2022 AMC 10B Problem 21

A shortcut for remainders. Since $x^2 = (x^2 + 1) - 1$, every $x^2$ is "$-1$ plus a multiple of $x^2 + 1$". So to find the remainder after dividing by $x^2 + 1$, you may replace $x^2$ by $-1$ (and so $x^3 = x \cdot x^2$ by $-x$).

For $x^2 + x + 1$ the same idea says replace $x^2$ by $-x - 1$. Also $x^3 - 1 = (x - 1)(x^2 + x + 1)$ is a multiple of $x^2 + x + 1$, so replace $x^3$ by $1$.

1. Try a cubic. Two remainder conditions, each with two numbers to match, give four equations, so a polynomial with four coefficients is the natural first guess:
$$P(x) = ax^3 + bx^2 + cx + d.$$
2. Divide by $x^2 + 1$. Replace $x^3$ by $-x$ and $x^2$ by $-1$:
$$-ax - b + cx + d = (c - a)x + (d - b).$$
This must be $2x + 1$, so $c - a = 2$ and $d - b = 1$.
3. Divide by $x^2 + x + 1$. Replace $x^3$ by $1$ and $x^2$ by $-x - 1$:
$$a - bx - b + cx + d = (c - b)x + (a + d - b).$$
This must be $x + 2$, so $c - b = 1$ and $a + d - b = 2$.
4. Solve. From $d - b = 1$ and $a + (d - b) = 2$ we get $a = 1$. Then $c = a + 2 = 3$, $b = c - 1 = 2$, and $d = b + 1 = 3$.
5. Nothing smaller works. A polynomial of degree $2$ or less is the case $a = 0$. But step 4 showed $a$ must be $1$, so no smaller polynomial fits. The least-degree polynomial is
$$P(x) = x^3 + 2x^2 + 3x + 3.$$
6. Check. Replace $x^2$ by $-1$: $-x - 2 + 3x + 3 = 2x + 1$. Replace $x^3$ by $1$ and $x^2$ by $-x - 1$: $1 - 2x - 2 + 3x + 3 = x + 2$. Both match.
7. Sum of squares. $1^2 + 2^2 + 3^2 + 3^2 = 1 + 4 + 9 + 9 = 23$.

The answer is $\boxed{\textbf{(E)}\ 23}$.

The remainder only cares about what is left after taking away multiples of the divisor, so you can swap $x^2$ (or $x^3$) for something simpler that differs from it by such a multiple. This turns each remainder condition into two plain equations. Two conditions give four equations, which is why a polynomial with four coefficients, a cubic, is the smallest that can satisfy both.

Build $P$ from the second condition. Any $P$ with remainder $2x + 1$ after dividing by $x^2 + 1$ has the form $P(x) = (x^2 + 1)(px + q) + 2x + 1$ for a cubic. Multiplying out gives $px^3 + qx^2 + (p + 2)x + (q + 1)$. Now replace $x^3$ by $1$ and $x^2$ by $-x - 1$: the remainder is $(p - q + 2)x + (p + 1)$. Matching $x + 2$ gives $p = 1$ and $q = 2$, so again $P(x) = x^3 + 2x^2 + 3x + 3$.

$P$ cannot have degree $\le 1$, since then both remainders would equal $P$ itself, but $x + 2 \ne 2x + 1$.

The second condition says $P(x) = (x^2 + 1)Q(x) + 2x + 1$. Try the smallest $Q$.

If $Q = c$ is constant, $P = cx^2 + 2x + c + 1$. Modulo $x^2 + x + 1$ we may replace $x^2$ by $-x - 1$: $P \equiv c(-x-1) + 2x + c + 1 = (2 - c)x + 1$. The constant term $1$ can never match the required $2$, so no quadratic works.

Let $Q = ax + b$:
$$P(x) = (x^2+1)(ax+b) + 2x + 1 = ax^3 + bx^2 + (a+2)x + (b+1).$$
Modulo $x^2 + x + 1$ use $x^2 \equiv -x - 1$ and $x^3 = x \cdot x^2 \equiv -x^2 - x \equiv (x + 1) - x = 1$:
$$P \equiv a + b(-x-1) + (a+2)x + (b+1) = (a - b + 2)x + (a + 1).$$
Matching with $x + 2$: $a + 1 = 2$ gives $a = 1$, then $1 - b + 2 = 1$ gives $b = 2$.

So $P(x) = x^3 + 2x^2 + 3x + 3$, and the sum of the squares of its coefficients is $1 + 4 + 9 + 9 = 23$.

The answer is $\boxed{\textbf{(E)}\ 23}$.