# If an integer is not sum of two cubes in integers, then the integer cannot be sum of two cubes modulo every integer.

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.

• 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".. Sep 17, 2019 at 3:56
• I think you have the question down now.. Sep 17, 2019 at 4:06
• 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. Sep 17, 2019 at 4:42
• any two numbers not congruent in every mod are not equal.
– user645636
Sep 17, 2019 at 14:50
• 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 Sep 19, 2019 at 1:32

$$\left({17\over21}\right)^3+\left({37\over21}\right)^3=6$$ $$x^3+y^3=6$$ 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 $$k$$.

$$n=20$$ 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.

• 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. Sep 17, 2019 at 18:33
• 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? Sep 17, 2019 at 18:55