Solve this system equation $\lfloor xy\rfloor=12,\lfloor x^2\rfloor+\lfloor y^2\rfloor=25$

Question

Let $x,y\in R$.solve this system equation $$\begin{cases} \lfloor xy\rfloor=12\\ \lfloor x^2\rfloor+\lfloor y^2\rfloor=25 \end{cases}$$

I have found integer solution $(x,y)=(3,4)$ How to Find the other

Answer 1

For all $(x,y)$ in the solution set $S$ one has $xy\geq12$, $(-x,-y)\in S$, and $(y,x)\in S$. It is therefore sufficient to determine the set $S':=S\cap\bigl\{(x,y)\>\bigm|\>0\leq x\leq y\bigr\}$.

When $(x,y)\in S'$ then $$0\leq k:=\lfloor x^2\rfloor\leq\lfloor y^2\rfloor=25-k\ ,$$ hence $0\leq k\leq 12$. It follows that $S'$ is contained in the union $U:=\bigcup_{0\leq k\leq 12} R_k$ of the rectangles $$R_k:=\bigl\{(x,y)\>\bigm|\>\sqrt{k}\leq x<\sqrt{k+1}, \quad \sqrt{25-k}\leq y<\sqrt{26-k}\bigr\}\qquad(0\leq k\leq 12)\ .$$ Now $S'$ is the intersection of $U$ with the "hyperbolical band" $$B:=\bigl\{(x,y)\>\bigm|\>x>0, \ y>0, \quad 12\leq xy<13\bigr\}\ .$$ For all $(x,y)\in R_k$ one has $x^2y^2<(k+1)(26-k)=26+25k-k^2$. Here the right hand side is maximal when $k=12.5$, and for $k=6$ one obtains $x^2y^2<140$. It follows that $R_k\cap B=\emptyset$ when $0\leq k\leq 6$. On the other hand the point $\bigl(\sqrt{8}-\epsilon, \sqrt{19}-\epsilon\bigr)$ lies in $R_7\cap B$, and the point $\bigl(\sqrt{12},\sqrt{13}\bigr)$ lies in $R_{12}\cap B$.

Summing it up, we can say that $$S'=\bigcup_{k=7}^{12} \bigl(R_k\cap B\bigr)\ .$$

Answer 2

There is a non-countable solution set $S$ and we expose here a subset $T$ of $S$. We have $$\begin{cases}x=\lfloor x\rfloor+\{x\}\\y=\lfloor y\rfloor+\{y\}\end{cases}\Rightarrow\begin{cases} xy=\lfloor x\rfloor\lfloor y\rfloor+\lfloor x\rfloor\{y\}+\lfloor y\rfloor\{x\}+\{x\}\{y\}\\x^2=\lfloor x\rfloor^2+2\lfloor x\rfloor\{x\}+\{x\}^2\\y^2=\lfloor y\rfloor^2+2\lfloor y\rfloor\{y\}+\{y\}^2\end{cases}$$ A clear solution is the Pythagorean $(x,y)=(3,4)$. Taking $x=3+\{x\}$ and $y=4+\{y\}$ one has $$\begin{cases}xy=12+3\{y\}+4\{x\}+\{x\}\{y\}\\x^2=9+6\{x\}+\{x\}^2\\y^2=16+8\{y\}+\{y\}^2\end{cases}$$ We still have the two given equations when $\{x\}$ and $\{y\}$ satisfy the system $$\begin{cases}0\lt3\{y\}+4\{x\}+\{x\}\{y\}\lt 1\\0\lt6\{x\}+\{x\}^2+ 8\{y\}+\{y\}^2\lt 1\end{cases}$$ In particular it is satisfied making $\{x\}=\{y\}$ which gives $7\{x\}+\{x\}^2\lt1$ and $14\{x\}+2\{x\}^2\lt1$ which reduces to $7\{x\}+\{x\}^2\lt \frac 12$ so we can choose $\{x\}\le0.07$. Hence our subset $T$ is formed by all the positive ordered pair $$(x,y)=(3+\{x\},4+\{y\})\text { with } \{x\}, \{y\}\le 0.07$$ Example.- $x=3.07$ and $y=4.07$ hence $7(0.07)+(0.07)^2=0.4949\lt \frac 12$ consequently we can choose arbitrarily $ \{x\}, \{y\}\le 0.07$.

Note.- There are other subsets without the integer part $3$ and $4$.

Answer 3

Form the first equation,

$$12\le xy<13$$ which is the area between to hyperbolas.

From the second equation,

$$\begin{cases}m\le x^2<m+1\\n\le y^2<n+1\end{cases},$$ with $m+n=25$. These are rectangular tiles that approximately fill a circle of radius $5$.

Looking at the solutions with a grapher, there are actually few tiles that intersect with the two hyperbolic stripes. They are such that

$$mn<13^2\land12^2<(m+1)(n+1).$$

There are $12$ solutions in the positives, from $m=7,n=18$ to $m=18,n=7$.

A tile is fully filled when

$$12^2\le mn\land(m+1)(n+1)<13^2$$ and this does not occur.

Hence the solutions are a few dented tiles.