An algebra problem by Aareyan Manzoor

A polynomial is always continuous,

Given,

$P(x)=x^{P(9)}+x^{P(8)}+x^{P(7)}+x^{P(6)}+x^{P(5)}+x^{P(4)}+x^{P(3)}+x^{P(2)}+x^{P(1)}+\dfrac{x}{P(x)}$

This implies the denominator should not be zero , thus zero solutions

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Please comment , Please correct me.

Clarification : $P(x)$ is assumed to be a Polynomial throughout.

Solution : Simplifying the given expression, we have $P^2(x)=P(x)\sum_{i=1}^{9}x^{P(i)}+x$ Let $\alpha$ be a root of $P(x)$ . Thus $P(\alpha)=0$ . Putting $x=\alpha$ in the above, we have $0=0+\alpha$ i.e., $\alpha=0$ Hence all roots of the polynomial $P(x)$ are zero. Thus $P(x)$ is of the form $x^d$ for some positive integer $d\geq 1$ ( $P(x)\equiv 0$ is not allowed). Hence we have the following identity $x^{2d}\equiv x^d\sum_{i=1}^{9}x^{i^d}+x$ Since $9^d +d> 2d$ , for any $d\geq 1$ , the above expression is impossible to satisfy. Hence no such polynomial $P(x)$ is possible. $\blacksquare$

We will assume that a nonconstant polynomial exists with n roots and show that this cannot be the case.

Consider any non-zero roots. Let the root be A.(Note: This could be imaginary as well) . So (A)^2= (A^P(9)+.....+A^P(1))P(A)+x. Since P(A) is zero, we see that the LHS gives us 0 and the RHS gives us A. So all roots must be zero. So P(x)= Bx^n where B is the coefficient of x^n.

Repeating the above argument, we can see that n=1. So P(X)=Bx. Setting x=1 gives B=9.109.... or, -0.1097... If we check the 2 values, we see that neither of them work as the degrees of the polynomials are not the same and it is a well known fact that the highest degree term will dominate for very large values of x meaning that the 2 sides cannot be equal.

So no polynomials exist and so the number of roots is 0.