# A Curious Property of 3 and 5

### Solution

We first treat the case $n=5.$ WLOG, let there be five numbers $a\ge b\ge c\ge d\ge e.$ Then

\displaystyle \begin{align} &a-b\ge -(b-a),\\ &a-c\ge b-c\ge 0,\\ &a-d\ge b-d\ge 0,\\ &a-e\ge b-e\ge 0. \end{align}

It follows that

$(a-b)(a-c)(a-d)(a-e)+(b-a)(b-c)(b-d)(b-e)\ge 0.$

Similarly,

$(d-a)(d-b)(d-c)(d-e)+(e-a)(e-b)(e-c)(e-d)\ge 0.$

And also

$(c-a)(c-b)(c-d)(c-e)=[(c-a)(c-b)]\cdot [(c-d)(c-e)]\ge 0.$

The case of $n=3$ is analogous. For other $n$ there are counterexamples:

For $n=4:$ $a_1=0,\,a_2=a_3=a_4=1.$

For $n\gt 5:$ $a_1=\cdots=a_{n-4}=0,\,a_{n-3}=a_{n-2}=a_{n-1}=2,\,a_n=1.$

### Remark

For $n=2k,$ $k\t 1,$ the claim is obviously true. Because, changing the sign of all $a's$ flips the sign of the whole expression.

### Acknowledgment

This is the original problem 1 from the 1971 IMO, see Djukic et al, The IMO Compendium, Springer, 2011 (Second edition).

The remark is due to Roland van Gaalen.