I saw this question somewhere, have a doubt whether it's correct.
Suppose $a_1, a_2 \cdots a_{2n}$ are distinct integers. The equation $(x-a_1)(x-a_2)...(x-a_{2n})-(-1)^n(n!)^2$=0
has an integer solution
$r=(a_1+a_2...+a_{2n})/2^n$.
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How do we know that $2^{n}|(a_1 + a_2 ... + a_{2n})$? – Indrayudh Roy Jan 21 '14 at 15:07
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If $a_1 = 1$, and $a_2 = 2$, then $r$ is not an integer, nor a solution to the equation. – fuglede Jan 21 '14 at 15:12
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setai.. some junour asked me.. Indrayudh Roy – UNM Jan 21 '14 at 15:31
1 Answers
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I tried using $a(i)=i$ for many values of $n$. I have not been able to find any case where this happens.
Please, take into account that I did not perform millions of runs.
Using the case given by fuglede, the roots are complex.
Claude Leibovici
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