5

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 $a = b$?

Bill Dubuque
  • 272,048
Ѕᴀᴀᴅ
  • 34,263

1 Answers1

5

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 $$a=k=b$$

N. S.
  • 132,525