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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!

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  • $\begingroup$ As a hint: $$x=n+p \text{ where } n=\lfloor x\rfloor\\ \{x\}=x-\lfloor x\rfloor=(n+p)-n=p$$ so $$x\lfloor x\rfloor \{x\}=56\\(n+p)np=56\\$$ $\endgroup$
    – Khosrotash
    Nov 28 '20 at 16:21
  • $\begingroup$ If you try to plot the function ... this equation has infinite solution...desmos.com/calculator/0wggfa1r53 $\endgroup$
    – Khosrotash
    Nov 28 '20 at 16:23
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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 $x=29/3$.

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  • $\begingroup$ Do you mean that os only solution ? $\endgroup$
    – Khosrotash
    Nov 28 '20 at 16:35
  • $\begingroup$ @Khosrotash Yes. The poster seeks rational solution. $\endgroup$
    – Zerox
    Nov 28 '20 at 16:36
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    $\begingroup$ @Khosrotash And your comments to the poster are invalid since the poster already did the job you "hint". $\endgroup$
    – Zerox
    Nov 28 '20 at 16:38
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    $\begingroup$ @NeatMath No, $k$ could be minus. $\endgroup$
    – Zerox
    Nov 28 '20 at 16:41
  • $\begingroup$ I could not understand the inequality part. $\endgroup$
    – MRupam
    Nov 28 '20 at 19:25

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