A number theory problem by Thành Đạt Lê

Find the smallest value of n Z n \in \mathbb Z such that ( 2 n + 1 ) ( 2 n 2 n + 2 ) (2n + 1) \space | \space (2n^{2} - n + 2) .


The answer is -2.

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1 solution

Marco Brezzi
Aug 11, 2017

We are given that 2 n + 1 2 n 2 n + 2 n Z 2n+1|2n^2-n+2 \quad n\in\mathbb{Z}

In other words:

2 n 2 n + 2 2 n + 1 Z \dfrac{2n^2-n+2}{2n+1}\in\mathbb{Z}

Let's simplify the fraction

2 n 2 n + 2 2 n + 1 = 2 n 2 + n 2 n + 2 2 n + 1 = n + 2 n + 2 2 n + 1 = n + 2 n 1 + 3 2 n + 1 = n 1 + 3 2 n + 1 \begin{aligned} \dfrac{2n^2-n+2}{2n+1}&=\dfrac{\mathbin{\color{#D61F06}2n^2+n}-2n+2}{2n+1}\\ &=\mathbin{\color{#D61F06}n}+\dfrac{-2n+2}{2n+1}\\ &=n+\dfrac{\mathbin{\color{#3D99F6} -2n-1}+3}{2n+1}\\ &=n\mathbin{\color{#3D99F6}-1}+\dfrac{3}{2n+1} \end{aligned}

So we need to have 2 n + 1 3 2n+1|3 . The smallest divisor of 3 3 is 3 -3

2 n + 1 = 3 n = 2 \Longrightarrow 2n+1=-3 \iff n=\boxed{-2}

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