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Find the sum of all positive integers $k$ for which $5x^2-2kx+1<0$ has exactly one integral solution.

My attempt is as follows:

$$\left(x-\dfrac{2k-\sqrt{4k^2-20}}{10}\right)\left(x-\dfrac{2k+\sqrt{4k^2-20}}{10}\right)<0$$ $$\left(x-\dfrac{k-\sqrt{k^2-5}}{5}\right)\left(x-\dfrac{k+\sqrt{k^2-5}}{5}\right)<0$$ $$x\in\left(\dfrac{k-\sqrt{k^2-5}}{5},\dfrac{k+\sqrt{k^2-5}}{5}\right)$$

As it is given that it has got only one integral solution, so there must be exactly one integer between $\dfrac{k-\sqrt{k^2-5}}{5}$ and $\dfrac{k+\sqrt{k^2-5}}{5}$

Let $x_1=\dfrac{k-\sqrt{k^2-5}}{5}$ and $x_2=\dfrac{k+\sqrt{k^2-5}}{5}$ , then $[x_2]-[x_1]=1$ where [] is a greater integer function.

But from here, how to proceed? Please help me in this.

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  • $\begingroup$ Do you mean: $$\text{Find the sum of all positive integers $k$ for which $5x^2-2kx+1<0$ has exactly one integral solution.}$$ ? $\endgroup$ – Servaes Mar 20 '20 at 18:56
  • $\begingroup$ yeah obviously. $\endgroup$ – user3290550 Mar 20 '20 at 18:57
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    $\begingroup$ No, not obviously, otherwise I wouldn't have asked. Your current phrasing makes no sense, this seemed like the nearest sensible interpretation. $\endgroup$ – Servaes Mar 20 '20 at 18:58
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    $\begingroup$ What you meant may be obvious. But what you said was dead wrong. $k$ is a constant. So the sum of "all" $k$ is ..... $k$. What you meant was the sum of all possible values of $k$ or as Servaes put it equivalently "Find the sum of all $k$ where...." $\endgroup$ – fleablood Mar 20 '20 at 21:18
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Your idea is good; you want to find all positive integers $k$ for which there is precisely on integer between the roots of $$5x^2-2kx+1=0.$$ Then the distance between the roots can be at most $2$, where the distance between the roots is precisely $$\frac{1}{5}\sqrt{(-2k)^2-4\cdot1\cdot5}=\frac25\sqrt{k^2-5},$$ as you already found. This is at most $2$ if and only if $\sqrt{k^2-5}\leq5$, or equivalently $k\leq5$. This leaves only $5$ values of $k$ to check.

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    $\begingroup$ you made the mistake while calculating discriminant, it should be k^2-5 $\endgroup$ – user3290550 Mar 20 '20 at 19:04
  • $\begingroup$ @user3290550 Thanks for spotting that, corrected now. $\endgroup$ – Servaes Mar 20 '20 at 19:05
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    $\begingroup$ one more thing you missed, k^2-5>=0 because otherwise discriminant would not be defined. $\endgroup$ – user3290550 Mar 20 '20 at 19:08
  • $\begingroup$ I hadn't missed that; it just leaves fewer values of $k$ to check. The discriminant is still defined though, it's just that the polynomial is then positive for every integer $x$. $\endgroup$ – Servaes Mar 20 '20 at 19:18
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Well one way of looking at the solution is for $n, k \in Z$ it will have only one integral solution if

$$n-1 \le \dfrac{k-\sqrt{k^2-5}}{5}<n< \dfrac{k+\sqrt{k^2-5}}{5} \le n+1$$

Now, $D \ge 0 \Rightarrow |k| \ge \sqrt5$ and $|\alpha -\beta| \le 2 \Rightarrow k\le \sqrt{30}$

Combining both the conditions we get $k \in ${3,4,5}.

If k=3, then we get $$n-1 \le \dfrac{1}{5}<n< 1 \le n+1 \Rightarrow n \notin Z$$

Wolfram alpha provides the following integral solutions of the problem

Wolfram image

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hint

Observe that

$$x_2=\frac{k+\sqrt{k^2-5}}{5}=\frac{1}{k-\sqrt{k^2-5}}$$

$ x_2 $ is integral if $ (k -\sqrt{k^2-5})=\pm 1$ ,

By the same, $ x_1 $ is integral if $ (k+\sqrt{k^2-5}) =\pm 1$.

This gives the possible values for $ k$ : $-4, -3, 3, 4$. the sum of positives values is $7$.

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