The equation $\{x^2\} + \{x\}=1$ has no solution over positive rationals

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.

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 $\gcd(p,q)=1$.