# For what integer values of k does sqrt(x)+sqrt(x+1)=k have a rational solution?

Aug 24, 2016

Any positive integer $k$ gives a rational solution.

#### Explanation:

Given any positive integer $k$, let $x = {\left({k}^{2} - 1\right)}^{2} / \left(4 {k}^{2}\right)$

Note that this value of $x$ is rational.

Then:

$\sqrt{x} + \sqrt{x + 1}$

$= \sqrt{{\left({k}^{2} - 1\right)}^{2} / \left(4 {k}^{2}\right)} + \sqrt{{\left({k}^{2} - 1\right)}^{2} / \left(4 {k}^{2}\right) + 1}$

$= \sqrt{{\left({k}^{2} - 1\right)}^{2} / \left(4 {k}^{2}\right)} + \sqrt{\frac{{\left({k}^{2} - 1\right)}^{2} + 4 {k}^{2}}{4 {k}^{2}}}$

$= \sqrt{{\left({k}^{2} - 1\right)}^{2} / \left(4 {k}^{2}\right)} + \sqrt{\frac{{\left({k}^{2} + 1\right)}^{2}}{4 {k}^{2}}}$

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

$= {k}^{2} / k$

$= k$

Aug 24, 2016

$x = {\left(\frac{{k}^{2} - 1}{2 k}\right)}^{2}$ for $k \in \mathbb{N} - \left\{0\right\}$ are the rational solutions

#### Explanation:

$\sqrt{x} + \sqrt{x + 1} = k$. If $x$ is rational, and $k$ is an integer then

$x = {\left(\frac{n}{m}\right)}^{2}$ so

$\frac{n}{m} + \sqrt{{\left(\frac{n}{m}\right)}^{2} + 1} = k$ and

$\sqrt{{\left(\frac{n}{m}\right)}^{2} + 1} = k - \frac{n}{m}$

so

${\left(\frac{n}{m}\right)}^{2} + 1 = {\left(k - \frac{n}{m}\right)}^{2}$ or

$1 = {k}^{2} - 2 k \frac{n}{m}$ then

$\frac{n}{m} = \frac{{k}^{2} - 1}{2 k}$

Finally

$x = {\left(\frac{{k}^{2} - 1}{2 k}\right)}^{2}$ for $k \in \mathbb{N} - \left\{0\right\}$ are the rational solutions