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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 \(P(x) = 2005\).

Note by Swapnil Das
3 years 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 \(P(x) = 2005\).

Satyajit Mohanty - 3 years ago

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Very nice and simple!

Adarsh Kumar - 3 years ago

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