
