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This is a follow up to a question that I posed a few days ago here: The product of n radii in an ellipse

Suppose we have an ellipse $x^2/a^2 + y^2/b^2 = 1$ (centered at the origin). Let $n>4$. There are $n$ rays going out of the origin, at angles $0, 2\pi/n, 4\pi/n, 6\pi/n,...,2\pi(n-1)/n$. Let $(x_0,y_0),...,(x_{n-1},y_{n-1})$ be the points of intersection of the rays and the ellipse. The product from $k=0$ to $n-1$ of ${x_k}^2 + {y_k}^2$ is equal to one. Can $a$ and $b$ be rational if $a \neq b$?

It was discovered here that the answer is no, because Fermat's Last Theorem is true. Notice that the points of intersection of the rays and the ellipse, $(x_k,y_k)=x_k+{y_k}i$, generate a subfield of the complex numbers, since the ellipse is symmetric about the real line and the product of the points is one and the sum of the points is zero. (If one takes the set of all elements that can be expressed as a combination of additions and multiplications of the points of intersection of the rays and the ellipse, this set will be a subfield of the complex numbers, since zero and one are in this set.)

Conjecture: If $a,b$ are rational and $a \neq b$, the set of all complex numbers that can be expressed as a combination of additions and multiplications of the points of intersection of the rays and the ellipse, $(x_k,y_k)=x_k+{y_k}i$, cannot be a subfield of the complex numbers.

Prove this conjecture.

Craig

This is a follow up to a question that I posed a few days ago here: The product of n radii in an ellipse

Suppose we have an ellipse $x^2/a^2 + y^2/b^2 = 1$ (centered at the origin). Let $n>4$. There are $n$ rays going out of the origin, at angles $0, 2\pi/n, 4\pi/n, 6\pi/n,...,2\pi(n-1)/n$. Let $(x_0,y_0),...,(x_{n-1},y_{n-1})$ be the points of intersection of the rays and the ellipse. The product from $k=0$ to $n-1$ of ${x_k}^2 + {y_k}^2$ is equal to one. Can $a$ and $b$ be rational if $a \neq b$?

It was discovered here that the answer is no, because Fermat's Last Theorem is true. Notice that the points of intersection of the rays and the ellipse, $(x_k,y_k)=x_k+{y_k}i$, generate a subfield of the complex numbers, since the ellipse is symmetric about the real line and the product of the points is one and the sum of the points is zero. (If one takes the set of all elements that can be expressed as a combination of additions and multiplications of the points of intersection of the rays and the ellipse, this set will be a subfield of the complex numbers, since zero and one are in this set.)

Conjecture: If $a,b$ are rational and $a \neq b$, the set of all complex numbers that can be expressed as a combination of additions and multiplications of the points of intersection of the rays and the ellipse, $(x_k,y_k)=x_k+{y_k}i$, cannot be a subfield of the complex numbers.

Prove this conjecture.

Craig

This is a follow up to a question that I posed a few days ago here: The product of n radii in an ellipse

Suppose we have an ellipse $x^2/a^2 + y^2/b^2 = 1$ (centered at the origin). Let $n>4$. There are $n$ rays going out of the origin, at angles $0, 2\pi/n, 4\pi/n, 6\pi/n,...,2\pi(n-1)/n$. Let $(x_0,y_0),...,(x_{n-1},y_{n-1})$ be the points of intersection of the rays and the ellipse. The product from $k=0$ to $n-1$ of ${x_k}^2 + {y_k}^2$ is equal to one. Can $a$ and $b$ be rational if $a \neq b$?

It was discovered here that the answer is no, because Fermat's Last Theorem is true. Notice that the points of intersection of the rays and the ellipse, $(x_k,y_k)=x_k+{y_k}i$, generate a subfield of the complex numbers, since the ellipse is symmetric about the real line and the product of the points is one and the sum of the points is zero. (If one takes the set of all elements that can be expressed as a combination of additions and multiplications of the points on the ellipse, this set will be a subfield of the complex numbers, since zero and one are in this set.)

Conjecture: If $a,b$ are rational and $a \neq b$, the set of all elements that can be expressed as a combination of additions and multiplications of the points on the ellipse cannot be a subfield of the complex numbers.

Prove this conjecture.

Craig

This is a follow up to a question that I posed a few days ago here: The product of n radii in an ellipse

Suppose we have an ellipse $x^2/a^2 + y^2/b^2 = 1$ (centered at the origin). Let $n>4$. There are $n$ rays going out of the origin, at angles $0, 2\pi/n, 4\pi/n, 6\pi/n,...,2\pi(n-1)/n$. Let $(x_0,y_0),...,(x_{n-1},y_{n-1})$ be the points of intersection of the rays and the ellipse. The product from $k=0$ to $n-1$ of ${x_k}^2 + {y_k}^2$ is equal to one. Can $a$ and $b$ be rational if $a \neq b$?

It was discovered here that the answer is no, because Fermat's Last Theorem is true. Notice that the points of intersection of the rays and the ellipse, $(x_k,y_k)=x_k+{y_k}i$, generate a subfield of the complex numbers, since the ellipse is symmetric about the real line and the product of the points is one and the sum of the points is zero. (If one takes the set of all elements that can be expressed as a combination of additions and multiplications of the points of intersection of the rays and the ellipse, this set will be a subfield of the complex numbers, since zero and one are in this set.)

Conjecture: If $a,b$ are rational and $a \neq b$, the set of all complex numbers that can be expressed as a combination of additions and multiplications of the points of intersection of the rays and the ellipse, $(x_k,y_k)=x_k+{y_k}i$, cannot be a subfield of the complex numbers.

Prove this conjecture.

Craig

This is a follow up to a question that I posed a few days ago here: The product of n radii in an ellipse

Suppose we have an ellipse $x^2/a^2 + y^2/b^2 = 1$ (centered at the origin). Let $n>4$. There are $n$ rays going out of the origin, at angles $0, 2\pi/n, 4\pi/n, 6\pi/n,...,2\pi(n-1)/n$. Let $(x_0,y_0),...,(x_{n-1},y_{n-1})$ be the points of intersection of the rays and the ellipse. The product from $k=0$ to $n-1$ of ${x_k}^2 + {y_k}^2$ is equal to one. Can $a$ and $b$ be rational if $a \neq b$?

It was discovered here that the answer is no, provided that Fermat's Last Theorem is true. Notice that the points of intersection of the rays and the ellipse, $(x_k,y_k)=x_k+{y_k}i$, generate a subfield of the complex numbers, since the ellipse is symmetric about the real line and the product of the points is one and the sum of the points is zero.

Also notice that if $a$ and $b$ are rational with $a \neq b$ and $n$ is odd (I won't worry about when $n$ is even), all of the points $(x_1,y_1),...,(x_{n-1},y_{n-1})$ are irrational and $(x_0,y_0)=(a,0)$.

Question: What does this subfield (generated by the points of intersection of the rays and the ellipse) of the complex numbers look like? (What interesting things can be said about it?) And why does the assumption that $a$ and $b$ are rational with $a \neq b$ make such a subfield impossible to exist?

Craig

This is a follow up to a question that I posed a few days ago here: The product of n radii in an ellipse

Suppose we have an ellipse $x^2/a^2 + y^2/b^2 = 1$ (centered at the origin). Let $n>4$. There are $n$ rays going out of the origin, at angles $0, 2\pi/n, 4\pi/n, 6\pi/n,...,2\pi(n-1)/n$. Let $(x_0,y_0),...,(x_{n-1},y_{n-1})$ be the points of intersection of the rays and the ellipse. The product from $k=0$ to $n-1$ of ${x_k}^2 + {y_k}^2$ is equal to one. Can $a$ and $b$ be rational if $a \neq b$?

It was discovered here that the answer is no, because Fermat's Last Theorem is true. Notice that the points of intersection of the rays and the ellipse, $(x_k,y_k)=x_k+{y_k}i$, generate a subfield of the complex numbers, since the ellipse is symmetric about the real line and the product of the points is one and the sum of the points is zero. (If one takes the set of all elements that can be expressed as a combination of additions and multiplications of the points on the ellipse, this set will be a subfield of the complex numbers, since zero and one are in this set.)

Conjecture: If $a,b$ are rational and $a \neq b$, the set of all elements that can be expressed as a combination of additions and multiplications of the points on the ellipse cannot be a subfield of the complex numbers.

Prove this conjecture.

Craig