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Are there infinitely many pairs of rational numbers $(a,b)$ such that $a^3+1$ is not a square in $\mathbf{Q}$, $b^3+2$ is not a square in $\mathbf{Q}$ and $b^3+2 = x^2(a^3+1)$ for some $x$ in $\mathbf{Q}$?

This question can be phrased also as follows:

Are there infinitely many rational numbers $(a,b)$ such that the extensions $\mathbf{Q}(\sqrt{a^3+1})$ and $\mathbf{Q}(\sqrt{b^3+2})$ are quadratic and equal?

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What motivated this question? It seems like this should have some relation to elliptic curves. –  Thom Tyrrell Nov 26 '12 at 22:35
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1 Answer 1

up vote 9 down vote accepted

Yes. Consider the two elliptic curves: $$7r^2 = a^3+1$$ and $$7s^2 = b^3+2.$$

One can check (using SAGE or MAGMA) that these two curves have infinitely many rational points (in fact they both have rank 1).

For rational points $(a,r)$ and $(b,s)$ on these curves we have that $a$ and $b$ satisfy the requirements of your question.

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I chose 7 because this was the smallest square-free positive integer for which both curves have positive rank. If you don't care that $a$ and $b$ vary, you could take $b=0$ and look at the curve $2r^2=a^3+1$, which has infinitely many rational points. –  Zach Nov 27 '12 at 18:38
    
Question: Do I understand correctly that we always have $\mathbf{Q}(\sqrt{a^3+1}) =\mathbf{Q}(\sqrt{7}) = \mathbf{Q}(\sqrt{b^3+2})$ if $(a,r)$ and $(b,s)$ lie on the curve? –  Harry Nov 27 '12 at 20:49
    
I like this answer a lot. It prompts the following question: Are there infinitely many square-free integers $n$ such that $nr^2=a^3+1$ has positive rank and $ns^2=b^3+2$ has positive rank? –  Harry Nov 27 '12 at 21:32
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