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

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.

• Do you mean: $$\text{Find the sum of all positive integers k for which 5x^2-2kx+1<0 has exactly one integral solution.}$$ ? – Servaes Mar 20 '20 at 18:56
• yeah obviously. – user3290550 Mar 20 '20 at 18:57
• No, not obviously, otherwise I wouldn't have asked. Your current phrasing makes no sense, this seemed like the nearest sensible interpretation. – Servaes Mar 20 '20 at 18:58
• 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...." – fleablood Mar 20 '20 at 21:18

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.

• you made the mistake while calculating discriminant, it should be k^2-5 – user3290550 Mar 20 '20 at 19:04
• @user3290550 Thanks for spotting that, corrected now. – Servaes Mar 20 '20 at 19:05
• one more thing you missed, k^2-5>=0 because otherwise discriminant would not be defined. – user3290550 Mar 20 '20 at 19:08
• 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 $x$. – Servaes Mar 20 '20 at 19:18

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}

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}

Wolfram alpha provides the following integral solutions of the problem

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 $$7$$.