Solving a system of equations involving the floor function.

Question

I have the following system of equations that I am stuggeling with:

$$ax\lfloor y\rfloor=k,by\lfloor x\rfloor=d$$

And I know that $x$ and $y$ are bigger than zero and all the other constants are natural numbers. Given is that $a$ divides $k$ and $b$ does not divide $d$.

What is the way to go on this problem?

Thanks for any advice

Answer 1

Assignment:

Solve:

$$ \begin{cases} \alpha x\left\lfloor\text{y}\right\rfloor=\gamma\\ \\ \beta\text{y}\left\lfloor x\right\rfloor=\Delta \end{cases} $$

For $x\space\wedge\space\text{y}\in\mathbb{R}_{>0}$, where we know that $\alpha\mid\gamma$ and $\beta\nmid\Delta$ it is also given that $\alpha\space\wedge\space\gamma\space\wedge\space\beta\space\wedge\space\Delta\in\mathbb{N}$.

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Solution:

Because we know that $\alpha\mid\gamma$, we can divide both sides of the first equation by $\alpha$ to get:

$$ \begin{cases} x\left\lfloor\text{y}\right\rfloor=\frac{\gamma}{\alpha}\\ \\ \beta\text{y}\left\lfloor x\right\rfloor=\Delta \end{cases}\tag1 $$

Now, I am going to decompose $x$ and $\text{y}$ in there integer part and there fractional part, as follows:

• $$x=\left\lfloor x\right\rfloor+\text{a}\tag2$$ Where we know that $\text{a}\in\left[0,1\right)$.
• $$\text{y}=\left\lfloor\text{y}\right\rfloor+\text{b}\tag3$$ Where we know that $\text{b}\in\left[0,1\right)$.

Now, we put $(2)$ and $(3)$ into $(1)$, in order to get:

$$ \begin{cases} \left(\left\lfloor x\right\rfloor+\text{a}\right)\left\lfloor\text{y}\right\rfloor=\frac{\gamma}{\alpha}\\ \\ \beta\left(\left\lfloor\text{y}\right\rfloor+\text{b}\right)\left\lfloor x\right\rfloor=\Delta \end{cases}\tag4 $$

Multiplying out gives:

$$ \begin{cases} \left\lfloor x\right\rfloor\left\lfloor\text{y}\right\rfloor+\text{a}\left\lfloor\text{y}\right\rfloor=\frac{\gamma}{\alpha}\\ \\ \beta\left\lfloor x\right\rfloor\left\lfloor\text{y}\right\rfloor+\beta\text{b}\left\lfloor x\right\rfloor=\Delta \end{cases}\tag5 $$

Now, we can solve:

$$ \begin{cases} \text{a}\left\lfloor\text{y}\right\rfloor=\frac{\gamma}{\alpha}-\left\lfloor x\right\rfloor\left\lfloor\text{y}\right\rfloor\\ \\ \beta\text{b}\left\lfloor x\right\rfloor=\Delta-\beta\left\lfloor x\right\rfloor\left\lfloor\text{y}\right\rfloor \end{cases}\tag6 $$

It is not hard to notice that the RHS of the first equation in $(6)$ is an integer and this also holds for the RHS of the second equation in $(6)$. Now, we can see that we can write $\text{a}$ as quotient involving the $\left\lfloor\text{y}\right\rfloor$ and this similarly holds for $\text{b}$. So I write:

$$\text{a}=\frac{\text{m}}{\left\lfloor\text{y}\right\rfloor}\space\space\space\wedge\space\space\space\text{b}=\frac{\text{n}}{\beta\left\lfloor x\right\rfloor}\tag7$$

So, we see that:

$$ \begin{cases} \text{a}\left\lfloor\text{y}\right\rfloor=\underbrace{\frac{\gamma}{\alpha}-\left\lfloor x\right\rfloor\left\lfloor\text{y}\right\rfloor}_{=\space\text{m}\space\ne0}\\ \\ \beta\text{b}\left\lfloor x\right\rfloor=\underbrace{\Delta-\beta\left\lfloor x\right\rfloor\left\lfloor\text{y}\right\rfloor}_{=\space\text{n}\space\ne0} \end{cases}\tag8 $$

Now, let's substitute $(7)$ into $(5)$, to get:

$$ \begin{cases} \left\lfloor x\right\rfloor\left\lfloor\text{y}\right\rfloor+\text{m}=\frac{\gamma}{\alpha}\\ \\ \beta\left\lfloor x\right\rfloor\left\lfloor\text{y}\right\rfloor+\text{n}=\Delta \end{cases}\tag9 $$

Now, multiply the first equation from $(9)$ with $-\beta$ and add the first and second equation from $(9)$, to get:

$$\text{n}-\beta\text{m}=\Delta-\frac{\beta\gamma}{\alpha}\space\Longleftrightarrow\space\text{n}=\Delta-\frac{\beta\gamma}{\alpha}+\beta\text{m}=\Delta+\beta\left(\text{m}-\frac{\gamma}{\alpha}\right)\tag{10}$$

Now, we know that $x>0$, so we know that $\left\lfloor x\right\rfloor\ge0$ but when dividing by it it can not be equal to $0$, this implies that $\left\lfloor x\right\rfloor\ge1$. We know that $\text{b}$ is positive so $\text{n}$ must also be positive, so:

$$\text{n}=\Delta+\beta\left(\text{m}-\frac{\gamma}{\alpha}\right)\ge0\space\Longleftrightarrow\space\text{m}\ge\frac{\gamma}{\alpha}-\frac{\Delta}{\beta}\tag{11}$$

But, because we know that $\text{m}$ is an integer, we also know that:

$$\text{m}>\left\lfloor\frac{\gamma}{\alpha}-\frac{\Delta}{\beta}\right\rfloor\tag{12}$$

Because we know that $\text{a}\in\left[0,1\right)$, we get:

$$\text{a}=\frac{\text{m}}{\left\lfloor\text{y}\right\rfloor}\space\Longrightarrow\space\left\lfloor\text{y}\right\rfloor>\left\lfloor\frac{\gamma}{\alpha}-\frac{\Delta}{\beta}\right\rfloor\tag{13}$$

Now, given that $x\left\lfloor\text{y}\right\rfloor=\frac{\gamma}{\alpha}$, we see that:

$$x<\frac{\frac{\gamma}{\alpha}}{\left\lfloor\frac{\gamma}{\alpha}-\frac{\Delta}{\beta}\right\rfloor}\tag{14}$$

Now, you know that:

$$1\le\left\lfloor x\right\rfloor<\left\lfloor\frac{\frac{\gamma}{\alpha}}{\left\lfloor\frac{\gamma}{\alpha}-\frac{\Delta}{\beta}\right\rfloor}\right\rfloor\tag{15}$$

So:

$$\left\lfloor x\right\rfloor\in\left\{1,\dots,\left\lfloor\frac{\frac{\gamma}{\alpha}}{\left\lfloor\frac{\gamma}{\alpha}-\frac{\Delta}{\beta}\right\rfloor}\right\rfloor\right\}\tag{16}$$

And you can test all possible values of that using $(1)$.