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Can two set of numbers(same size) have same arithmetic , geometric, and harmonic mean ? When I say different set they must differ by at-least $1$ element and also what if set is not be of distinct elements?

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not be of distinct elements i.e. multiset. I imagine consideration of degrees of freedom should do the trick, depending on cardinality.. –  anon Jul 13 '12 at 2:49

3 Answers 3

Yes. For example, say $r_1, r_2, r_3, r_4$ are solutions to the equation $x^4+ax^3+bx^2+cx+d=0$. Then $$(x-r_1)(x-r_2)(x-r_3)(x-r_4)=x^4+ax^3+bx^2+cx+d \, ;$$ by expanding the left-hand side, you can see that \begin{eqnarray} a &=& -(r_1+r_2+r_3+r_4) \\ c &=& -(r_1r_2r_3+r_1r_2r_4+r_1r_3r_4+r_2r_3r_4)\\ d &=& r_1r_2r_3r_4 \end{eqnarray} (You can also get a similar expression for $b$, but we don't care about it at the moment.)

So the arithmetic mean is $\frac{r_1+r_2+r_3+r_4}{4}=\frac{-a}{4}$, the geometric mean is $\sqrt[4]{r_1r_2r_3r_4}=\sqrt[4]{d}$, and the harmonic mean is $\frac{4r_1r_2r_3r_4}{r_1r_2r_3+r_1r_2r_4+r_1r_3r_4+r_2r_3r_4}=\frac{-4d}{c}$.

That is, the arithmetic, geometric, and harmonic means of $r_1$, $r_2$, $r_3$, and $r_4$ don't depend on $b$. On the other hand, the set $\{r_1,r_2,r_3,r_4\}$ itself depends very much on $b$. So you can get an infinite family of 4-element sets with the same three means just by choosing different $b$s.

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+1 I find this answer very elegant. –  Alex Becker Jul 13 '12 at 3:09
    
I think, this can be generalized to any polynomial equation of degree n≥4. –  lab bhattacharjee Jul 13 '12 at 6:36
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Very nice solution. –  Ben Alpert Jul 13 '12 at 8:09
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The Vieta formulae are a wonderful thing. :) –  J. M. Jul 13 '12 at 12:39
    
@labbhattacharjee: It absolutely can (using the general Vieta formulae as J. M. suggests); I just restricted myself to quartics for simplicity. –  Micah Jul 13 '12 at 18:17

Yes. In fact, you can find solutions in the integers.

$\{-4, -3, 1, 3, 4\}$

$\{-6, -2, 1, 2, 6\}$

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Yes. For instance, $\{1,2,3,4\}$ and $\{1.00025,\sim 1.997016, \sim 3.006751, \sim 3.995982\}$.

In general, take $n>3$ distinct positive numbers and perturb $n-3$ of them slightly. You will then have $3$ constraints in $3$ unknowns, which you should be able to solve for.

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