Here is what I tried - I assumed 3 quadratics $f(x) , g(x)$ and $h(x)$ with these I formed the gigantic equation $f(g(h(x)))$ it had a degree 8 therefore was satisfying the 8 degree criterion but I'm stuck at finding is it possible that it has roots $1,2,3,4,5,6,7,8.$
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Thanks @YouKnowMe for the edit . – AdhirajSinghBrar Nov 09 '20 at 03:35
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I suggest that you start by solving a simpler problem of the same type: find just $p$ and $q$ where $p(q(x))$ has the four roots $1,2,3,4$. – MJD Nov 09 '20 at 03:39
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@MJD will this help me ? – AdhirajSinghBrar Nov 09 '20 at 03:39
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I think it might. – MJD Nov 09 '20 at 03:40
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Ok then i will try – AdhirajSinghBrar Nov 09 '20 at 03:40
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But if anyone knows the solution please post it for some reference – AdhirajSinghBrar Nov 09 '20 at 03:41
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The polynomial $h$ must be $2$ to $1$ on the set $\{1,2,3,4,5,6,7,8\}$. Thus its stationary point must occur at $4.5$ (considering real and imaginary parts of $h$ separately). Thus $$h(x)=a(x(9-x))+b,$$ and the values of $h$ applied to that set are $8a+b,14a+b,18a+b,20a+b$ for constants $a,b$.
By the same reasoning $g$ must be $2$ to $1$ on this set, so its stationary point must occur at $14a+b$ so that $g(8a+b)=g(20a+b)$. However then $g(18a+b)\neq g(14a+b)$. Thus we cannot have $3$ such quadratics.
tkf
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@AdhirajSinghBrar It is where the derivative of the quadratic is $0$. That is the minima or the maxima of the quadratic. About this point the function is symmetric. – tkf Nov 09 '20 at 04:08
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Ok and also one more thing what did u mean by h must be 2 to 1 – AdhirajSinghBrar Nov 09 '20 at 04:11
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@AdhirajSinghBrar As a quartic can have at most $4$ roots, $h(1),h(2),h(3,h(4),h(5),h(6),h(7),h(8)$ can take at most $4$ values. As $h$ is a quadratic, no $3$ of these can be equal, so they must split up into $4$ pairs of equal values. – tkf Nov 09 '20 at 04:14
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