Problem 2954 from REVISTA MATEMATICĂ a ELEVILOR din TIMIȘOARA

 You can see issue 1-1977 from which I took the issue (page 62), here.

A larger collection is here.

             2 954.  Determine the real numbers  $x$  for which the number

$\frac{x}{x^2-5x+7}$  is an integer.

(Matematika v Șkole)

ANSWER CiP

$x\in \{0,\;2,\;\frac{7}{3},\;3,\;\frac{7}{2}\}$

                    Solution CiP

          If  $\frac{x}{x^2-5x+7}=k\in\mathbb{Z}$  then  $x$  is the root of the quadratic equation

$kx^2-(5k+1)x+7k=0 \tag{1}$

The equation (1) has the discriminant  $\Delta_x=(5k+1)^2-28k^2=1+10k-3k^2$.

     But  $1+10k-3k^2=\frac{28}{3}-\left (\frac{25}{3}-10k+3k^2\right )=\frac{28}{3}-3\cdot \left(\frac{25}{9}-\frac{10}{3}k+k^2\right)=$

$=\frac{28}{3}-3\cdot \left (\frac{5}{3}-k\right)^2\leqslant \frac{28}{3}$,  so  

$\Delta_x\leqslant \frac{28}{3}\tag{2}$

     For the equation (1) to have rational roots, it must be  $\sqrt{\Delta_x}\in\mathbb{Q}$,  in fact  $\sqrt{\Delta_x}\in\mathbb{N}$. This happens if  (cf. (2))

$\sqrt{\Delta_x}\in\{0,\;1,\;2,\;3\} \tag{3}$

     Solving the equations  $1+10k-3k^2=0\;or\;1\;or\;2\;or\;3$  we obtain the integer values

$k\in\{0,\;2,\;3\}$

for which, we obtain from the equation (1) the values ​​of  $x$  in the answer.

$\blacksquare$

          Correction

          My answer is incomplete, as the commenter noted. 

          For  $x=3\pm\sqrt{2}$, the number  $\frac{x}{x^2-5x+7}$  takes the value  $1$, also an integer.

I wrongly assumed that  $x$  is a rational number. Let's analyze the sign of  $\Delta_x$.

          Setting the condition  $\Delta_x \geqslant 0\;\Rightarrow\; 1+10k-3k^2\geqslant 0$  which gives us the admissible values

$k\in\left ( \frac{5-\sqrt{28}}{3}\;,\;\frac{5+\sqrt{28}}{3}\right )\cap \mathbb{Z}=\{0,\;\color{Red}1,\;2,\;3\}$

Considering that the roots of the equation (1) are  $x_{1,2}=\frac{5k+1\pm\sqrt{\Delta_x}}{2k}$, for  $k=1$  we find  $x=3\pm\sqrt{2}$, so the correct answer is

$x\in \{0,\;3-\sqrt{2},\;2,\;\frac{7}{3},\;3,\;\frac{7}{2},\;3+\sqrt{2}\}$

$\color{Red}{!!!}\square \color{Red}{!!!}$