Say some integer $n$ is not sum of two cubes in the integers, then I want to show that there exists $k$ in positive integers such that $x^{3} + y^{3} \equiv n\pmod{k}$ is not solvable.
All I seem to know regarding even vicinity of such problems are following:
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If $n$ is not sum of two cubes in integers, then exists an integer $\theta(n)$ such that $n\theta(n)$ is sum of two cubes.
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Any number $n$ is sum of two cubes in integers if and only if following condition is satisfied:
$$\exists m \mid n ,\quad n^{1/3} \leq m \leq 4^{1/3} n^{1/3}$$ such that $( m^{2} – \frac{n}{m} ) = 3l$ and $(m^{2} – 4l)$ is a perfect square.
May be there is counterexample here, i.e. exists $n$ which is not sum of two cubes yet the congruence equation $x^{3} + y^{3} \equiv n\pmod{k}$ is solvable $\forall k \in\mathbb{N}$. Any help is appreciated.
Best Answer
$$\left({17\over21}\right)^3+\left({37\over21}\right)^3=6$$ $x^3+y^3=6$ has no solution in integers, positive, negative, or zero (exercise for the reader), but the displayed equation shows there's a solution to $x^3+y^3\equiv6\bmod k$ for every $k$ relatively prime to $21$.
Now $x^3+y^3\equiv6\bmod3$ has the solution $x=y=0$, and $x^3+y^3\equiv6\bmod7$ has the solution $x=3$, $y=0$.
This almost takes care of things, but $x^3+y^3\equiv6\bmod9$ has no solution, so this is really close-but-not-quite.
BUT here's one that works. $$\left({7\over3}\right)^3+\left({11\over3}\right)^3=62$$ $x^3+y^3=62$ has no solution in integers, positive, negative or zero, but the display shows there's a solution to $x^3+y^3\equiv62\bmod k$ for every $k$ relatively prime to $3$. And $2^3+0^3\equiv62\bmod{27}$, together with an application of Hensel's Lemma, takes care of values of $k$ that are powers of $3$. Then the Chinese Remainder Theorem gives solutions for all $k$.