2022 AMC 10B Problem 21

P21-25~5 minPrint

Polynomials Algebraic Manipulation Set up the equation/formula and compute; no special trick needed Substitute to simplify (u = x+1/x, shifting, scaling)

Let $P(x)$ be a polynomial with rational coefficients such that when $P(x)$ is divided by the polynomial $x^2 + x + 1$ , the remainder is $x + 2$ , and when $P(x)$ is divided by the polynomial $x^2 + 1$ , the remainder is $2x + 1$ . There is a unique polynomial of least degree with these two properties. What is the sum of the squares of the coefficients of that polynomial?

Show solution without answering

Key insight

For the remainder on dividing by x² + 1, replace x² by −1. For x² + x + 1, replace x² by −x − 1 and x³ by 1.

Solution

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

Why this works

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.

Alternative approach

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 $P(x) = x^3 + 2x^2 + 3x + 3$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Intermediate Algebra, Chapter 6: Polynomial Division (§6.2, §6.4)
Introduction to Algebra, Chapter 11: Special Factorizations (§11.2, §11.3, §11.5)
Intermediate Algebra, Chapter 20: Some Advanced Strategies (§20.2)
Intermediate Algebra, Chapter 7: Polynomial Roots Part I (§7.1, §7.2, §7.3)