Diophante A2856
February 27, 2023
Find all rational solutions to the system: \[\begin{align*} 2x \lfloor y \rfloor & = 2022 \\ 3\lfloor x \rfloor y & = 2023, \end{align*}\] where \(\lfloor u \rfloor\) denotes the integer part of \(u\). For example \(\lfloor 2.49 \rfloor =2\), \(\lfloor -1.876 \rfloor = -2\).
We claim that \(y \neq \lfloor y \rfloor\), otherwise \[ 3\lfloor x \rfloor y = 3\lfloor x \rfloor \lfloor y \rfloor = 2023 \] but that is not possible since \(2023\) is not divisible by 3.
Now we first consider the special case where \(x = \lfloor x \rfloor\). In this case, we have \[ 2x \lfloor y \rfloor = 2\lfloor x \rfloor \lfloor y \rfloor = 2022\] which implies \(\lfloor x \rfloor \lfloor y \rfloor = 1011 = 3 \times 337\).
Let \(y = \lfloor y \rfloor + \{y\}\) where \(0 < \{ y \} < 1\), then \[ 3\lfloor x \rfloor y = 3 \lfloor x \rfloor (\lfloor y \rfloor + \{y\}) = 3033 + 3 \lfloor x \rfloor \{y\} = 2023 \] \[ 1010 + 3 \lfloor x \rfloor \{y\} = 0\] Therefore \(- \lfloor x \rfloor = \frac{1010} { 3 \{ y \} } > 336\). The only two possible values for \(\lfloor x \rfloor\) are \(-337\) and \(-1011\).
If \(\lfloor x \rfloor = -337\), it follows that \(\lfloor y \rfloor = -3\), \(\{y\} = \frac{1010}{1011}\); If \(\lfloor x \rfloor = -1011\), it follows that \(\lfloor y \rfloor = -1\), \(\{y\} = \frac{1010}{3033}\). Hence there are at least two solutions, \((x,y) = (-337, -3 + \frac{1010}{1011})\) and \((x,y) = (-1011, -1 + \frac{1010}{3033})\) which both satisfy the given system.
Then we consider the general case where \(x \neq \lfloor x \rfloor\). We first consider the subcase where \(x > 0\) and \(y>0\). Since neither \(\lfloor x \rfloor\) nor \(\lfloor y \rfloor\) can be \(0\), we actually have \(x > 1\) and \(y>1\). From \(2x \lfloor y \rfloor = 2022\), we have \[2x (y-1) < 2022 < 2xy ;\] From \(3\lfloor x \rfloor y = 2023\), we have \[3(x-1) y < 2023 < 3xy. \] It follows that \[3(x-1) y - 2xy < 2023 - 2022 = 1 \implies y(x-3) < 1 \implies x < 4.\] Therefore \(\lfloor x \rfloor\) can only be \(1\), \(2\) or \(3\). If \(\lfloor x \rfloor = 1\), solving original system gives \((x,y) = (3/2, 2023/3)\). If \(\lfloor x \rfloor = 2\), then \(y = 2023/6\), \(\lfloor y \rfloor = 337\), but then \(x = 3\) which contradicts with \(\lfloor x \rfloor = 2\).
If \(\lfloor x \rfloor = 3\), then \(y = 2023/9\), \(\lfloor y \rfloor = 224\), but then \(x = 1011/224 > 4\). Hence there is only one solution \((x,y) = (3/2, 2023/3)\) when both \(x\) and \(y\) are positive.
Finally we consider the subcase where \(x < 0\) and \(y<0\). From \(2x \lfloor y \rfloor = 2022\), we have \[2x (y-1) > 2022 > 2xy ;\] From \(3\lfloor x \rfloor y = 2023\), we have \[3(x-1) y > 2023 > 3xy. \] It follows that \[3xy - 2x(y-1) < 2023 - 2022 = 1 \implies x(y+2) < 1 \implies y> \frac{1}{x} - 2.\]
We observe that \(x < -1\), otherwise \(\lfloor x \rfloor = -1\), \(y = -2023/3\), \(\lfloor y \rfloor = -675\), \(x = -1011/675 < -1\) which is a contradiction. Therefore \(\frac{1}{x} > -1\) and \(y > -3\). Therefore \(\lfloor y \rfloor\) can only be \(-1\), \(-2\) or \(-3\). If \(\lfloor y \rfloor = -1\), solving original system gives \((x,y) = (-1011, -2023/3033)\), which is one of the solutions found in the special case where \(x = \lfloor x \rfloor\). If \(\lfloor y \rfloor = -2\), solving original system gives \((x,y) = (-1011/2, -2023/1518)\), which is a valid solution. If \(\lfloor y \rfloor = -3\), solving original system gives \((x,y) = (-337, -2023/1011)\), which we note is another solution found in the special case where \(x = \lfloor x \rfloor\).
In summary, there are \(4\) rational solutions to the system: \[\begin{align*} 2x \lfloor y \rfloor & = 2022 \\ 3\lfloor x \rfloor y & = 2023, \end{align*}\] and they are \(\boxed{(-1011, -2023/3033), (-337, -2023/1011), (-1011/2, -2023/1518), (3/2, 2023/3)}\).