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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:

  1. 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.

  2. 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.

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    $\begingroup$ all your previous sentence says that any number is sum of two cubes modulo 3. My last sentence is asking for a counter-example to my first statement which is " If n is not a sum of two integer cubes, then there must be a integer k such that n is not sum of two cubes modulo k".. $\endgroup$ – HumbleStudent Sep 17 at 3:56
  • $\begingroup$ I think you have the question down now.. $\endgroup$ – HumbleStudent Sep 17 at 4:06
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    $\begingroup$ Now that you have started using this notation, lets avoid $\eta$(n) because that is taken by something else regarding this problem. Another thing is that $\eta$(n) is not unique. $\endgroup$ – HumbleStudent Sep 17 at 4:42
  • $\begingroup$ any two numbers not congruent in every mod are not equal. $\endgroup$ – Roddy MacPhee Sep 17 at 14:50
  • $\begingroup$ If you are satisfied with one of the answers, Humble, I encourage you to "accept" it by clicking in the check mark next to it $\endgroup$ – Gerry Myerson Sep 19 at 1:32
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$$\left({17\over21}\right)^3+\left({37\over21}\right)^3=6$$ $x^3+y^3=6$ 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 $k$.

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$n=20$ is a counterexample (turns out to be the smallest positive one, assuming negative cubes allowed).

Theorem 2.1 in the paper [found by @Mason] states that $x^3+y^3\equiv 20\pmod{k}$ is solvable for each $k$. [UPDATE: Similarly to the answer by Gerry Myerson, we have $20=(1/7)^3+(19/7)^3$, so we're left to deal with $k$ a power of $7$, which is done using Hensel's lemma and the solution $x=6,y=0$ for $k=7$.]

It remains to show that $20$ is not a sum of two integer cubes. Here is an algorithmic recipe. Suppose $n=x^3+y^3=(x+y)(x^2-xy+y^2)$. The second factor is positive, hence $d=x+y$ is a positive divisor of $n$, and we have $n/d=3x^2-3dx+d^2$. This has an integer solution $x$ if and only if $n/d-d^2$ is a multiple of $3$ and the discriminant is a square, i.e. iff $(4n/d-d^2)/3$ is a square of an integer. Examining the divisors of $20$ this way, we're done.

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  • $\begingroup$ I don't see where Mason states that thing you were mentioning. I am sure existence of rational solution (like in example below) will do it for us. $\endgroup$ – HumbleStudent Sep 17 at 18:33
  • $\begingroup$ I deleted the original comment where I brought up this paper. It seemed like a duplication of data (now that it's in the answer above) and I want to make an effort to keep the website uncluttered. I really dislike having to read long comment threads hunting for valuable content. If you delete your comment above then I'll delete this one: It doesn't seem to add much value to this page, right? $\endgroup$ – Mason Sep 17 at 18:55

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