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Define the polynomial $P$ with the least possible degree which satisfies the following conditions:

(a) All coefficients of $P$ are integers,
(b) All roots of $P$ are integers,
(c) $P(0)=-1$,
(d) $P(3)=128$.
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Let's analyze the given conditions:
(a) All coefficients of $P$ are integers.
(b) All roots of $P$ are integers. Let the integer roots be $r_1, r_2, \ldots, r_n$. Then, we can express $P(x)$ as:
$P(x)=a\left(x-r_1\right)\left(x-r_2\right) \cdots\left(x-r_n\right)$, where $a$ is an integer constant.
(c) $P(0)=-1$. When we plug in $x=0$, we get:
$-1=P(0)=a\left(-r_1\right)\left(-r_2\right) \cdots\left(-r_n\right)$
(d) $P(3)=128$. When we plug in $x=3$, we get:
$128=P(3)=a\left(3-r_1\right)\left(3-r_2\right) \cdots\left(3-r_n\right)$
Now, let's find the least degree polynomial that satisfies these conditions. Since $P(0)=$ -1 , we know that the product of the roots must be -1 , which means there must be an odd number of roots. The simplest case would be a polynomial with one integer root.
Let's assume the polynomial has a degree of 1 . In this case, it would be a linear polynomial of the form:
$P(x)=a\left(x-r_1\right)$

From condition (c), we know that:
$-1=P(0)=a\left(-r_1\right)$
We also know that $P(3)=128$, so:
$128=P(3)=a\left(3-r_1\right)$
Now, we need to find the integers $a$ and $r_1$ that satisfy these equations. From the first equation, we have:
$a=\frac{-1}{-r_1}=\frac{1}{r_1}$
However, since $a$ must be an integer, this implies that $r_1=1$, so we have:
$P(x)=a(x-1)$
Now, we plug in $x=3$ into the polynomial and set it equal to 128 :
$128=a(3-1) \Rightarrow 128=2 a \Rightarrow a=64$
Thus, we find the polynomial with the least possible degree that satisfies the conditions:
$P(x)=64(x-1)$
by Diamond (89,043 points)

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