A pretty much simple number theory problem

Question

Let $x$ be an irrational number, and $n$ be a positive integer. Will there ever be a set of $(n,x)$ which satisfies $x(n-x) \in \mathbb{Z}$ ?

If so, could you suggest those numbers? And, if not, could you prove why?

Answer 1

Hint:

• Suppose $x(n-x)=m$. Then express $x$ in terms of $n$ and $m$.
• Can you find an example where $n$ and $m$ are integers (and $n$ positive) but $x$ irrational?
• Remember that square roots of non-square positive integers are irrational

Answer 2

Here's an example $$n=3$$ $$m=1$$ $$x=\frac12\left(3+\sqrt{5}\right)$$

This is what happens if you simplify it. $$\frac12\left(3+\sqrt{5}\right)\left(3-\frac12\left(3+\sqrt{5}\right)\right)$$ $$\frac32\left(3+\sqrt{5}\right)-\left(\frac12\left(3+\sqrt{5}\right)\right)^2$$ $$\frac32\left(3+\sqrt{5}\right)-\frac14\left(3+\sqrt{5}\right)^2$$ $$\frac32\left(3+\sqrt{5}\right)-\frac14\left(14+6\sqrt{5}\right)$$ $$\frac32\left(3+\sqrt{5}\right)-3.5-\frac32\sqrt{5}$$ $$\frac92+\frac32\sqrt{5}-3.5-\frac32\sqrt{5}$$ $$1+\frac32\sqrt{5}-\frac32\sqrt{5}$$ $$1$$

Answer 3

I already posted an answer with an example, but I decided to make a different answer which shows how you can find these values.

$$x(n-x)=m$$ Lets solve for $x$ $$ \begin{eqnarray} x(n-x)=m\\ -x^2+nx=m\\ x^2-nx=-m\\ x^2-nx+\left(\frac{n}{2}\right)^2=\frac{n^2}{4}-m\\ \left(x-\frac{n}{2}\right)^2=\frac{n^2}{4}-m\\ 4\left(x-\frac{n}{2}\right)^2=n^2-4m\\ 2x-n=\pm\sqrt{n^2-4m}\\ 2x=n\pm\sqrt{n^2-4m}\\ x=\frac12\left(n\pm\sqrt{n^2-4m}\right)\\ \end{eqnarray} $$ This gives us 2 values of $x$ for each $(n,m)$
This $x$ is going to be irrational when $\sqrt{n^2-4m}$ is irrational.
This implies that if $x$ is irrational then $n^2-4m$ is not a perfect square.

Conclusion:

$x(n-x)=m$ has 2 irrational values of $x$ which solves the equation for each pair of $(m,n)$ where $n^2-4m$ is not a perfect square.