<s>[INST] If $a,b\in\mathbb N_+$ satisfy $\sqrt{ab},\sqrt{\frac{a^2+b^2}2}\in\mathbb N_+$, is it necessarily true that $a=b$? This question is related to another question where it asks to find pairs of positive integers $(a, b)$ such that

$$
\frac{a + b}{2},\ \sqrt{ab},\ \frac{2ab}{a + b} \in \mathbb{N}_+.
$$

In that question I gave a set of solutions, which is actually all the possible solutions. When I was trying to further incorporate the requirement that $\sqrt{\dfrac{a^2 + b^2}{2}} \in \mathbb{N}_+$, it seems, but I failed to prove, that all pairs such that$$
\sqrt{ab},\ \sqrt{\frac{a^2 + b^2}{2}} \in \mathbb{N}_+ \tag{1}
$$ 
are trivial, i.e. $a = b$, whereas the requirements$$
\frac{2ab}{a + b},\ \sqrt{\frac{a^2 + b^2}{2}} \in \mathbb{N}_+ \tag{2}
$$
still yields non-trivial solutions (See below).
Here is what I have done so far: I first derived that all positive integer solutions to $ab = c^2$ are$$
(a, b, c) = (km^2, kn^2, kmn), \tag{3}
$$
all positive integer solutions to $2ab = c(a + b)$ are$$
(a, b, c) = (km(m + n), kn(m + n), 2kmn) \tag{4}
$$
or$$
(a, b, c) = (k(2m - 1)(m + n - 1), k(2n - 1)(m + n - 1), k(2m - 1)(2n - 1)), \tag{4'}
$$
and all positive integer solutions to $a^2 + b^2 = 2c^2$ are$$
(a, b, c) = (k\,|m^2 + 4mn + 2n^2|, k\,|m^2 - 2n^2|, k\,|m^2 + 2mn + 2n^2|), \tag{5}
$$
or the positions of $a$ and $b$ swapped.
Thus\begin{align*}
a &= k\,|m^2 + 4mn + 2n^2| (|m^2 + 4mn + 2n^2| + |m^2 - 2n^2|)\\
b &= k\,|m^2 - 2n^2| (|m^2 + 4mn + 2n^2| + |m^2 - 2n^2|)
\end{align*}
are non-trivial solutions to (2). To get solutions to (1), I combined (3) and (5) to get\begin{align*}
k_1 m_1^2 &= k\,|m^2 + 4mn + 2n^2|\\
k_1 n_1^2 &= k\,|m^2 - 2n^2|
\end{align*}
Without loss of generality, assume that $(m, n) = (m_1, n_1) = 1$, then it can be proved that$$
(|m^2 + 4mn + 2n^2|, |m^2 - 2n^2|) = 1 \text{ or } 2.
$$
Now it reduces to two cases:$$
\begin{cases}
|m^2 + 4mn + 2n^2| = m_1^2\\
|m^2 - 2n^2| = n_1^2
\end{cases} \text{ or } \begin{cases}
|m^2 + 4mn + 2n^2| = 2m_1^2\\
|m^2 - 2n^2| = 2n_1^2
\end{cases}.
$$
Now I am not sure if it is the right way to proceed. If there were to be only one equation, it could be transformed to a Pell's equation and explicit solutions are known. However, here it is a system of quadratic equations and I have no idea how to deal with it.
Incidentally, for $|m|, |n| \leqslant 30000$, all pairs of $(m, n)$ satisfying either system of equations are of the form $m + n = 0$, $m + 2n = 0$, $m = 0$, or $n = 0$, all of which lead to $a = b$.
Thus, is it necessarily true that all positive integer solutions to (1) satisfy $a = b$?
 [/INST] First, it is easy to argue that, since $\sqrt{ab} \in \mathbb N$, there exists a square free integer $k$ and integers $m,n$ such that 
$$a=m^2k\\
b=n^2k$$ 
($k$ is simply the product of the primes which in the prime decomposition of $a$ appear at an odd power).
Then,
$$\sqrt{\frac{a^2+b^2}{2}}=\sqrt{k^2\frac{m^4+n^4}{2}} \in \mathbb N$$
It follows from here that $\sqrt{2( m^4+n^4) }$ is rational and hence integer. Thus there must exists some integer $l_1$ such that 
$$2(m^4+n^4)=l_1^2$$
Now, $l_1=2l$ for some integer $l$, and hence
$$m^4+n^4=2l^2$$
Your claim follows now from this post:
$x^4+y^4=2z^2$ has only solution, $x=y=z=1$ .
Indeed, since $m=n=1$, we have 
$$a=k=b$$
</s>