Because $Q(2022)=P(2022)$, let $Q(x)-P(x)=(x-2022)(b_m.x^m+b_{m-1}x^{m-1}+...+b_0)=b_m x^{m+1}+(-2022b_m+b_{m-1})x^m+...+(-2022b_1+b0)x+(-2022b_0)$ The coffients of $P(x)$ is not larger than $2021$ and $Q(x)$ has at least one coffient larger than $2021$, we get $Q(x)-P(x)$ is not constant (which means $m > 0$) and all of its coffients $\ge -2021$. Then, we have $b_m \ge-2021$ $-2022b_m+b_{m-1} \ge-2021$ $...$ $-2022b_1+b0\ge-2021$ $-2022b_0 \ge-2021$ Assume $b_i \ge1$ for some $i$, because $-2022b_i+b_{i-1}\ge-2021$ $=>$$ b_{i-1}\ge b_i+2021( b_i-1)\ge b_i \ge 1$ Keep doing that, we get $b_0 \ge 1$, which contradicts to $-2022b_0 \ge-2021$ Thus, $b_i \le 0$ for all $i=0,m$ $=> b_m.x^m+b_{m-1}x^{m-1}+...+b_0 < 0$ for all $x > 0$ $=> Q(n)-P(n) >0 $ for all $n=1,2,...,2021$ Now, because $P(x)$ has nonnegative coffients, so its root $\frac p q$ must $< 0$, assume $p > 0$, we get $pn-q = |p|+n|q| > 0$ for all $n=1,2,...,2021$ Since $P(\frac p q)=Q(\frac p q)=0$, let $Q(x)-P(x)=(px-q) H(x)$, $H(x)$ has interger coffients. We have $(|p|+n|q|)H(n)=Q(n)-P(n) > 0$ for all $n=1,2,...,2021$ $=> H(n) >0$ $=> H(n) \ge 1$ for all $n=1,2,...,2021$. Therefore $Q(n)-P(n)=(|p|+n|q|)H(n) \ge |p|+n|q| $ for all $n=1,2,...,2021$

Well, @above, you proved that $b_i<1$ for all $i$, but that doesn't give $b_i \leq 0$ for all $i$ since the coefficients are not integers. Correct me if I am missing something, but either the statement is missing the condition for integer coefficients, or the above solution doesn't work for non-integer coefficients.

VicKmath7 wrote: Well, @above, you proved that $b_i<1$ for all $i$, but that doesn't give $b_i \leq 0$ for all $i$ since the coefficients are not integers. Correct me if I am missing something, but either the statement is missing the condition for integer coefficients, or the above solution doesn't work for non-integer coefficients. @above Actually, the problem missed the condition "integer coefficients"

@above the original problem give the condition non-negative coefficients