Postby **sudarshan** » 28 Sep 2020 09:32

Just thinking out loud.

Let's say that we have n odd numbers with the property, that the AM of any pair, can be represented as the GM of a set of 1 or more numbers from the same set. If all n numbers are the same number, we have no problem. We want to see if we can find a set of n numbers, where all numbers are not the same value, which will satisfy the condition. So in such a set, by definition, one number will be the largest, and one number will be the second largest (there could be multiple instances of the largest and/ or second largest number in the set, of course).

Let's say the largest value in the set is y, and the second largest value is x. If there are other unique numbers in the set, then they are all smaller than x.

1. There is no number in the set larger than y.

2. There is also no number in the set, which lies between x and y (since x is the second largest number).

3. The arithmetic mean of x and y is a = (x+y)/2.

3.a. Since x < y, that means x < a < y (i.e., x is strictly less than a, and a is strictly less than y, meaning, a lies between x and y).

3.b. Since x is at least 1, y cannot be a multiple of a (1*a < y, and 2*a = x+y, which is already larger than y).

4. There has to be some combination of one or more numbers in the set, whose GM equals a = (x+y)/2.

4.a. The GM of any k numbers g1, g2, ... gk is (g1*g2*...gk)^(1/k).

5. If the set <g1, g2, ... gk> (which is a subset of the full set of n numbers) consists of just one number g1 - then by definition, its GM is g1 itself. So by 4., g1 will have to be the same as a, which lies between x and y. But g1 is a member of the full set of n numbers, and by 2., there is no number in the set between x and y.

So there have to be at least two numbers in the set, such that their GM equals the value a = (x+y)/2.

But x is the second largest value in the set, and y is the largest. If the two numbers are y and y (assuming multiple occurrences of y in the set), the GM will be y itself, which is too large (3.a). If the two numbers are x and x (assuming multiple occurrences of x in the set), the GM will be x itself, which is too small (3.a).

How about x and y? Recall, the AM of two unequal positive numbers will *always* be larger than its GM. So even x and y won't work. And the other numbers in the set of n are smaller than x (x is the second largest, remember?) so they definitely won't work.

6. We need a way to "pull up" the value of the GM to be a = (x+y)/2. So we need to include more numbers in the GM calculation, which are large enough to pull the GM up. x, however, is already smaller than a, and will pull *down* the GM value. So these additional numbers can only be further occurrences of y (since y is the only number in the set of n, which is larger than a). Thus, it is necessary to have multiple occurrences of the largest number, y in the set.

Of course, if we include many instances of y in the GM set, we can include a few x's (or even numbers smaller than x) since the many instances of y will compensate for these smaller numbers.

If the GM calculation set <g1, g2, ... gk> has "k" numbers, then, a^k has to be the product of all the numbers in the GM calculation set. This means the numbers in the GM calculation set can only be composed of the factors of a, and can have no other factors. From 3.b., y is also not a multiple of a.

So y is a number larger than a, all of whose factors are also factors of a, and the set <g1, g2, ... gk> also has multiple occurrences of y in it (i.e., many of the values of g1, g2, ... are actually y itself). <g1, g2, ... gk> also consists of at least 3 numbers, of which at least two are equal to y (see 6.).

Still thinking from here on....