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

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.

  • $\begingroup$ Clearly, it has solutions. A trivial solution is $x=0$. $\endgroup$ – Prasun Biswas Oct 5 '17 at 19:01
  • $\begingroup$ Resultant[2x2-3,3x-1,x] gives 25. $\endgroup$ – lhf Oct 5 '17 at 19:05
  • 2
    $\begingroup$ 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 $b$. $\endgroup$ – Prasun Biswas Oct 5 '17 at 19:05

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.

  • $\begingroup$ 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 $0$? $\endgroup$ – sequence Oct 24 '19 at 0:36
  • $\begingroup$ @sequence, see en.wikipedia.org/wiki/Factor_theorem $\endgroup$ – lhf Oct 24 '19 at 0:52
  • $\begingroup$ I know this theorem, but I see no connection with this problem. Can you please clarify? @lhf $\endgroup$ – sequence Oct 24 '19 at 1:00

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