TITLE: Solve in rational numbers, the equation, $x\lfloor x\rfloor\{x\}=58$
QUESTION [5 upvotes]: Solve in rational numbers, the equation,
$$x\lfloor x\rfloor\{x\}=58$$
where $\lfloor x\rfloor$ and $\{x\}$ are the greatest integer less than or equal to $x$ and the fractional part of $x$ respectively.

I tried to make an equation like:
$$(a+b)ab=58$$
where $a=\lfloor x\rfloor$ and $b=\{x\}$ and giving the restrictions that $a$ is an integer and $b$ is a rational number such that $0≤b<1$.


Then I made a quadratic equation over $b$ and used the famous quadratic formula. But everything became complex (not the mathematical 'complex') and so please help me out. Thanks in advance!

REPLY [7 votes]: Let $b=p/q$ where $\gcd(p,q)=1$ and $0 \le p<q$. The equation becomes $$(qa+p)ap=58q^2$$
Since $\gcd((qa+p)p,q)=\gcd(p^2,q)=1$ we have $q^2|a$. Write $a=kq^2(k \in \Bbb{Z})$(apparently $k \ne 0$) and we get $$(kq^3+p)kp=58$$
Then $p|58$. Moreover, $|kq^3+p| \ge |k|q^3-p \ge (p+1)^3-p>p$, thus $p$ is the smaller positive factor of $58 \Rightarrow p=1,2$. A simple discussion will conclude that $p=2,q=3,k=1,a=9$, so the solution is $x=29/3$.