AMC 10 · 2007 · #20
Grade 8 algebraSuppose that the number a satisfies the equation 4=a+a−1. What is the value of a4+a−4?
Pick an answer.
AMC 10 2007 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A number $a$ satisfies $a + a^{-1} = 4$. Find the value of $a^4 + a^{-4}$.
Givens: The number $a$ obeys the equation $a + a^{-1} = 4$; Here $a^{-1}$ means $\frac{1}{a}$, so $a$ and its reciprocal add to $4$; Answer choices: (A) $164$, (B) $172$, (C) $192$, (D) $194$, (E) $212$
Unknowns: The value of $a^4 + a^{-4}$, the fourth power of $a$ plus the fourth power of its reciprocal
Understand
Restated: A number $a$ satisfies $a + a^{-1} = 4$. Find the value of $a^4 + a^{-4}$.
Givens: The number $a$ obeys the equation $a + a^{-1} = 4$; Here $a^{-1}$ means $\frac{1}{a}$, so $a$ and its reciprocal add to $4$; Answer choices: (A) $164$, (B) $172$, (C) $192$, (D) $194$, (E) $212$
Plan
Primary tool: #16 Change Focus / Count the Complement
Secondary: #7 Identify Subproblems, #5 Look for a Pattern
The direct route, solving $a + a^{-1} = 4$ for $a$, gives the ugly irrational $a = 2 \pm \sqrt{3}$, and raising that to the fourth power by hand is painful. Tool #16 (Change Focus) redirects the goal: never find $a$ at all, and instead work only with the symmetric quantities $a^n + a^{-n}$. The engine that makes this work is squaring, because $a \cdot a^{-1} = 1$ makes the middle term of every square a clean constant $2$. Tool #7 (Identify Subproblems) turns the leap from power $1$ to power $4$ into two easy rungs, $a + a^{-1} \to a^2 + a^{-2} \to a^4 + a^{-4}$. Tool #5 (Look for a Pattern) recognizes that both rungs are the exact same move — square, then subtract $2$ — so the second step needs no new idea.
Execute — Answer: D
8.EE.A.1 Step 1 Aim at the powers, not at a
- Do not solve for $a$.
- Instead notice what happens when you square a quantity of the form $x + x^{-1}$: the outer terms become $x^2$ and $x^{-2}$, and the middle term is $2 \cdot x \cdot x^{-1} = 2 \cdot 1 = 2$, a fixed constant, because a number times its reciprocal is $1$.
- So squaring $x + x^{-1}$ produces $x^2 + x^{-2}$ plus a tidy $+2$, lifting the exponent from $1$ to $2$.
- Repeating this squaring will lift $2$ to $4$, which is exactly the ladder we need.
💡 Because $a$ times $a^{-1}$ is $1$, squaring always leaves a clean $+2$ and doubles the exponent, so the messy value of $a$ never has to appear.
6.EE.A.3 Step 2 Square once to reach power 2
- Square both sides of $a + a^{-1} = 4$.
- By the identity from Step 1 the left side becomes $a^2 + a^{-2} + 2$, and the right side becomes $4^2 = 16$.
- So $a^2 + a^{-2} + 2 = 16$.
- Subtracting $2$ from both sides isolates the power-2 sum: $a^2 + a^{-2} = 14$.
💡 One squaring converts what we know about power $1$ into a fact about power $2$, giving the stepping stone $14$ for the next jump.
6.EE.B.7 Step 3 Square again to reach power 4
- Apply the very same move to $a^2 + a^{-2} = 14$.
- Squaring the left side gives $a^4 + a^{-4} + 2$ (the middle term is again $2 \cdot a^2 \cdot a^{-2} = 2$), and squaring the right side gives $14^2 = 196$.
- So $a^4 + a^{-4} + 2 = 196$, and subtracting $2$ leaves $a^4 + a^{-4} = 194$.
- That is choice (D).
💡 The second rung is identical to the first — square, then subtract $2$ — so no new idea is needed to climb from power $2$ to power $4$.
8.EE.A.1 Do not solve for $a$. Instead notice what happens when you square a quantity of 6.EE.A.3 Square both sides of $a + a^{-1} = 4$. By the identity from Step 1 the left side 6.EE.B.7 Apply the very same move to $a^2 + a^{-2} = 14$. Squaring the left side gives $a Review
Reasonableness: Check with the actual value of $a$. The equation $a + a^{-1} = 4$ becomes $a^2 - 4a + 1 = 0$, so $a = 2 + \sqrt{3}$ (its reciprocal is $2 - \sqrt{3}$). Then $a^2 = 7 + 4\sqrt{3}$ and $a^{-2} = 7 - 4\sqrt{3}$, which sum to $14$, matching Step 2. Squaring again, $a^4 = 97 + 56\sqrt{3}$ and $a^{-4} = 97 - 56\sqrt{3}$, which sum to $194$. The irrational parts cancel exactly, confirming $a^4 + a^{-4} = 194$. The number is also comfortably inside the answer range and just below $14^2 = 196$, as expected since $a^4 + a^{-4}$ is $14^2$ minus $2$.
Alternative: Use a running recurrence for $p_n = a^n + a^{-n}$. Multiplying $a + a^{-1} = 4$ through shows $p_n = 4\,p_{n-1} - p_{n-2}$, starting from $p_0 = a^0 + a^0 = 2$ and $p_1 = 4$. Then $p_2 = 4(4) - 2 = 14$, $p_3 = 4(14) - 4 = 52$, and $p_4 = 4(52) - 14 = 194$. This reaches the same $194$ using only additions and multiplications, with no squaring at all.
CCSS standards used (min grade 8)
8.EE.A.1Know and apply the properties of integer exponents (Reading $a^{-1}$ and $a^{-4}$ as reciprocal powers and using $a \cdot a^{-1} = 1$ to see that the cross term of each square is the constant $2$.)6.EE.A.3Apply the properties of operations to generate equivalent expressions (Expanding $\left(a + a^{-1}\right)^2$ into the equivalent expression $a^2 + a^{-2} + 2$ when squaring both sides.)6.EE.B.7Solve real-world problems by writing and solving equations of the form px = q (Treating each power sum as a single unknown and solving the one-step equations $a^2 + a^{-2} + 2 = 16$ and $a^4 + a^{-4} + 2 = 196$ to isolate $14$ and then $194$.)
⭐ To find a high power of a number plus its reciprocal, don't solve for the number — just square the equation, because the reciprocals multiply to $1$ and leave a clean $+2$ while the exponent doubles.
⭐ To find a high power of a number plus its reciprocal, don't solve for the number — just square the equation, because the reciprocals multiply to $1$ and leave a clean $+2$ while the exponent doubles.
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