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Difference between revisions of "2023 AIME I Problems/Problem 9"
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<math>P(x) = x^3 + ax^2 + bx + c</math> is a polynomial with integer coefficients between <math>-20</math> and <math>20</math>, inclusive. There is exactly one integer <math>m</math> such that <math>P(m) = P(2)</math>. How many possible values are there for the ordered triple <math>(a, b, c)</math>?
 
<math>P(x) = x^3 + ax^2 + bx + c</math> is a polynomial with integer coefficients between <math>-20</math> and <math>20</math>, inclusive. There is exactly one integer <math>m</math> such that <math>P(m) = P(2)</math>. How many possible values are there for the ordered triple <math>(a, b, c)</math>?
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==Solution==
 
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If <math>m</math> is the only integral value that satisfies <math>P(m) = P(2)</math>, we can show that <math>m</math> is the only real value that satisfies <math>P(m) = P(2)</math>.
 
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Next, we have <math>P(m) = P(2)</math>, so therefore
 
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<cmath>m^3 + am^2 + bm + c - 8 - 4a - 2b - c = 0.</cmath>
 
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We can now simplify:
 
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<cmath>m^3 - 8 + am^2 - 4a + bm - 2b = 0</cmath>
 
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<cmath>(m-2)(m^2 + 2m + 4) + a(m-2)(m+2) + b(m-2) = 0</cmath>
 
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<cmath>(m-2)(m^2 + 2m + 4 + am + 2a + b) = 0.</cmath>
 
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Since <math>m \neq 2</math>,
 
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<cmath>m^2 + (a+2)m + 2a + b + 4 = 0.</cmath>
 
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We can now apply the quadratic formula, yielding
 
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<cmath>m = \frac{-a-2 \pm \sqrt{a^2 - 4a - 4b - 12}}{2}.</cmath>
 
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For this to have exactly <math>1</math> solution, we must have <math>a^2 - 4a - 4b - 12 = 0</math>, and thus <math>(a-2)^2 = 4(b+2)</math>. This means that <math>|a| \leq 9</math>, yielding <math>9</math> solutions for <math>a</math>. For any solution of <math>a</math>, <math>b</math> can only attain one value. And, the value of <math>c</math> doesn't matter. Our answer is thus <math>9 \cdot 41 = \boxed{369}</math>.
 
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~mathboy100
 
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I believe this solution is wrong. The answer is <math>738</math>. ~r00tsOfUnity
 
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??? I checked with like 5 people ~mathboy100
 
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they probably all sillied it ~r00tsOfUnity
 

Revision as of 12:43, 8 February 2023

Problem (Unofficial, please update when official one comes out):

$P(x) = x^3 + ax^2 + bx + c$ is a polynomial with integer coefficients between $-20$ and $20$, inclusive. There is exactly one integer $m$ such that $P(m) = P(2)$. How many possible values are there for the ordered triple $(a, b, c)$?

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