Weird Polynomial Relation

I will try to post a general solution that doesn't require the specific values of $P(0)$ and $P(1)$ .

Assume $P$ is nonconstant (it can't be constant due to the differing values of $P(0)$ and $(P(1)$ .

Let $P(x)=\displaystyle \sum_{i=0}^n a_ix^i$ with $n \geq 0$ and $a_n \neq 0$ .

Suppose $n>2$ . Thus, if we apply the substitution $\left ( x, \frac{1}{x}, 0 \right )$ (noting that this satisfies $xy+yz+zx=1$ , we get

$P(x)+P \left ( \frac{1}{x} \right ) + P(0) = P \left (x+\frac{1}{x} \right )$

Thus, $a_n \left ( x^n+\frac{1}{x^n} \right ) = a_n \left (x+\frac{1}{x} \right )^n$ , which can only occur if $a_n=0$ for $n>2$ . This is a contradiction. Thus, $n \leq 2$ .

Plugging in $P(x)=ax^2+bx+c$ in $P(x)+P \left ( \frac{1}{x} \right ) + P(0) = P \left (x+\frac{1}{x} \right )$ gives $\begin{aligned} ax^2+bx+c + \frac{a}{x^2}+\frac{b}{x} + c + c &= a\left ( x+\frac{1}{x} \right )^2 + b \left (x+\frac{1}{x} \right ) + c\\ &= ax^2 + 2a + \frac{a}{x^2} + bx + \frac{b}{x} + c\\ 2c &= 2a\\ a&=c \end{aligned}$ Thus, we have $P(x)=ax^2+bx+a$ .

To check this works, we substitute back into the original expression to get $\begin{aligned} ax^2+bx+a+ay^2+by+a+az^2+bz+a &= a(x+y+z)^2+b(x+y+z)+a\\ &= ax^2+ay^2+az^2+2axy+2ayz+2azx+bx+by+bz+a\\ 2a &= 2a(xy+yz+zx)\\ xy+yz+zx &=1 \end{aligned}$ Thus, we have verified, so this polynomial satisfies.

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Now, since $P(0)=1$ , $P$ isn't constant. Thus, $P(x)=ax^2+bx+a$ , so $a=1$ . $P(1)=4$ gives us $2+b=4$ , so $b=2$ . Thus, $P(x)=x^2+2x+1=(x+1)^2$ . Thus, $\sqrt{P(2017)} = 2018$ .