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Find the number of cubic polynomials <math>p(x) = x^3 + ax^2 + bx + c</math>, where <math>a</math>, <math>b</math>, and <math>c</math>
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==Problem==
are integers in <math>{−20, −10, −18, . . . , 18, 19, 20}</math>, such that there is a unique integer
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Find the number of cubic polynomials <math>p(x) = x^3 + ax^2 + bx + c,</math> where <math>a, b,</math> and <math>c</math> are integers in <math>\{-20,-19,-18,\ldots,18,19,20\},</math> such that there is a unique integer <math>m \not= 2</math> with <math>p(m) = p(2).</math>
<math>m \neq 2</math> with <math>p(m) = p(2)</math>.
 
  
Find the number of cubic polynomials <math>p(x) = x^3 + ax^2 + bx + c</math>, where <math>a</math>, <math>b</math>, and <math>c</math>
+
==Solution 1 (bash) ==
are integers in <math>{-20, -10, -18, . . . , 18, 19, 20}</math>, such that there is a unique integer
 
<math>m \neq 2</math> with <math>p(m) = p(2)</math>.
 
  
==Solution==
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Plugging <math>2</math> and <math>m</math> into <math>P(x)</math> and equating them, we get <math>8+4a+2b+c = m^3+am^2+bm+c</math>. Rearranging, we have <cmath>(m^3-8) + (m^2 - 4)a + (m-2)b = 0.</cmath> Note that the value of <math>c</math> won't matter as it can be anything in the provided range, giving a total of <math>41</math> possible choices for <math>c.</math> So what we just need to do is to just find the number of ordered pairs <math>(a, b)</math> that work, and multiply it by <math>41.</math>
 +
We can start by first dividing both sides by <math>m-2.</math> (Note that this is valid since <math>m\neq2:</math> <cmath>m^2 + 2m + 4 + (m+2)a + b = 0.</cmath> We can rearrange this so it is a quadratic in <math>m</math>: <cmath>m^2 + (a+2)m + (4 + 2a + b) = 0.</cmath> Remember that <math>m</math> has to be unique and not equal to <math>2.</math> We can split this into two cases: case <math>1</math> being that <math>m</math> has exactly one solution, and it isn't equal to <math>2</math>; case <math>2</math> being that <math>m</math> has two solutions, one being equal to <math>2,</math> but the other is a unique solution not equal to <math>2.</math>
  
Define <math>q \left( x \right) = p \left( x \right) - p \left( 2 \right)</math>.
 
Hence, for <math>q \left( x \right)</math>, beyond having a root 2, it has a unique integer root that is not equal to 2.
 
  
We have
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<math>\textbf{Case 1:}</math>
<cmath>
 
\begin{align*}
 
q \left( x \right) & = p \left( x \right) - p \left( 2 \right) \
 
& = \left( x - 2 \right)
 
\left( x^2 + \left( 2 + a \right) x + 4 + 2a + b \right) .
 
\end{align*}
 
</cmath>
 
  
Thus, the polynomial <math>x^2 + \left( 2 + a \right) x + 4 + 2a + b</math> has a unique integer root and it is not equal to 2.
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There is exactly one solution for <math>m,</math> and that solution is not <math>2.</math> This means that the discriminant of the quadratic equation is <math>0,</math> using that, we have <math>(a+2)^2 = 4(4 + 2a + b),</math> rearranging in a neat way, we have <cmath>(a-2)^2 = 4(4 + b)\Longrightarrow a = 2\pm2\sqrt{4+b}.</cmath> Using the fact that <math>4+b</math> must be a perfect square, we can easily see that the values for <math>b</math> can be <math>-4, -3, 0, 5,</math> and <math>12.</math> Also since it's a "<math>\pm</math>" there will usually be <math>2</math> solutions for <math>a</math> for each value of <math>b.</math> The two exceptions for this would be if <math>b = -4</math> and <math>b = 12.</math> For <math>b=-4</math> because it would be a <math>\pm0,</math> which only gives one solution, instead of two. And for <math>b=12</math> because then <math>a = -6</math> and the solution for <math>m</math> would equal to <math>2,</math> and we don't want this. (We can know this by putting the solutions back into the quadratic formula).  
  
Following from Vieta' formula, the sum of two roots of this polynomial is <math>- 2 - a</math>.
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So we have <math>5</math> solutions for <math>b,</math> each of which give <math>2</math> values for <math>a,</math> except for <math>2,</math> which only give one. So in total, there are <math>5*2 - 2 = 8</math> ordered pairs of <math>(a,b)</math> in this case.
Because <math>a</math> is an integer, if a root is an integer, the other root is also an integer.
 
Therefore, the only way to have a unique integer root is that the determinant of this polynomial is 0.
 
Thus,
 
<cmath>
 
\[
 
\left( 2 + a \right)^2 = 4 \left( 4 + 2a + b \right) . \hspace{1cm} (1)
 
\]
 
</cmath>
 
  
In addition, because two identical roots are not 2, we have
 
<cmath>
 
\[
 
2 + a \neq - 4 .
 
\]
 
</cmath>
 
  
Equation (1) can be reorganized as
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<math>\textbf{Case 2:}</math>
<cmath>
 
\[
 
4 b = \left( a - 2 \right)^2 - 16 .  \hspace{1cm} (2)
 
\]
 
</cmath>
 
  
Thus, <math>2 | a</math>. Denote <math>d = \frac{a-2}{2}</math>.
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<math>m</math> has two solutions, but exactly one of them isn't equal to <math>2.</math> This ensures that <math>1</math> of the solutions is equal to <math>2.</math>
Thus, (2) can be written as
 
<cmath>
 
\[
 
b = d^2 - 4 . \hspace{1cm} (3)
 
\]
 
</cmath>
 
  
Because <math>a \in \left\{ -20, -19, -18, \cdots , 18, 19, 20 \right\}</math>, <math>2 | a</math>, and <math>2 + a \neq -4</math>, we have <math>d \in \left\{ - 11, - 10, \cdots, 9 \right\} \backslash \left\{ 4 \right\}</math>.
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Let <math>r</math> be the other value of <math>m</math> that isn't <math>2.</math> By Vieta:
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<cmath>\begin{align*}
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r+2 &= -a-2\
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2r &= 4+2a+b.
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\end{align*}</cmath> From the first equation, we subtract both sides by <math>2</math> and double both sides to get <math>2r = -2a - 8</math> which also equals to <math>4+2a+b</math> from the second equation. Equating both, we have <math>4a + b + 12 = 0.</math> We can easily count that there would be <math>11</math> ordered pairs <math>(a,b)</math> that satisfy that.
  
Therefore, we have the following feasible solutions for <math>\left( b, d \right)</math>: <math>\left( -4 , 0 \right)</math>, <math>\left( -3 , \pm 1 \right)</math>, <math>\left( 0 , \pm 2 \right)</math>, <math>\left( 5, \pm 3 \right)</math>, <math>\left( 12 , 4 \right)</math>.
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However, there's an outlier case in which <math>r</math> happens to also equal to <math>2,</math> and we don't want that. We can reverse engineer and find out that <math>r=2</math> when <math>(a,b) = (-6, 12),</math> which we overcounted. So we subtract by one and we conclude that there are <math>10</math> ordered pairs of <math>(a,b)</math> that satisfy this case.
Thus, the total number of <math>\left( b, d \right)</math> is 8.
 
  
Because <math>c</math> can take any value from <math>\left\{ -20, -19, -18, \cdots , 18, 19, 20 \right\}</math>, the number of feasible <math>c</math> is 41.
 
  
Therefore, the number of <math>\left( a, b, c \right)</math> is <math>8 \cdot 41 = \boxed{\textbf{(328) }}</math>.
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This all shows that there are a total of <math>8+10 = 18</math> amount of ordered pairs <math>(a,b).</math> Multiplying this by <math>41</math> (the amount of values for <math>c</math>) we get <math>18\cdot41=\boxed{738}</math> as our final answer.
  
 +
~s214425
  
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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==Solution 2 (factor the difference)==
  
==Solution==
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<math>p(x)-p(2)</math> is a cubic with at least two integral real roots, therefore it has three real roots, which are all integers.
  
It can be easily noticed that <math>c</math> is independent of the condition <math>P(m) = P(2)</math>, and can thus safely take all <math>41</math> possible values between <math>-20</math> and <math>20</math>.
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There are exactly two distinct roots, so either <math>p(x)=p(2)+(x-2)^2(x-m)</math> or <math>p(x)=p(2)+(x-2)(x-m)^2</math>, with <math>m\neq 2</math>.
  
There are two possible ways for <math>m\ne2</math> to be the only integer satisfying <math>P(m) = P(2)</math>: <math>P</math> has a double root at <math>2</math> or a double root at <math>m</math>.
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In the first case <math>p(x)=x^3-(4+m)x^2+(4+4m)x-4m+p(2)</math>, with <math>|4+4m|\leq 20</math> (which entails <math>|4+m|\leq 20</math>), so <math>m</math> can be <math>-6,-5,-4,-3,-2,-1,0,1, (\textbf{not 2}!), 3,4</math> and <math>-4m+p(2)</math> can be any integer from <math>-20</math> to <math>20</math>, giving <math>410</math> polynomials (since the coefficients are given by linear functions of <math>m</math> and thus are distinct).  
  
Case 1: <math>P</math> has a double root at <math>2</math>:
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In the second case <math>p(x)=x^3-(2+2m)x^2+(4m+m^2)x-2m^2+p(2)</math>, and <math>m</math> can be <math>-6,-5,-4,-3,-2,-1,0,1</math> and <math>-4m+p(2)</math> can be any integer from <math>-20</math> to <math>20</math>, giving <math>328</math> polynomials.
  
In this case, <math>\frac{dP}{dx}(2) = 0</math>, or <math>12 + 4a + b = 0</math>. Thus <math>a</math> ranges from <math>-8</math> to <math>2</math>. One of these values, <math>(a,b) = (-6,-12)</math> corresponds to a triple root at <math>2</math>, which means <math>m=2</math>. Thus there are <math>10</math> possible values of <math>(a,b)</math>. (It can be verified that <math>m</math> is an integer).
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The total is <math>\boxed{738}</math>.
  
Case 2: <math>P</math> has a double root at <math>m</math>:
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~EVIN-
  
See the above solution. This yields <math>8</math> possible combinations of <math>a</math> and <math>b</math>.
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==Video Solution==
 +
https://youtu.be/-Asb_5nTgSg
  
Thus, in total we have <math>41*18 = \boxed{738}</math> combinations of <math>(a,b,c)</math>.
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~MathProblemSolvingSkills.com
  
  
  
-Alex_Z
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==See also==
 +
{{AIME box|year=2023|num-b=8|num-a=10|n=I}}
 +
 
 +
[[Category:Intermediate Algebra Problems]]
 +
{{MAA Notice}}

Revision as of 19:40, 27 December 2023

Problem

Find the number of cubic polynomials $p(x) = x^3 + ax^2 + bx + c,$ where $a, b,$ and $c$ are integers in $\{-20,-19,-18,\ldots,18,19,20\},$ such that there is a unique integer $m \not= 2$ with $p(m) = p(2).$

Solution 1 (bash)

Plugging $2$ and $m$ into $P(x)$ and equating them, we get $8+4a+2b+c = m^3+am^2+bm+c$. Rearranging, we have \[(m^3-8) + (m^2 - 4)a + (m-2)b = 0.\] Note that the value of $c$ won't matter as it can be anything in the provided range, giving a total of $41$ possible choices for $c.$ So what we just need to do is to just find the number of ordered pairs $(a, b)$ that work, and multiply it by $41.$ We can start by first dividing both sides by $m-2.$ (Note that this is valid since $m\neq2:$ \[m^2 + 2m + 4 + (m+2)a + b = 0.\] We can rearrange this so it is a quadratic in $m$: \[m^2 + (a+2)m + (4 + 2a + b) = 0.\] Remember that $m$ has to be unique and not equal to $2.$ We can split this into two cases: case $1$ being that $m$ has exactly one solution, and it isn't equal to $2$; case $2$ being that $m$ has two solutions, one being equal to $2,$ but the other is a unique solution not equal to $2.$


$\textbf{Case 1:}$

There is exactly one solution for $m,$ and that solution is not $2.$ This means that the discriminant of the quadratic equation is $0,$ using that, we have $(a+2)^2 = 4(4 + 2a + b),$ rearranging in a neat way, we have \[(a-2)^2 = 4(4 + b)\Longrightarrow a = 2\pm2\sqrt{4+b}.\] Using the fact that $4+b$ must be a perfect square, we can easily see that the values for $b$ can be $-4, -3, 0, 5,$ and $12.$ Also since it's a "$\pm$" there will usually be $2$ solutions for $a$ for each value of $b.$ The two exceptions for this would be if $b = -4$ and $b = 12.$ For $b=-4$ because it would be a $\pm0,$ which only gives one solution, instead of two. And for $b=12$ because then $a = -6$ and the solution for $m$ would equal to $2,$ and we don't want this. (We can know this by putting the solutions back into the quadratic formula).

So we have $5$ solutions for $b,$ each of which give $2$ values for $a,$ except for $2,$ which only give one. So in total, there are $5*2 - 2 = 8$ ordered pairs of $(a,b)$ in this case.


$\textbf{Case 2:}$

$m$ has two solutions, but exactly one of them isn't equal to $2.$ This ensures that $1$ of the solutions is equal to $2.$

Let $r$ be the other value of $m$ that isn't $2.$ By Vieta: \begin{align*} r+2 &= -a-2\\ 2r &= 4+2a+b. \end{align*} From the first equation, we subtract both sides by $2$ and double both sides to get $2r = -2a - 8$ which also equals to $4+2a+b$ from the second equation. Equating both, we have $4a + b + 12 = 0.$ We can easily count that there would be $11$ ordered pairs $(a,b)$ that satisfy that.

However, there's an outlier case in which $r$ happens to also equal to $2,$ and we don't want that. We can reverse engineer and find out that $r=2$ when $(a,b) = (-6, 12),$ which we overcounted. So we subtract by one and we conclude that there are $10$ ordered pairs of $(a,b)$ that satisfy this case.


This all shows that there are a total of $8+10 = 18$ amount of ordered pairs $(a,b).$ Multiplying this by $41$ (the amount of values for $c$) we get $18\cdot41=\boxed{738}$ as our final answer.

~s214425

Solution 2 (factor the difference)

$p(x)-p(2)$ is a cubic with at least two integral real roots, therefore it has three real roots, which are all integers.

There are exactly two distinct roots, so either $p(x)=p(2)+(x-2)^2(x-m)$ or $p(x)=p(2)+(x-2)(x-m)^2$, with $m\neq 2$.

In the first case $p(x)=x^3-(4+m)x^2+(4+4m)x-4m+p(2)$, with $|4+4m|\leq 20$ (which entails $|4+m|\leq 20$), so $m$ can be $-6,-5,-4,-3,-2,-1,0,1, (\textbf{not 2}!), 3,4$ and $-4m+p(2)$ can be any integer from $-20$ to $20$, giving $410$ polynomials (since the coefficients are given by linear functions of $m$ and thus are distinct).

In the second case $p(x)=x^3-(2+2m)x^2+(4m+m^2)x-2m^2+p(2)$, and $m$ can be $-6,-5,-4,-3,-2,-1,0,1$ and $-4m+p(2)$ can be any integer from $-20$ to $20$, giving $328$ polynomials.

The total is $\boxed{738}$.

~EVIN-

Video Solution

https://youtu.be/-Asb_5nTgSg

~MathProblemSolvingSkills.com


See also

2023 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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