Integer means, Help me!

Here's a proof that uses some descent ideas. First suppose $m=\sqrt{ab}, n = \sqrt{\frac{a^2+b^2}2}$ with $a,b,m,n$ all integers. Now, if $\gcd(a,b) = g,$ then $g|m$ and $g|n,$ and replacing $a,b,m,n$ with $a/g,b/g,m/g,n/g$ gives another solution. So we can assume that $a,b$ are relatively prime.

Now, $m^2=ab$ with $a,b$ relatively prime implies that $a$ and $b$ are squares. So let $a=c^2, b=d^2.$ The remaining equation is $2n^2 = c^4 + d^4.$ This is a typical application of infinite descent, which I will just give a reference to. The unique solution for coprime $c,d$ is $c=d=1,$ which implies $a=b.$

Hi Gerrit, this seems like a very interesting problem. It could probably be generalised (by someone smarter than me) that a result holds for the $n$ variable means, not just with $2$ variables. It is obvious that for $a=b$, we can find a solution, since the equality of the means all hold. For distinct integers, I have had a think and come up with the reasoning below. It could be (and probably is!) not true and I've made a slip, but this is why I think there is no $a$ and $b$:

Assume that $a$ and $b$ satisfy the conditions. $m = \sqrt{ab}$ $n = \sqrt{\frac{a^2 + b^2}{2}}$ Squaring the first one and then multiplying by two, as well as squaring the second one gives us, $2m^2 = 2ab$ $2n^2 = a^2 + b^2$ Subtracting, $2n^2 - 2m^2 = a^2 - 2ab + b^2$ $2(n-m)(n+m) = (a-b)^2$ Now from the bit that says $2n^2 = a^2 + b^2$, we see that the LHS is even, meaning that either both $a$ and $b$ are odd, or they are both even. This means that whatever the case, $a-b$ is even. Let $2j = a-b$ where $j$ is an integer. $2(n-m)(n+m) = (2j)^2$ $(n-m)(n+m) = 2j^2$ As the RHS is now even, the LHS must also be. However, if one of the brackets is even, the other must be as well (a reason why the difference of two squares is never in the form $4p + 2$). This makes the LHS a multiple of 4, and dividing by 2, we get that $j^2$ must be even, meaning $j$ is even. Now let $2k = j$ such that $a-b=4k$. $2(n-m)(n+m) = (4k)^2$ $(n-m)(n+m) = 8k^2$ Since we know that the LHS is a multiple of 4, we can let $(n-m)(n+m) = 4p$. $p = 2k^2$ This makes $p$ even, so we can express it as $2q$ and so on and so on. We can repeat this logic infinitely, showing that either side is a multiple of $2$. Since we can't keep producing integers infinitely, we have a contradiction, and no such $a$ and $b$ exist.