Find all integers $x$ such that $\frac{2x^2-3}{3x-1}$ is an integer.

Question

Well if this is an integer then $3x-1 \mid 2x^2-3$ so $2x^2-3=3x-1(k)$ such that $k\in \mathbb{Z}$ from here not sure where to go I know that it has no solutions I just can't see the contradiction yet.

Use long division to get $2x^2-3=q(x)(3x-1)+r(x)$ where $r(x)$ is a constant polynomial, say $r(x)=b$. Then, note that $3x-1\mid r(x)=b$. Thus the values of $x$ required would be such that $3x-1$ is a factor of $b$. — Prasun Biswas, Oct 05 '17 at 19:05

lhf · Accepted Answer · 2017-10-06T13:25:46.010

4

Using the extended Euclidean algorithm in $\mathbb Q[x]$, we get $$ 25 = (-9)(2x^2-3)+(6x+2)(3x-1) $$ Therefore, if $3x-1$ divides $2x^2-3$, then it divides $25$.

So, $3x-1 \in \{\pm 1, \pm 5, \pm 25 \}$ and there is not much to test.

An easier (but equivalent) argument is $2x^2-3 = (3x-1)q(x) + r$, where $r$ is the value of $2x^2-3$ at $x=1/3$, the root of $3x-1$. This gives $r=-25/9$. The rest follows as above.

edited Oct 06 '17 at 13:25

answered Oct 05 '17 at 19:09

lhf

221,500

Can you please clarify why does one have to set $r(x)$ to be a constant, and why is it evaluated at the root of the polynomial in the original denominator? If that polynomial is $0$ then how can there be a remainder after division by $0$? – sequence Oct 24 '19 at 00:36
@sequence, see https://en.wikipedia.org/wiki/Factor_theorem – lhf Oct 24 '19 at 00:52
I know this theorem, but I see no connection with this problem. Can you please clarify? @lhf – sequence Oct 24 '19 at 01:00

score 1 · Answer 2 · answered May 02 '21 at 09:17

1

Let $k=3x-1$ then $$3x\equiv_k 1$$ and $$2x^2-3 \equiv _k0$$ Multiplying last equation with 9 we get $$0\equiv _k2(3x)^2-27\equiv _k -25$$ So $$3x-1\mid 25 \implies 3x-1\in\{\pm 1,\pm5,\pm 25\}$$ and thus $$x\in\{0,2,-8\}$$

answered May 02 '21 at 09:17

nonuser

91,557

Find all integers $x$ such that $\frac{2x^2-3}{3x-1}$ is an integer.

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