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Selmer's curve is the equation $3x^3 +4y^3 +5z^3=0$. This equation is famous for having non-trivial solutions in every completion of $\mathbb{Q}$ but only having the trivial solution in the rationals. This curve has been discussed on Mathoverflow before such as here and here. A nice proof that the curve always have solutions for all $p$-adics is in this writeup by Kevin Buzzard. I have two questions related to this curve.

Question I: How much worse can we get for cubics if the number of variable is increased. That is:

For every $n \geq 3$ is there a list of non-zero integers $a_1, a_2 \cdots a_n$ such that the equation $$a_1x_1^3 +a_2x_2^3 \cdots a_n x_n^3 =0$$ has solutions in every completion of $\mathbb{Q}$ but no non-trivial integer solutions?

Question II: can we make a family of such equations which is nested? That is is there a sequence of non-zero integers $a_1, a_2, a_3 \cdots $ such that for any $n \geq 3$ the equation $$a_1x_1^3 +a_2x_2^3 \cdots a_n x_n^3 =0$$ has solutions in every completion of $\mathbb{Q}$ but no non-trivial integer solutions? And if so, can we take $a_1=3$, $a_2=4$ and $a_3=5$ (that is using Selmer's curve as the start of our family).

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  • $\begingroup$ In your formulas the indices go $x_1,x_2,x^n$, some downstairs and some upstairs. I suppose that's a typo. $\endgroup$
    – Wojowu
    Commented Aug 1, 2021 at 16:53
  • $\begingroup$ @Wojowu Fixing now. Thanks. $\endgroup$
    – JoshuaZ
    Commented Aug 1, 2021 at 17:06

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The answer is no. Heath-Brown has shown every cubic form over the integers in at least 14 variables represents zero nontrivially. The Wikipedia article on Hasse principle contains references as well as related results for forms in higher odd degrees.

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