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$P(x)$ is a polynomial with integral coefficients. $P(x) = 4010$ for $5$ different integral values of $x$ . Prove that there is no integer $x$ such that $P(x) = 2005$ .

#Algebra

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Comments

Let $P(x) = (x-a_1)(x-a_2)(x-a_3)(x-a_4)(x-a_5)Q(x)$ for $a_1 < a_2 < a_3 < a_4 < a_5, \quad a_i \in \mathbb Z$ , and where coefficients of $Q(x)$ are also integral.

Suppose for an integer $x= \alpha$ , let $P(x) = P(\alpha) = 4010$ .

Now We can represent $4010 = (-1)(1)(2)(5)(401) \cdot (-1)$ as product of five different integers, and that extra $-1$ can be plugged into $Q(x)$ to make it as $P(\alpha) = (\alpha-a_1)(\alpha-a_2)(\alpha-a_3)(\alpha-a_4)(\alpha-a_5) \cdot R(\alpha)$ where $R(x) = -Q(x)$ having integral coefficients as well.

Thus, you can very well compare the five distinct integral factors of $4010$ with five $(\alpha - a_i)$ (s) to see that.

Now, incase of 2005, it can be expressed as $(-1)(1)(5)(401)$ , product of four distinct integers only. Thus $(x-a_1)(x-a_2)(x-a_3)(x-a_4)(x-a_5) = (-1)(1)(5)(401)$ cannot satisfy as there are five integers multiplied in the left whereas four integers are multiplied in the right. Hence, there is no integer $x$ such that $P(x) = 2005$ .

Satyajit Mohanty - 5 years, 9 months ago

Very nice and simple!

Adarsh Kumar - 5 years, 9 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$