Find all real solutions $x$ to the equation $\lfloor x^2 - 2x \rfloor + 2\lfloor x \rfloor = \lfloor x \rfloor^2$.
Problem
Source: Pan African MO 2012 Day 1 Q3
Tags: function, floor function, algebra proposed, algebra
Farenhajt
14.10.2012 16:19
Put $x=n+a$ where $n=[x], a=\{x\}$ and discuss the necessary condition $0\leqslant a<1$. Answer: $x\in\mathbb{Z}^-_0\cup [1,2)\cup\bigcup_{n=1}^\infty[n+1,1+\sqrt{n^2+1})$
ahmedosama
13.09.2015 18:23
[x]^2 or [x^2] ????
ahmedosama
13.09.2015 18:24
Farenhajt wrote: Put $x=n+a$ where $n=[x], a=\{x\}$ and discuss the necessary condition $0\leqslant a<1$. Answer: $x\in\mathbb{Z}^-_0\cup [1,2)\cup\bigcup_{n=1}^\infty[n+1,1+\sqrt{n^2+1})$ how did you get your answer ?
JoelBinu
29.05.2017 18:06
Wouldn't the equation hold for any real value of $x$.
jelena_ivanchic
15.04.2020 08:06
a required bump /.
starchan
15.04.2020 09:01
Big hint Add 1 to both sides and rearrange You get the equation $[(x-1)^2] = [x-1]^2$
Imajinary
23.06.2020 21:57
We notice that all integers satisfy the equation, so we will now ignore any integer possibilities for $x$. Set $\lfloor x \rfloor = k$. Rearranging yields $\lfloor x^2 - 2x \rfloor = k^2 - 2k$ . From this we deduce the inequality (ignoring integer case) $k^2 - 2k < x^2 - 2x < k^2 - 2k + 1$ or equivalently $(k-1)^2 < (x-1)^2 < (k-1)^2 + 1$. We now take cases.
$1 \leq k$: Under this case, we can convert the above inequality to $k-1 < x-1 < \sqrt{(k-1)^2 + 1}$. Notice that $k-1 < x-1$ is always true in this case (no equal sign since we are ignoring integers), we so ignore this inequality. We are left with $x-1 < \sqrt{(k-1)^2 + 1} \Leftrightarrow x < \sqrt{(k-1)^2 + 1} + 1$. The only other restriction $x$ needs to satisfy is $k < x < k+1$, (because of the floor function), so we need to find intersections of the solutions to $k < x < \sqrt{(k-1)^2 + 1} + 1$ and $x < k+1$, where $k$ is an integer greater than or equal to 1 .
We then see $\sqrt{(k-1)^2 + 1} + 1 \leq k + 1 \Leftrightarrow k^2 - 2k + 2 \leq k^2 \Leftrightarrow 1 \leq k$, which is always true under our assumption of $1 < x$. Therefore, we have $k < x < \sqrt{(k-1)^2 + 1} + 1 \leq k+1$. The union of $k < x < \sqrt{(k-1)^2 + 1} + 1$ and $x < k+1$ is then just $k < x < \sqrt{(k-1)^2 + 1} + 1$. Therefore we have $x \in \bigcup_{n = 2} ^{\infty} (n, \sqrt{(n-1)^2 + 1} + 1)$ in this case.
$k < 1$: Under this case, we can convert the initial inequality to $k-1 > x - 1 > \sqrt{(k-1)^2 + 1} + 1$. However, $k-1 > x-1$ is never true, and hence there are no solutions for this case.
Our solutions are then $\boxed{x \in \mathbb{Z} \cup \bigcup_{n = 1} ^{\infty} (n, \sqrt{(n-1)^2 + 1} + 1)}$.
PCChess
06.09.2020 00:41
Rearrange the equation to $\lfloor x \rfloor ^2-2 \lfloor x \rfloor - \lfloor x^2-2x \rfloor =0$. This is a quadratic in $\lfloor x \rfloor$, so using the quadratic formula, we get that $$\lfloor x \rfloor=\frac{2 \pm \sqrt{4-4(- \lfloor x^2 - 2x \rfloor)}}{2}=1 \pm \sqrt{1+ \lfloor x^2-2x \rfloor}.$$$1+ \lfloor x^2-2x \rfloor$ needs to be a perfect square, so let it equal some nonnegative integer $y$. We have $$1+ \lfloor x^2-2x \rfloor =y^2 \implies \lfloor x^2-2x \rfloor =y^2-1$$$$\implies y^2-1 \leq x^2-2x < y^2 \implies y^2 \leq (x-1)^2 < y^2+1.$$The possible solutions for $x$ (by considering if $x-1$ is positive or negative) are $$x \in [y+1, \sqrt{y^2+1}+1) \text{ or } (1-\sqrt{y^2+1}, -y+1] \qquad \qquad \qquad (1)$$for a given $y$. Now, for the same $y$ in the previous line, going back to the equation of the quadratic formula, since $y=1 + \lfloor x^2-2x \rfloor$, $$\lfloor x \rfloor = 1 \pm y$$$$\implies y+1 \leq x <y+2 \text{ or } -y+1 \leq x <-y+2. \qquad \qquad \qquad (2)$$In order for $x$ to be a possible solution, it must satisfy equations $(1)$ and $(2)$. One can easily check that $y+2 \geq \sqrt{y^2+1}+1$ for all nonnegative integers $y$, so the possible solutions for $x$ are $[y+1, \sqrt{y^2+1}+1) \cup -y+1, y \in \mathbb{N}_0.$
Williamgolly
17.09.2020 16:56
Let $k = \lfloor x \rfloor,$ $m = \{x\}.$ Plugging these into our equation and rearranging (if you add 1 to both sides), we see that $$0 \leq 2mk+m^2-2m <1.$$ Case 1: $k \geq 0.$ If $k =1,$ all values of $m$ work. If $k = 0,$ $m = 0.$ If $k \geq 2,$ we have the LHS of the inequality to be always true, and the RHS of the inequality we have $$m^2+2(k-1)m <1 \implies m^2+2(k-1)m+(k-1)^2<1+(k-1)^2.$$Thus, $m \in [0, \sqrt{(k-1)^2+1}-(k-1))$ which means $x \in [k, 1+\sqrt{(k-1)^2+1}).$ Case 2: $k<0$ We see the LHS of the inequality, $m^2+2(k-1)m \geq 0$ is not satisfied for $k \leq -1$ since $m \in [0, 1).$ Therefore, the only value of $m$ is $m=0.$ Thus, the solutions for $x$ are all nonpositive integers, $[1, 2),$ and $\bigcup_{k=1}^\infty[k,1+\sqrt{(k-1)^2+1})$
Zorger74
31.12.2020 00:11
Rearrange into $\lfloor x^2-2x+1\rfloor=\lfloor x\rfloor^2 -2\lfloor x\rfloor +1$, or equivalently $\lfloor (x-1)^2\rfloor=(\lfloor x -1\rfloor)^2$. Clearly, this equation is satisfied for all $x\in\mathbb{Z}$.
First we look at positive non-integers. As $x$ increases, the RHS only changes when $x$ reaches an integer. Thus, we want the maximum value of $\{x\}$ such that if $\lfloor x\rfloor=a$, then the equation is satisfied, because every value less than that maximum will also work.
\begin{align*}
\lfloor x^2-2x+1\rfloor &= \lfloor x\rfloor^2 -2\lfloor x\rfloor +1\\
a^2-2a+1 &\leq x^2-2x+1<a^2-2a+2\\
a-1 &\leq x-1 <\sqrt{(a-1)^2+1}\\
a &\leq x <\sqrt{(a-1)^2+1}+1\\
\end{align*}
Next, we look at negative non-integers. The equation is true for $x=0$, but when $x$ decreases, the RHS decreases immediately, so the LHS cannot be equal until it reaches that decreased value. However, this only occurs when $x$ decreases into the next integer, so there are no negative non-integer solutions.