Struggling to solve an equation containing a floor function

Question

I am trying to solve the following equation ($x\in \mathbb R)$ $$ 2\lfloor{(x+1)^2+8}\rfloor=(x+1)(2x+3) \tag{1} $$ I have tried substituting $x$ with its split form (integer part ($n$) + decimal part ($\alpha$)), i.e., $x=n+\alpha$ with $n=\lfloor x \rfloor.$

The substitution yields: $$ \lfloor 2\alpha^2 + 4\alpha n + 4\alpha\rfloor=2\alpha^2+4\alpha n+5\alpha + 3n -15\tag{2} $$ or equivalently $$ \lfloor 2\alpha(\alpha + 2n +2)\rfloor= \alpha(2\alpha + 4n + 5) + 3n -15\tag{3} $$

and here I am stuck and I don't know how to proceed, or whether I've already complicated matters more than needed. Any hints would be much appreciated.

Answer 1

$$ 2\lfloor{(x+1)^2+8}\rfloor=(x+1)(2x+3) \tag{1} $$

Let $~~(x+1) = P + r ~: ~P \in \Bbb{Z}, ~0 \leq r < 1.$

Then, in (1) above, the LHS equals
$\displaystyle 2\lfloor (P + r)^2 + 8\rfloor = 2P^2 + 16 + 2\lfloor 2Pr + r^2\rfloor$

and the RHS equals
$\displaystyle (P+r)(2P+2r+1) = 2P^2 + 4Pr + 2r^2 + P + r.$

Therefore
$~~\displaystyle (P - 16) + (4Pr + 2r^2 + r) = 2\lfloor 2Pr + r^2\rfloor.$

Therefore
$$2(2Pr + r^2) - 2\lfloor 2Pr + r^2\rfloor = (16 - P - r). \tag2$$

Note
For any Real Number $s$, you must have that
$0 \leq \left(s - \lfloor s \rfloor \right) < 1.$
This implies that $0 \leq 2s - 2\lfloor s\rfloor < 2.$

Therefore, in (2) above, $0 \leq LHS < 2.$
Therefore, either $~P = 15, ~$ or $~P = 14$.

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Edit
$P = 16, r = 0$ also satisfies the constraints.

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Further, in (2) above,
$\displaystyle \lfloor 2Pr + r^2\rfloor = \lfloor r(2P + r)\rfloor \leq \lfloor 2P + r\rfloor = 2P.$
Therefore, $~~2\lfloor 2Pr + r^2\rfloor \leq 4P$.

Therefore, there must exist some $k \in \{0,1,2,\cdots,4P\}$ such that $k = 2\lfloor 2Pr + r^2\rfloor.$

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$\underline{\text{Case 1: }~P = 15}$
Examining (2) above, $~~P = 15 \implies ~\exists ~k \in \{0,1,2,\cdots,60\}$ such that
$\displaystyle 2(30r + r^2) - k = 1-r.$

This implies that
$\displaystyle 2r^2 + 61r - (k+1) = 0 \implies $
$\displaystyle r = \frac{1}{4} \left[-61 \pm \sqrt{61^2 + 8k + 8}\right].$

Since $r \geq 0$, this implies that
$\displaystyle r = \frac{1}{4} \left[\sqrt{61^2 + 8k + 8} - 61\right] = \frac{1}{4} \left[\sqrt{3729 + 8k} - 61\right] ~: ~k \in \{0,1,\cdots, 60\}.$
Here, since $(x + 1) = P + r = 15 + r$, you have that $x = 14 + r.$

The Case 1 candidate solutions are

• $x = 14 + r$.
• $r = \frac{1}{4} \left[\sqrt{3729 + 8k} - 61\right]$.
• $k \in \{0,1,2,\cdots,60\}$.

On the RHS of (1) above, you have
$(15 + r) (31 + 2r) = 465 + 61r + 2r^2 = 465 + (k + 1) = 466 + k.$

Therefore, values of $k \in \{0,1,2,\cdots, 60\}$ must be inspected with respect to the LHS of equation (1) above, to see if they match the RHS of equation (1) above.

On the LHS of equation (1) above,
$(15 + r)^2 + 8 = 233 + 30r + r^2 \implies $
the LHS of equation (1) equals
$(2 \times 233) + 2\lfloor r^2 + 30r\rfloor.$

As $k$ ranges from $0$ through $60$, $r$ will be strictly between $0$ and $1$.
From the analysis, $2r^2 + 61r = (k+1)$.
Further, $~-1 < -r < 0 \implies k < 2r^2 + 60r < k+1.$

Therefore, $\displaystyle \frac{k}{2} < r^2 + 30r < \frac{k + 1}{2}$.

Therefore $\displaystyle\left\lfloor \frac{k}{2} \right\rfloor = \lfloor r^2 + 30r\rfloor.$

Therefore, if $k$ even, you have that $2\lfloor r^2 + 30r\rfloor = 2 \times \lfloor\frac{k}{2}\rfloor = 2 \times \frac{k}{2} = k.$

However, if $k$ odd, you have that $2\lfloor r^2 + 30r\rfloor = 2 \times \lfloor\frac{k}{2}\rfloor = 2 \times \frac{k-1}{2} = k-1.$

Therefore, in Case 1, the LHS of equation (1) above will match the RHS of equation (1) above only if $k$ is even.

Therefore, the Case 1 solutions are

• $x = 14 + r$
• $r = \frac{1}{4} \left[\sqrt{3729 + 8k} - 61\right]$
• $k \in \{0,2,4,6,\cdots,60\}$

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$\underline{\text{Case 2: }~P = 14}$
Examining (2) above, $~~P = 14 \implies ~\exists ~k \in \{0,1,2,\cdots,56\}$ such that
$\displaystyle 2(28r + r^2) - k = 2-r.$

This implies that
$\displaystyle 2r^2 + 57r - (k+2) = 0 \implies $
$\displaystyle r = \frac{1}{4} \left[-57 \pm \sqrt{57^2 + 8k + 16}\right].$

Since $r \geq 0$, this implies that
$\displaystyle r = \frac{1}{4} \left[\sqrt{57^2 + 8k + 16} - 57\right] = \frac{1}{4} \left[\sqrt{3265 + 8k} - 57\right] ~: ~k \in \{0,1,\cdots, 56\}.$
Here, since $(x + 1) = P + r = 14 + r$, you have that $x = 13 + r.$

The Case 2 candidate solutions are

• $x = 13 + r$.
• $r = \frac{1}{4} \left[\sqrt{3265 + 8k} - 57\right]$.
• $k \in \{0,1,2,\cdots,56\}$.

On the RHS of (1) above, you have
$(14 + r) (29 + 2r) = 406 + 57r + 2r^2 = 406 + (k + 2) = 408 + k.$

Therefore, values of $k \in \{0,1,2,\cdots, 56\}$ must be inspected with respect to the LHS of equation (1) above, to see if they match the RHS of equation (1) above.

On the LHS of equation (1) above,
$(14 + r)^2 + 8 = 204 + 28r + r^2 \implies $
the LHS of equation (1) equals
$(2 \times 204) + 2\lfloor r^2 + 28r\rfloor.$

As $k$ ranges from $0$ through $56$, $r$ will be strictly between $0$ and $1$.
From the analysis, $2r^2 + 57r = (k+2)$.
Further, $~-1 < -r < 0 \implies k+1 < 2r^2 + 56r < k+2.$

Therefore, $\displaystyle \frac{k+1}{2} < r^2 + 28r < \frac{k + 2}{2}$.

Therefore $\displaystyle\left\lfloor \frac{k+1}{2} \right\rfloor = \lfloor r^2 + 28r\rfloor.$

Therefore, if $k$ even, you have that $2\lfloor r^2 + 28r\rfloor = 2 \times \lfloor\frac{k+1}{2}\rfloor = 2 \times \frac{k}{2} = k.$

However, if $k$ odd, you have that $2\lfloor r^2 + 28r\rfloor = 2 \times \lfloor\frac{k+1}{2}\rfloor = 2 \times \frac{k+1}{2} = k+1.$

Therefore, in Case 2, the LHS of equation (1) above will match the RHS of equation (1) above only if $k$ is even.

Therefore, the Case 2 solutions are

• $x = 13 + r$
• $r = \frac{1}{4} \left[\sqrt{3265 + 8k} - 57\right]$
• $k \in \{0,2,4,\cdots,56\}.$

Answer 2

There are a lot of solutions (at most $60$), so we will describe a general way of finding them all as well as give an example of the method described.

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Define the function

$$f(x)=2\lfloor{(x+1)^2+8}\rfloor-(x+1)(2x+3)$$

Note that

$$f(x)\leq 2(x+1)^2+16-(x+1)(2x+3)=15-x$$

$$f(x)\geq 2(x+1)^2+14-(x+1)(2x+3)=13-x$$

So if we have $x_0$ such that $f(x_0)=0$ then it must be the case that

$$15-x_0\geq 0\Rightarrow15\geq x_0$$

$$13-x_0\leq 0 \Rightarrow 13\leq x_0$$

Thus $x_0\in [13,15]$. It is easily seen that $f(13)=2$ and $f(15)=0$. Let $x=13+r$ for some $r\in(0,2)$. Then

$$f(13+r)=2\lfloor r(r+28)\rfloor -2r^2-57r+2$$

Now, note that

$$0(0+28)=0$$

$$2(2+28)=60$$

Thus, the portion inside the floor function will travel through the integers $[1,2,...,59]$. Define

$$r_k=\sqrt{k+196}-14$$

and note that $r_k(r_k+28)=k$. Now, consider the intervals given by

$$S_0=(0,r_1)$$

$$S_1=[r_1,r_2)$$

$$S_2=[r_2,r_3)$$

$$\vdots$$

$$S_{59}=[r_{59},2)$$

For $r\in S_k$ we have

$$k=r_k(r_k+28)\leq r(r+28)<r_{k+1}(r_{k+1}+28)=k+1$$

and therefore

$$\lfloor r(r+28)\rfloor=k$$

(with a similar behavior at $S_0$ and $S_{59}$). Then for these $r\in S_k$ we know

$$f(13+r)=2k-2r^2-57r+2$$

This is a quadratic in $r$ which we can then solve to get

$$r_0=\frac{\sqrt{16k+3265}-57}{4}$$

We just need to check each $k\in [0,1,...,59]$ and see if the $r_0$ defined above falls in the interval $S_k$. If it does, then we have a root. If not, then we do not have a root. For example, for $k=5$ we have

$$r_0=\frac{\sqrt{3345}-57}{4}=0.21$$

and

$$S_5=[r_5,r_6)=[\sqrt{207}-14,\sqrt{208}-14)=[0.38,.42)$$

Thus, there is no root corresponding to $k=5$. However, if we instead chose $k= 8$ then

$$r_0=\frac{3\sqrt{377}-57}{4}=0.3123$$

$$S_8=[0.2829,0.3178)$$

Since $r_0\in S_8$, we conclude $f(13+r_0)=0$ (this can be checked if one wishes). Just do this for all $k\in [0,1,...,59]$ and you will have all the solutions.

Answer 3

Let $x$ be an integer. Set $y = x+1$, then we have $2\lfloor y^{2}+8 \rfloor=y(2y+1)=2y^{2}+y. $

Since $\lfloor a+c \rfloor = \lfloor a \rfloor +c$ for all integers $a$, $2\lfloor y^{2}+8 \rfloor=2(\lfloor y^{2} \rfloor + 8)=2\lfloor y^{2} \rfloor +16=2y^{2}+y$ from above.

If $y > 16$, then $2\lfloor y^{2} \rfloor = 2y^{2} + c$ for some $c > 0 \implies 2y^{2}<2 \lfloor y^{2} \rfloor \implies y^{2}<\lfloor y^2 \rfloor$ which is a contradiction as $\lfloor a \rfloor \leq a$ for all $a$ by definition.

Hence, we must check the finite number of cases of the form $y \leq 16 \implies x \leq 15$.

Since $x$ is an integer ($\implies y$ is an integer), then $2y^{2}+16=2\lfloor y^{2} \rfloor +16=2y^{2}+y \implies$

$y = 16 \implies x = 15.$

Therefore $15$ is the only integer solution.