2022 AMC 10B Problem 21

$P$ 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 $\boxed{\textbf{(E)}\ 23}$.

A remainder condition is a congruence: $P \equiv 2x + 1 \pmod{x^2 + 1}$. Parametrize all such $P$ with an unknown quotient, then impose the second congruence by reducing powers of $x$ with the relation the modulus provides ($x^2 = -x - 1$, hence $x^3 = 1$). Two linear equations in the quotient's coefficients fall out. This is the polynomial version of the Chinese Remainder Theorem, and the least degree is $\deg(\text{divisor}_1) + \deg(\text{divisor}_2) - 1 = 3$ generically.

Reduce $P(x) = ax^3 + bx^2 + cx + d$ modulo each divisor directly. Mod $x^2 + 1$ ($x^2 = -1$, $x^3 = -x$): $(c - a)x + (d - b) = 2x + 1$. Mod $x^2 + x + 1$ ($x^3 = 1$): $(c - b)x + (a + d - b) = x + 2$. Solving the four equations gives $(a, b, c, d) = (1, 2, 3, 3)$.

• Trying a quadratic P first and not recognizing the contradiction, or expanding P and forgetting the constant term 3 when summing squares.
• Mis-reducing $x^3$ modulo $x^2 + x + 1$ (it equals $1$, not $-1$), which flips signs and leads to a non-integer or wrong $a$.