# Solve in rational numbers, the equation, $x\lfloor x\rfloor\{x\}=58$

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!

• As a hint: $$x=n+p \text{ where } n=\lfloor x\rfloor\\ \{x\}=x-\lfloor x\rfloor=(n+p)-n=p$$ so $$x\lfloor x\rfloor \{x\}=56\\(n+p)np=56\\$$ Nov 28 '20 at 16:21
• If you try to plot the function ... this equation has infinite solution...desmos.com/calculator/0wggfa1r53 Nov 28 '20 at 16:23

Let $$b=p/q$$ where $$\gcd(p,q)=1$$ and $$0 \le p. The equation becomes $$(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$$.
• @NeatMath No, $k$ could be minus. Nov 28 '20 at 16:41