Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
you mean H^2 + G^2 = A^2 – Emanuele Paolini Jul 13 '14 at 10:39
Right. Excuse the typo. – EsperantoSpeaker1 Jul 13 '14 at 10:55
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,

$$\frac{4a^2 b^2}{(a+b)^2} + ab = \frac{(a + b)^2}{4}$$


$$\frac{4a^2 b^2}{(a + b)^2} = \frac{(a - b)^2}{4}$$

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.

share|cite|improve this answer
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 '14 at 10:42
That's quite nice, @DanielFischer. – user61527 Jul 13 '14 at 10:46

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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