Find all integers such that $2n^2+1$ divides $n^3+9n-17$

Question

Find all integers such that $2n^2+1$ divides $n^3+9n-17$.

Answer : $n=(2 \ and \ 5)$

I did it.

As $2n^2+1$ divides $n^3+9n-17$, then $2n^2+1 \leq n^3+9n-17 \implies n \geq 2$

So $n =2$ is solution and doesn't exist solution when n<2. How can I do now to find 5 ? Or better, how can you solve this with another good method ?

Thanks

Answer 1

HINT:

If integer $d$ divides $n^3+9n-17,2n^2+1$

$d$ must divide $2(n^3+9n-17)-n(2n^2+1)=17n-34$

$d$ must divide $17(2n^2+1)-2n(17n-34)=68n+17$

$d$ must divide $68n+17-4(17n-34)=153$

So the necessary condition is $2n^2+1$ must divide $153$

$\implies2n^2+1\le153\iff n^2\le76\iff-9<n<9$

Answer 2

The elimination in Lab's answer essentially employs the extended Euclidean algorithm to compute a Bezout identity. As usual, when we need only the result, such calculations can be simplified using modular arithmetic (so simple they can often be done mentally with practice). For example

$$\begin{eqnarray} 2n^2+1\ &\,\mid &\, n^3+9n-17\\[.2em] \Rightarrow\ {\rm mod}\,\ \color{#0c0}{2n^2}\!+\color{#c00}1\!:\ \ 0 &\equiv&\, 2\,(n^3+\ 9n\,-\,17)\\[.2em] &\equiv&\, \color{#0a0}{2n^2}n + 18n -34\\[.2em] &\equiv&\, 17(n\!-\!2)\ \ \ {\rm by}\ \ \ \color{#0a0}{2n^2}\equiv -\color{#c00}1\\[.2em] 2(\color{#90f}{n\!+\!2})\times{\rm prior}\,\Rightarrow\ \ 0 &\equiv&\, 34(n\!-\!2)(\color{#90f}{n\!+\!2})\\[.1em] &\equiv&\, 17\cdot\color{#0a0}{2n^2}-34\cdot 4\equiv -153 \end{eqnarray}\qquad\qquad$$

Remark $\ $ The multiplication by the factor $\,2(\color{#90f}{n\!+\!2})\,$ in the 2nd last step is not ad-hoc. Rather, it corresponds to taking the "norm" of a quadratic number, since here we have that $\, \color{#0a0}{n^2} \equiv -1/2\,$ so, ring-theoretically, $\,n\,$ behaves like $\,\sqrt{-1/2} = \sqrt{-2}/2,\,$ so we can do things like taking "norms" of expressions involving the "surd" $\,n,\,$ in order to get a simpler multiple, by mapping "irrational" objects into "rational" objects - as I did above, viz. $\,0\equiv \alpha\,\Rightarrow\, 0\equiv 2 N(\alpha) = 2\alpha\alpha'\equiv -153.$

We can also do the above computation using a slight generalization of the Euclidean algorithm that allows us to scale by integers coprime to the gcd.

The reason why this works will become clearer when one learns about various universal viewpoints of ring-theoretic objects. To me this is one of the most beautiful aspects of algebra - dramatically illustrating the success and power of the abstraction of algebraic structures such as rings.

Answer 3

hint: the quotient $$\frac{n^3+9n-17}{2n^2+1}$$ can be written as $$n/2+1/2\,{\frac {17\,n-34}{2\,{n}^{2}+1}}$$

Answer 4

If $2n^2 +1$ divides $n^3 + 9n -17$, then it also divides $2\cdot \left(n^3 + 9n -17\right)$, and thus, it also divides $2\cdot \left(n^3 + 9n -17\right) - n \cdot \left(2n^2 + 1\right)$ which is equal to $17n-34$.

If $17n - 34 < 2n^2+1$, then we are already done as $17n-34$ would be the remainder when $2n^2 +1$ divides $2\cdot \left(n^3 + 9n -17\right)$, and thus, $17n - 34 = 0$ or $n=2$. Checking shows that for $n=2$, $2n^2+1 = 9$ and $n^3+9n-17 = 9$; therefore, the condition is satisfied.

In case $17n - 34 \geq 2n^2+1$, we can solve the inequality to find that $n \in \left[\frac{7}{2}, 5\right]$. As $n$ is integer, $n \in \left\{4, 5 \right\}$. This is a sufficiently small set of numbers to test whether the resulting values satisfy the conditions. On checking, we find that for $n = 4$, $2n^2 + 1 = 33$, and $n^3 + 9n - 17 = 83$, and for $n=5$, $2n^2+1 = 51$, and $n^3 + 9n - 17 = 153$. Thus, $n = 5$ is the only other solution to satisfy the condition.