Let $s$ be a real number greater than $\frac {1}{4}$. Find the possible number of all real roots of $$Q(x)=x^{2022}-x^{2020}+x^{2018}-x^{1011}+x^{1009}+s$$
I tried to simplify the polynomial but failed. For example,
$$Q(x)=x^{1009}(x^{1013}-x^{1011}+x^{1009}-x^2+1)+s$$ or $$Q(x)=x^{2020}(x^2-1)+x^{1009}(x^{1009}-x^2+1)+s$$
I don't have a good idea how to proceed.
Source: 9th grade competition problem .
Best Answer
I don't know if this is the intended solution, but it works:
Suppose $x=c$ is a real root of $$Q(x) = x^{2018}(x^4 - x^2 + 1) - x^{1009}(x^2-1) + s$$
Then in particular, $x=c$ is a root of the real polynomial $$P(x) = x^{2018}(c^4-c^2+1) - x^{1009}(c^2-1) + s$$ which is quadratic in $x^{1009}$, and since it had a real root, it has a non-negative discriminant: $$(c^2-1)^2-4s(c^4-c^2+1) \geq 0.$$
Now, note that $c^4-c^2+1 > 0$ [show this if you need to] and $s > \frac14$, so $$(c^2-1)^2 \geq 4s(c^4-c^2+1) > c^4-c^2+1$$
Finally, solving this gives $c^2 < 0$, which is a contradiction.
Therefore there are no real roots of $Q$.
Much more concise method: Note that $$Q(x) = \left(x^{1010}\right)^2 + \left(x^{1011}-x^{1009}-\frac12\right)^2 + \left(s - \frac14\right)$$ When $x$ is real, the first two terms are $\geq 0$ and the last term is $> 0$ by assumption, so $Q$ has no real zeros.