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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.

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you mean H^2 + G^2 = A^2 –  Emanuele Paolini Jul 13 at 10:39
    
Right. Excuse the typo. –  EsperantoSpeaker1 Jul 13 at 10:55

1 Answer 1

up vote 6 down vote accepted

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.

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Perhaps still simpler, add $b^2 + 4a^2$ to obtain $(2a+b)^2 = 5a^2$. The right hand side is not a square. –  Daniel Fischer Jul 13 at 10:42
    
That's quite nice, @DanielFischer. –  T. Bongers Jul 13 at 10:46

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