2022 AMC 10B Problem 21

The first remainder condition gives $$\begin{aligned}P(x)&=(x^2+x+1)Q(x)\\&\quad+x+2.\end{aligned}$$ for some polynomial $Q.$ Modulo $x^2+1,$ we have $x^2\equiv-1,$ so $$P(x)\equiv xQ(x)+x+2.$$ If $Q(x)=c$ is constant, this remainder is $(c+1)x+2,$ which cannot equal $2x+1.$ Thus $Q$ must have degree at least $1.$ Now let $Q(x)=ax+b.$ Reducing modulo $x^2+1$ gives $$P(x)\equiv(b+1)x+(2-a).$$ Matching this with $2x+1$ yields $a=b=1.$ This constructs a degree-$3$ polynomial, and the failed constant case proves that degree $3$ is minimal. Therefore, $$\begin{aligned}P(x)&=(x+1)(x^2+x+1)\\&\quad+x+2\\&=x^3+2x^2+3x+3.\end{aligned}$$ The sum of the squares of its coefficients is $1^2+2^2+3^2+3^2=23.$ Thus, the answer is E .