Checking when quotient of polynomials is integer

Question

I'm learning some elementary number theory, and seek integers $n$ for which $\frac{n^3 - 3n^2 + 4}{2n-1}$ is an integer. My instinct was to let such an integer be denoted by $k$ and get the condition $n^3 - 3n^2 - 2kn + (4+k)=0$ then try to factorize to get a condition on $n$. However, I couldn't see how to factorise. In the solutions, instead, they start by observing that multiplying the fraction by $8$ means its still an integer, and seeing that $8 \cdot \frac{n^3 - 3n^2 + 4}{2n-1} = 4n^2 - 10n -5 + \frac{27}{2n-1}$ and checking when $2n-1$ divides $27$. My question is: why was my instinct to factorize incorrect, and how on Earth would one think to observe that multipling by eight still gives an integer. I understand their solution, I'm particularly looking for motivation or a perspective that makes their approach obvious so I can apply similar techniques in future.

Answer 1

I'm particularly looking for motivation or a perspective that makes their approach obvious so I can apply similar techniques in future.

Easy general way: $ $ via a $ $ fractional $ $ generalization of the polynomial remainder / factor theorems. If $\,a,b,n,m\,$ are integers and $\,f(x)\,$ is a polynomial with integer coefficients then

$$\begin{align}\,n\!-\!a\mid f(n) &\iff \ \ n\!-\!a\mid f(a),\ \ \ {\rm by}\ \ \ f(n)\equiv f(a)\!\pmod{\!n\!-\!a}\\[.6em] \leadsto\ \ bn\!-\!a\mid f(n) &\iff bn\!-\!a\mid b^k f\left(\frac{a}b\right),\ \ {\rm if}\ \ \color{#90f}{(a,b)=1},\ \ k\ge \deg f \end{align}\qquad\qquad$$

since $\,(bn\!-\!a,b) = \color{#90f}{(-a,b)}=1,\ $ so $\ bn\!-\!a\mid b^k m \iff bn\!-\!a\mid m,\,$ by Euclid's Lemma.

so $\,2n\!-\!1\mid n^3\!-\!3n^2\!+\!4 \iff \color{#0a0}{2n}\!-\!\color{#c00}1\mid 2^3(n^3\!-\!3n^2\!+\!4) = \underbrace{(\color{#0a0}{2n})^3\!-\!6(\color{#0a0}{2n})^2\!+\!32}_{\large\equiv\ \color{#c00}1^3-6(\color{#c00}1)^2+32\ \equiv\ 27}$

Answer 2

Well, if the fraction is an integer, the GCD of numerator and denominator is large.... the outcome of extended GCD is, after multiplying by 27,

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$$ 8 \left( x^{3} - 3 x^{2} + 4 \right) - \left( 2 x - 1 \right) \left( 4 x^{2} - 10 x - 5 \right) = 27 $$ then divide through by $(2x-1) $ to get to your $\frac{27}{2x-1}$

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$$ \left( x^{3} - 3 x^{2} + 4 \right) $$

$$ \left( 2 x - 1 \right) $$

$$ \left( x^{3} - 3 x^{2} + 4 \right) = \left( 2 x - 1 \right) \cdot \color{magenta}{ \left( \frac{ 4 x^{2} - 10 x - 5 }{ 8 } \right) } + \left( \frac{ 27}{8 } \right) $$ $$ \left( 2 x - 1 \right) = \left( \frac{ 27}{8 } \right) \cdot \color{magenta}{ \left( \frac{ 16 x - 8 }{ 27 } \right) } + \left( 0 \right) $$ $$ \frac{ 0}{1} $$ $$ \frac{ 1}{0} $$ $$ \color{magenta}{ \left( \frac{ 4 x^{2} - 10 x - 5 }{ 8 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ 4 x^{2} - 10 x - 5 }{ 8 } \right) }{ \left( 1 \right) } $$ $$ \color{magenta}{ \left( \frac{ 16 x - 8 }{ 27 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ 8 x^{3} - 24 x^{2} + 32 }{ 27 } \right) }{ \left( \frac{ 16 x - 8 }{ 27 } \right) } $$ $$ \left( x^{3} - 3 x^{2} + 4 \right) \left( \frac{ 8}{27 } \right) - \left( 2 x - 1 \right) \left( \frac{ 4 x^{2} - 10 x - 5 }{ 27 } \right) = \left( 1 \right) $$