Raven Gardner

2022-04-06

Find the sum of all positive integers k for which $5{x}^{2}-2kx+1<0$ has exactly one integral solution.
My attempt is as follows:
$\left(x-d\frac{2k-\sqrt{4{k}^{2}-20}}{10}\right)\left(x-d\frac{2k+\sqrt{4{k}^{2}-20}}{10}\right)<0$
$\left(x-d\frac{k-\sqrt{{k}^{2}-5}}{5}\right)\left(x-d\frac{k+\sqrt{{k}^{2}-5}}{5}\right)<0$
$x\in \left(d\frac{k-\sqrt{{k}^{2}-5}}{5},d\frac{k+\sqrt{{k}^{2}-5}}{5}\right)$
As it is given that it has got only one integral solution, so there must be exactly one integer between $d\frac{k-\sqrt{{k}^{2}-5}}{5}$ and $d\frac{k+\sqrt{{k}^{2}-5}}{5}$.
Let ${x}_{1}=d\frac{k-\sqrt{{k}^{2}-5}}{5}$ and ${x}_{2}=d\frac{k+\sqrt{{k}^{2}-5}}{5}$, then $\left[{x}_{2}\right]-\left[{x}_{1}\right]=1$ where [] is a greater integer function.

Frain4i62

Your idea is good; you want to find all positive integers k for which there is precisely on integer between the roots of
$5{x}^{2}-2kx+1=0.$
Then the distance between the roots can be at most 2, where the distance between the roots is precisely
$\frac{1}{5}\sqrt{{\left(-2k\right)}^{2}-4\cdot 1\cdot 5}=\frac{25}{\sqrt{{k}^{2}-5}},$
as you already found. This is at most 2 if and only if $\sqrt{{k}^{2}-5}\le 5$, or equivalently $k\le 5$. This leaves only 5 values of k to check.

Do you have a similar question?