What is the number of solution of the equation $\ x^3 - \lfloor x\rfloor = 3$?

Question

What is the number of solution of the equation $\ x^3 - \lfloor x\rfloor = 3$ ? (where $\lfloor x \rfloor\ $ is the greatest integer $\le x$)

I tried plotting the graphs of these equations on Desmos graph calculator and that they intersect each other in between x = 1 and 2 but I couldn't figure out a way to get to this conclusion on my own.

Is there any way by which I can determine where these functions intersect?

Answer 1

Notice that $x-1 < \lfloor x \rfloor \leq x$, so $x^3-x+1 > x^3- \lfloor x \rfloor \geq x^3-x$, and $x^3-x+1 > 3 \geq x^3-x$. The first part of the inequality shows that $x<2$, and the second shows that $x \geq 1$. This means that $\lfloor x \rfloor=1$, so $x^3-1=3$ and $x=\sqrt[3]4$.

Answer 2

First of all $x > 0$.

If $x \le 0$ then $x^3 < 0$ so $[x] = x^3 -3 \le -3$ so $[x] \le -3$ and $x \le -2$. (Obviously I'm being overly cautious. $[x]\ne x$ unless $x$ is on integer so $x \le -3$ and obviously $x$ can't be $0$ so $[x]\le -4$ and ... so on but... this just stick with $x \le -2$.)

This means that $|x^3| > |x|$ and that $x^3 - x < 0$. So $x^3 - [x] \ge x^3 -x > x^3 -[x] -1$ so $x^3 -[x] < 1< 3$.

So $x > 0$.

....

Now $[x] \le x$ so $[x]^3 - [x] \le x^3-[x] = 3$.

We must have $[x]^3 -[x] = [x]([x]^2 - 1) = [x]([x]-1)([x]+1) \le 3$.

If $[x] \ge 2$ then $[x]([x]-1)([x]+1)\ge 2\cdot 1\cdot 3 > 3$ so we must have

$[x]=1$

Okay. So $1\le x < 2$. Let $r = x-1$ and $x = 1+ r$.

Then $x^3 -[x] = (1+r)^3 =1 = 3r + 3r^2 + r^3 = 3$.

Now $r$ is monotonically increasing on $r\in [0,1)$ so there is at most one solution.

And $3\cdot 0 + 3\cdot 0^2 + 0^3 < 3$ and $\lim_{r\to 1} 3r + 3r^2 + r^3 = 7 > 3$.

So by continuity there is one solution.

We might be able to solve for $r$ but we don't have to. $r$ is the unique and only solution to $3r + 3r^2 + r^3 = 3$ and $x = 1+r$.

And... oh, googly moogly......

If $[x]=1$ then $x^3 - 1 =3$ and..... $x =\sqrt[3] 4$ and $r = \sqrt[3] 4 -1$ is the unique solution to $3r+3r^2 +r^3 = 3$ and ....

Ah... geeze.... that was trivial to solvi

$3r + 3r^2 + r^3 = 3$

$1+3r + 3r^2 + r^3 = 3+1 = 4$.

$(1+r)^3 = 4$.

.... forest for the trees, forest for the trees.

Answer 3

Here is another answer I came up with:

Notice that $\lfloor x\rfloor$ and $3$ are integers, so $x^3$ is an integer, say $x=\sqrt[3]{y^3+z}$, where $0\leq z<(y+1)^3-y^3$, and y and z are integers. Now we have $y^3+z-y=3$, or $(y-1)y(y+1)=3-z.$ Since $z\geq0$, $(y-1)y(y+1)\leq 3$ and $y\leq1$. However, notice that if $y<-1$, the inequality $z<(y+1)^3-y^3$ fails, so $y\geq-1$, $(y-1)y(y+1)=0$, $z=3$, and $y=1$ to make the inequality true, so $x=\sqrt[3]4$.