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

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 $4$.
