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| | <math>m \neq 2</math> with <math>p(m) = p(2)</math>. | | <math>m \neq 2</math> with <math>p(m) = p(2)</math>. |
| | | | |
| − | ==Solution==
| + | 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> |
| − | | + | are integers in <math>{-20, -10, -18, . . . , 18, 19, 20}</math>, such that there is a unique integer |
| − | Define <math>q \left( x \right) = p \left( x \right) - p \left( 2 \right)</math>.
| + | <math>m \neq 2</math> with <math>p(m) = p(2)</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
| |
| − | <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.
| |
| − | | |
| − | Following from Vieta' formula, the sum of two roots of this polynomial is <math>- 2 - a</math>.
| |
| − | 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
| |
| − | <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>.
| |
| − | 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>.
| |
| − | | |
| − | 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>.
| |
| − | 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>.
| |
| − | | |
| − | | |
| − | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
| |
| | | | |
| | ==Solution== | | ==Solution== |
Find the number of cubic polynomials 
, where 
, 
, and 
are integers in ${−20, −10, −18, . . . , 18, 19, 20}$ (Error compiling LaTeX. Unknown error_msg), such that there is a unique integer
with 
.
Find the number of cubic polynomials , where , , and
are integers in 
, such that there is a unique integer
with .
Find the number of cubic polynomials , where , , and
are integers in , such that there is a unique integer
with .
Solution
It can be easily noticed that is independent of the condition 
, and can thus safely take all 
possible values between 
and 
.
There are two possible ways for 
to be the only integer satisfying : 
has a double root at 
or a double root at 
.
Case 1: has a double root at :
In this case, 
, or 
. Thus ranges from 
to . One of these values, 
corresponds to a triple root at , which means 
. Thus there are 
possible values of 
. (It can be verified that is an integer).
Case 2: has a double root at :
See the above solution. This yields 
possible combinations of and .
Thus, in total we have 
combinations of 
.
-Alex_Z
, where 
, 
, and 
are integers in ${−20, −10, −18, . . . , 18, 19, 20}$ (Error compiling LaTeX. Unknown error_msg), such that there is a unique integer
with 
.
, such that there is a unique integer
, and can thus safely take all 
possible values between 
and 
.
to be the only integer satisfying 
has a double root at 
or a double root at 
.
, or 
. Thus 
to 
corresponds to a triple root at 
. Thus there are 
possible values of 
. (It can be verified that 
possible combinations of 
combinations of 
.
