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Joachim König
Joachim König
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A simultaneous solution of $ax^2-c_1=b_1y^2$ and $ax^2-c_2=b_2z^2$ as demanded gives rise to an integral point on $b_1b_2T^2 = (ax^2-c_1)(ax^2-c2)$$b_1b_2T^2 = (ax^2-c_1)(ax^2-c_2)$, and since $a\ne 0$ and $c_1\ne c_2$, the degree-$4$ polynomial on the right is separable, meaning that this equation defines a curve of genus $1$. There are only finitely many integral points by Siegel's theorem, and since there were also only finitely many possible values of $a,b_1,b_2,c_1,c_2$, one has finiteness in total.

(Of course the gist of this argument is mentioned in some of the sources on simultaneous Pell equations given in the comments.)

A simultaneous solution of $ax^2-c_1=b_1y^2$ and $ax^2-c_2=b_2z^2$ as demanded gives rise to an integral point on $b_1b_2T^2 = (ax^2-c_1)(ax^2-c2)$, and since $a\ne 0$ and $c_1\ne c_2$, the degree-$4$ polynomial on the right is separable, meaning that this equation defines a curve of genus $1$. There are only finitely many integral points by Siegel's theorem, and since there were also only finitely many possible values of $a,b_1,b_2,c_1,c_2$, one has finiteness in total.

(Of course the gist of this argument is mentioned in some of the sources on simultaneous Pell equations given in the comments.)

A simultaneous solution of $ax^2-c_1=b_1y^2$ and $ax^2-c_2=b_2z^2$ as demanded gives rise to an integral point on $b_1b_2T^2 = (ax^2-c_1)(ax^2-c_2)$, and since $a\ne 0$ and $c_1\ne c_2$, the degree-$4$ polynomial on the right is separable, meaning that this equation defines a curve of genus $1$. There are only finitely many integral points by Siegel's theorem, and since there were also only finitely many possible values of $a,b_1,b_2,c_1,c_2$, one has finiteness in total.

(Of course the gist of this argument is mentioned in some of the sources on simultaneous Pell equations given in the comments.)

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Joachim König
Joachim König
  • 3k
  • 11
  • 27

A simultaneous solution of $ax^2-c_1=b_1y^2$ and $ax^2-c_2=b_2z^2$ as demanded gives rise to an integral point on $b_1b_2T^2 = (ax^2-c_1)(ax^2-c2)$, and since $a\ne 0$ and $c_1\ne c_2$, the degree-$4$ polynomial on the right is separable, meaning that this equation defines a curve of genus $1$. There are only finitely many integral points by Siegel's theorem, and since there were also only finitely many possible values of $a,b_1,b_2,c_1,c_2$, one has finiteness in total.

(Of course the gist of this argument is mentioned in some of the sources on simultaneous Pell equations given in the comments.)

A simultaneous solution of $ax^2-c_1=b_1y^2$ and $ax^2-c_2=b_2z^2$ as demanded gives rise to an integral point on $b_1b_2T^2 = (ax^2-c_1)(ax^2-c2)$, and since $a\ne 0$ and $c_1\ne c_2$, the degree-$4$ polynomial on the right is separable, meaning that this equation defines a curve of genus $1$. There are only finitely many integral points by Siegel's theorem, and since there were also only finitely many possible values of $a,b_1,b_2,c_1,c_2$, one has finiteness in total.

(Of course the gist of this argument is mentioned in the sources on simultaneous Pell equations given in the comments.)

A simultaneous solution of $ax^2-c_1=b_1y^2$ and $ax^2-c_2=b_2z^2$ as demanded gives rise to an integral point on $b_1b_2T^2 = (ax^2-c_1)(ax^2-c2)$, and since $a\ne 0$ and $c_1\ne c_2$, the degree-$4$ polynomial on the right is separable, meaning that this equation defines a curve of genus $1$. There are only finitely many integral points by Siegel's theorem, and since there were also only finitely many possible values of $a,b_1,b_2,c_1,c_2$, one has finiteness in total.

(Of course the gist of this argument is mentioned in some of the sources on simultaneous Pell equations given in the comments.)

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Joachim König
Joachim König
  • 3k
  • 11
  • 27

A simultaneous solution of $ax^2-c_1=b_1y^2$ and $ax^2-c_2=b_2z^2$ as demanded gives rise to an integral point on $b_1b_2T^2 = (ax^2-c_1)(ax^2-c2)$, and since $a\ne 0$ and $c_1\ne c_2$, the degree-$4$ polynomial on the right is separable, meaning that this equation defines a curve of genus $1$. There are only finitely many integral points by Siegel's theorem, and since there were also only finitely many possible values of $a,b_1,b_2,c_1,c_2$, one has finiteness in total.

(Of course the gist of this argument is mentioned in the sources on simultaneous Pell equations given in the comments.)