Let a and b be positive integers with a =/= b. I checked, if the four most common Hölder-means (AM, GM, QM, HM) of a and b could then also be an integer. As expected this is the case for each of them seperately. I went one step further and checked, if two of these means could be integers at the same time (For a,b<5000). And for every combination of these Hölder means, except!!! for GM-QM there were plenty of solutions. I tried to find a proof, why there shouldn't be distinct integers a and b, for which the geometric and quadratic means aren't both integers, but I didn't succeed.

So, please help me and either find these tupels, or proof, that they can't exist.

Thanks in advance

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## Comments

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TopNewestHere'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 \(a=b.\)

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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.

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I don't think this quite works. Or at least I don't see how it rules out, say, \(n=3, m=1, a-b=4.\) When you get to the bottom you get \(k=1,\) but I don't see why there is an argument that \(k\) must be even. I guess the sentence I don't understand is "we can repeat this logic infinitely."

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It probably doesn't. If I have made mistake then maybe my method could be adapted to get either an answer or a proof as to why there aren't solutions

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