Find all real number(s) $x$ satisfying the equation $\{(x +1)^3\}$ = $x^3$ , where $\{y\}$ denotes the fractional part of $y$ , for example $\{3.1416\ldots\}=0.1416\ldots$.

I am trying all positive real numbers from $1,2,\dots$ but I didn't get any decimals.

Is there a smarter way to solve this problem? ... Please advise.

  • $\begingroup$ Use $\{ x\}$ for $\{ x\}$. $\endgroup$ – Shaun Jul 18 '18 at 9:47
  • 3
    $\begingroup$ You're trying all positive real numbers? xD $\endgroup$ – principal-ideal-domain Jul 18 '18 at 9:52
  • $\begingroup$ To begn with, the left side is a number between $0$ and $1$ , so the right side must be the same. This tells you that all solutions must fall in the range $[0,1)$ at best, if there are any solutions. $\endgroup$ – MPW Jul 18 '18 at 9:55

Fractional part is always in $[0,1)$. So the domain of $g(x)=x^3$ you are looking for is such that $R_g\in[0,1)$, which happens to be $[0,1)$.

$y=\{(x+1)^3\}$ is basically $y=(x+1)^3$ chopped into appropriate pieces and translated down by some $k\in\mathbb{Z}$ where $k=\lfloor(x+1)^3\rfloor$

So we solve $$(x+1)^3-k=x^3$$ which gives $$k=3x^2+3x+1$$

Since for $[0,1)$, $0\le(x+1)^3<8$, we check $$3x^2+3x+1=0,1,2,3,4,5,6,7$$ whose positive roots give the answers.

  • $\begingroup$ In fact the fractional part is always in $[0,1)$ $\endgroup$ – MPW Jul 18 '18 at 9:55
  • $\begingroup$ @MPW ugh my bad! edited. thanks a lot $\endgroup$ – Karn Watcharasupat Jul 18 '18 at 9:57


  • $0 \le \{y\} \lt 1$
  • so any solution to $\{(x +1)^3\} = x^3$ has $0 \le x^3 \lt 1$ and thus $0 \le x \lt 1$
  • so $1 \le x+1 \lt 2$ and $1 \le (x+1)^3 \lt 8$
  • any solution has $(x+1)^3 = x^3 +n$ for $n \in \{1,2,3,4,5,6,7\}$, which gives you seven quadratic equations to check
  • for example, $x=0$ is a solution when $n=1$

Well, first of all you're checking only 'all' integers and not all reals which include also $1/2$ or $\pi$ for example. I will try to give you an outline of how I would approach it.

Second of all, note that the left-hand side is always between $0$ and $1$. So, what is implied about the range $x$ might be in?
Once, you determined that you should be splitting up the range of possible $x$ into smaller segments such that for each segment you can find a natural number $n$ such that $$\{(x+1)^3\} = (x+1)^3 - n.$$ Hint: there are $7$ segments that you should consider. Then for each segment it boils down in solving $$ (x+1)^3 - n = x^3.$$ This looks difficult at first sight but I guarantee it will simplify to a quadratic polynomial (which you can solve). Please don't forget that the solutions of this equation should still be in the range on which we consider solutions to be part of.

I hope this helps and if you have any questions about any of these steps, feel free to ask them.

  • $\begingroup$ Thanks for the explanation ... This really helped. ... $\endgroup$ – Math Tise Jul 18 '18 at 10:26
  • $\begingroup$ I get the following answers : $\endgroup$ – Math Tise Jul 18 '18 at 10:27
  • $\begingroup$ for n=0, x is imaginary, for n=1 to 6, i get $x$ = $0$, ${\frac{\sqrt{21}-3} {6}}$, ${\frac{\sqrt{33}-3} {6}}$, ${\frac{\sqrt{5}-1} {2}}$, ${\frac{\sqrt{57}-3} {6}}$, ${\frac{\sqrt{69}-3} {6}}$ . $\endgroup$ – Math Tise Jul 18 '18 at 10:37
  • $\begingroup$ for n=7, x is a whole number, for n=8 to 17, i get ${\frac{\sqrt{93}-3} {6}}$, ${\frac{\sqrt{105}-3} {6}}$, ${\frac{\sqrt{13}-1} {2}}$, ${\frac{\sqrt{129}-3} {6}}$, ${\frac{\sqrt{141}-3} {6}}$, ${\frac{\sqrt{17}-1} {2}}$ . $\endgroup$ – Math Tise Jul 18 '18 at 10:39
  • $\begingroup$ and the list goes on ... are there only particular solutions to this problem?, or unlimited solutions... please advise @Stan Tendijck ... Thanks again. $\endgroup$ – Math Tise Jul 18 '18 at 10:40

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