The equation $\{x^2\} + \{x\}=1$ has no solution over positive rationals 
Prove there is no positive rational $x$ so that $$\{x^2\} + \{x\}=1 \tag1 $$


Let $x=\frac p q$ and $p=qc+r, p, q, c, r \in \mathbb{N}, 0 \le r \lt q$
From (1) $\{ 2c \frac r q  + (\frac r q)^2\} + \{\frac r q\}=1$ and here I've got stuck.
Any help is appreciated.
 A: Note that if $\{x^2\} + \{x\} = 1$, then $x^2 + x$ is an integer. Solve the equation $x^2 + x = n$ for an arbitrary $n$, and see that if it has rational solutions, then those rationals must be integers, which means that $\{x^2\} + \{x\} = 0$.
To see that $x^2 + x$ must be an integer, note that for any $y$ we may use the floor function to write $y = \lfloor y\rfloor + \{y\}$. This gives
$$
x^2 + x = \lfloor x^2\rfloor + \{x^2\} + \lfloor x \rfloor + \{x\} = \lfloor x^2\rfloor + \lfloor x \rfloor + 1
$$
which is an integer.
A: Suppose that there exists such a positive rational number.
We have
$$x^2-\lfloor x^2\rfloor+x-\lfloor x\rfloor =1,$$
i.e.
$$x^2+x=\lfloor x^2\rfloor +\lfloor x\rfloor +1$$
We can set $x:=p/q$ where $p,q$ are positive integer with $\gcd(p,q)=1$, then
$$x^2+x=\frac{p}{q}\left(\frac pq+1\right)=m\tag1$$
where $m\in\mathbb Z$. Then,
$$(1)\implies mq^2=p(p+q)\tag2$$
so, there exists an integer $k$ such that $m=pk$, and so we have
$$(2)\implies q(kq-1)=p$$
which contradicts that $\gcd(p,q)=1$.
A: EDIT I accidentally tried solving the wrong problem. I did not know about the $\{\cdot\}$ notation for fractional part. This is an attempt to show that $x^2+x=1$ has no solutions for $x\in\mathbb{Q}$.

Here's my attempt:
Assume there are $p,q \in {\mathbb Z}$ so that they are relative prime.
$$\left(\frac{p}{q}\right)^2 + \left(\frac{p}{q}\right) = 1\Leftrightarrow\\ \frac{p^2+pq-q^2}{q^2} = 0\Leftrightarrow\\(p+q)(p-q) + pq = 0$$
For this to be true $p$ or $q$ must share prime factors with $p+q$ or $p-q$. But that can only be true if $p$ and $q$ aren't relative prime. But we can demand them to be from the definition of a rational number.
