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The 4-distance problem is an open question(as far as I know it is still open) that asks if there exists a point P on the Euclidean plane such that its distances to all four points of a unit square are rational numbers.

I have reduced this problem(see here) to a family of elliptic curves $E_r$, parametrized by a rational number $r$. It is not hard to prove that as long as for any $r$, if one of the curves $E_r$ and $E_{1-r}$ has rank $0$, then the 4-distance problem has a negative answer, i.e. there is no such point exists.(but this is not necessary)

Roughly speaking, if the unit square is $A=(0,0),B=(0,1),C=(1,0),D=(1,1)$, then $E_r$ is a equation for points $P=(x,y)$ such that $x=r$ and $PA$ and $PB$ are rational numbers.

The family $E_r$ is represented by $y^2=x^3+(\frac{1}{r^2}-1)x^2-\frac{2}{r^2}x+\frac{1}{r^2}$. For $r=0, \frac{1}{2}, \frac{1}{3},\frac{1}{4},\frac{1}{5},\frac{2}{5}$, I have verified that at least one of the curve is rank $0$ so I conjecture this might be true for any $r$. Are there any methods to prove/disprove this fact?

The 4-distance problem is an open question(as far as I know it is still open) that asks if there exists a point P on the Euclidean plane such that its distances to all four points of a unit square are rational numbers.

I have related this problem(see here) to questions about a family of elliptic curves $E_r$, parametrized by a rational number $r$. One can easily prove that as long as for any $r$, if one of the curves $E_r$ and $E_{1-r}$ has rank $0$, then the 4-distance problem has a negative answer, i.e. there is no such point exists.(however this is not true according to Joachim‘s comment.)

Roughly speaking, if the unit square is $A=(0,0),B=(0,1),C=(1,0),D=(1,1)$, then $E_r$ is a equation for points $P=(x,y)$ such that $x=r$ and $PA$ and $PB$ are rational numbers.

The family $E_r$ is represented by $y^2=x^3+(\frac{1}{r^2}-1)x^2-\frac{2}{r^2}x+\frac{1}{r^2}$. For $r=0, \frac{1}{2}, \frac{1}{3},\frac{1}{4},\frac{1}{5},\frac{2}{5}$, I have verified that at least one of the curve is rank $0$. So I wonder for what set of rational numbers the rank of $E_r$ is greater than 0?

The 4-distance problem is an open question(as far as I know it is still open) that asks if there exists a point P on the Euclidean plane such that its distances to all four points of a unit square are rational numbers.

I have reduced this problem to a family of elliptic curves $E_r$, parametrized by a rational number $r$. It is not hard to prove that as long as for any $r$, if one of the curves $E_r$ and $E_{1-r}$ has rank $0$, then the 4-distance problem has a negative answer, i.e. there is no such point exists.(but this is not necessary)

Roughly speaking, if the unit square is $A=(0,0),B=(0,1),C=(1,0),D=(1,1)$, then $E_r$ is a equation for points $P=(x,y)$ such that $x=r$ and $PA$ and $PB$ are rational numbers.

The family $E_r$ is represented by $y^2=x^3+(\frac{1}{r^2}-1)x^2-\frac{2}{r^2}x+\frac{1}{r^2}$. For $r=0, \frac{1}{2}, \frac{1}{3},\frac{1}{4},\frac{1}{5},\frac{2}{5}$, I have verified that at least one of the curve is rank $0$ so I conjecture this might be true for any $r$. Are there any methods to prove/disprove this fact?

The 4-distance problem is an open question(as far as I know it is still open) that asks if there exists a point P on the Euclidean plane such that its distances to all four points of a unit square are rational numbers.

I have reduced this problem(see here) to a family of elliptic curves $E_r$, parametrized by a rational number $r$. It is not hard to prove that as long as for any $r$, if one of the curves $E_r$ and $E_{1-r}$ has rank $0$, then the 4-distance problem has a negative answer, i.e. there is no such point exists.(but this is not necessary)

Roughly speaking, if the unit square is $A=(0,0),B=(0,1),C=(1,0),D=(1,1)$, then $E_r$ is a equation for points $P=(x,y)$ such that $x=r$ and $PA$ and $PB$ are rational numbers.

The family $E_r$ is represented by $y^2=x^3+(\frac{1}{r^2}-1)x^2-\frac{2}{r^2}x+\frac{1}{r^2}$. For $r=0, \frac{1}{2}, \frac{1}{3},\frac{1}{4},\frac{1}{5},\frac{2}{5}$, I have verified that at least one of the curve is rank $0$ so I conjecture this might be true for any $r$. Are there any methods to prove/disprove this fact?

The 4-distance problem is an open question(as far as I know it is still open) that asks if there exists a point P on the Euclidean plane such that its distances to all four points of a unit square are rational numbers.

I have reduced this problem to a family of elliptic curves $E_r$, parametrized by a rational number $r$. It is not hard to prove that as long as for any $r$, if one of the curves $E_r$ and $E_{1-r}$ has rank $0$, then the 4-distance problem has a negative answer, i.e. there is no such point exists.(but this is not necessary)

Roughly speaking, if the unit square is $A=(0,0),B=(0,1),C=(1,0),D=(1,1)$, then $E_r$ is a equation for points $P=(x,y)$ such that $x=r$ and $PA$ and $PB$ are rational numbers.

The family $E_r$ is represented by $y^2=x^3-(\frac{1}{r^2}-1)x^2-\frac{2}{r^2}x+\frac{1}{r^2}$. For $r=0, \frac{1}{2}, \frac{1}{3},\frac{1}{4},\frac{1}{5},\frac{2}{5}$, I have verified that at least one of the curve is rank $0$ so I conjecture this might be true for any $r$. Are there any methods to prove/disprove this fact?

The 4-distance problem is an open question(as far as I know it is still open) that asks if there exists a point P on the Euclidean plane such that its distances to all four points of a unit square are rational numbers.

I have reduced this problem to a family of elliptic curves $E_r$, parametrized by a rational number $r$. It is not hard to prove that as long as for any $r$, if one of the curves $E_r$ and $E_{1-r}$ has rank $0$, then the 4-distance problem has a negative answer, i.e. there is no such point exists.(but this is not necessary)

Roughly speaking, if the unit square is $A=(0,0),B=(0,1),C=(1,0),D=(1,1)$, then $E_r$ is a equation for points $P=(x,y)$ such that $x=r$ and $PA$ and $PB$ are rational numbers.

The family $E_r$ is represented by $y^2=x^3+(\frac{1}{r^2}-1)x^2-\frac{2}{r^2}x+\frac{1}{r^2}$. For $r=0, \frac{1}{2}, \frac{1}{3},\frac{1}{4},\frac{1}{5},\frac{2}{5}$, I have verified that at least one of the curve is rank $0$ so I conjecture this might be true for any $r$. Are there any methods to prove/disprove this fact?