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# Need help!

$$P(x)$$ 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 $$P(x) = 2005$$.

Note by Swapnil Das
2 years, 4 months ago

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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 $$P(x) = 2005$$.

- 2 years, 4 months ago

Very nice and simple!

- 2 years, 4 months ago