Prove that at least 3 of the numbers are equal

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Prove that at least 3 of the numbers are equal

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#1 h Nov 12, 2008, 2:08 PM • 5 Y

Y by andreiromania, viterick, YusupTKM, Adventure10, Mango247

Consider integers $a_i,i=\overline{1,2002}$ such that
$a_1^{ - 3} + a_2^{ - 3} + \ldots + a_{2002}^{ - 3} = \frac {1}{2}$
Prove that at least 3 of the numbers are equal.

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#2 h Nov 5, 2017, 7:30 AM • 2 Y

Y by Adventure10, Mango247

BUMP.////

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#3 h Nov 5, 2017, 8:51 AM • 2 Y

Y by Adventure10, Mango247

As idea.
We will prove for $n$
If $a_1=1$ then $-a_2^{-3}+...-a_n^{-3}=\frac{1}{2}$ - problem for $n-1$
So let there are not $a_i=1$
Let there are no more than $2$ same elements. Then $\frac{1}{2}=a_1^{-3}+...+a_n^{-3} \leq 2(\frac{1}{2^3}+\frac{1}{3^3}+...\frac{1}{(n+1)^3})$

But $2(\frac{1}{2^3}+\frac{1}{3^3}+...\frac{1}{(n+1)^3})<2(\frac{1}{2^3}+\frac{1}{3^3}+...\frac{1}{(n+1)^3}+...)<2(\frac{1}{2^3}+\frac{1}{3^3}+...\frac{1}{15^3}+ 16*\frac{1}{16^3}+32*\frac{1}{32^3}+....)<2(\frac{1}{2^3}+\frac{1}{3^3}+...+\frac{1}{15^3}+\frac{3}{64})<\frac{1}{2}$ - contradiction

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#4 h May 28, 2021, 4:19 AM

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From RagvaloD, we may assume that all integers $a_i$ are greater than $1$ . We now show that the maximum possible sum $2\sum_{n=2}^{1001}\frac{1}{n^3}$ must be less than $\frac{1}{2}$ . Notice that $\sum_{n=a}^b \frac{1}{n^3} < \int_{a-1}^b \frac{1}{x^3} dx$ . Then
$2\sum_{n=2}^{1001}\frac{1}{n^3} < \frac{2}{2^3} + 2\int_2^{1001} \frac{1}{x^3} dx=\frac{1}{4}-\frac{1}{x^2}\biggr|_2^{1001}=\frac{1}{4}+\frac{1}{4}-\frac{1}{1001^2}<\frac{1}{2}.$ One possible set of integers is $a_k=2$ for $1\leq k \leq 4$ and $a_k=0$ for $5 \leq k \leq 2002$ , and we are done.

It is interesting to note that there must be three copies of the same number even when $i > 2002$ . Also, the approximate sum in question rounds to about $0.4041$ .

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