First of all, I am no number theorist, so this question may be a little dummy.

The two squares theorem imply that $m = x^2 + y^2$ for some (possible zero) integer numbers $x,y$ iff $m$ factors as $m = a b^2$, where $a$ has no prime factor $\equiv 3 \mod 4$ (see, eg., these notes)

The following question seems natural to me: Is there such a simple characterization for numbers expressed as $m = x^n + y^n$ for $n > 2$? Since this seems related to Fermat's Last Theorem, maybe there is only partial (necessary or sufficient) conditions?

I am also interested in possible generalizations (characterizations of numbers that may be expressed as $m = x_1^n + \ldots + x_k^n$ ). What is proven/conjectured about this and is there any good reference with these statements?

  • IIRC there's some specific small integer (around $21$?) for which it's not known whether or not it's a sum of three cubes, or something like that. So it seems very little is known. Matiyasevich's theorem is also relevant. – Qiaochu Yuan Feb 26 '14 at 1:50
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    Some discussion (gives $33$ as a number which is not known to be a sum of three cubes): – Qiaochu Yuan Feb 26 '14 at 1:57
  • You may be thinking of $33$. Not $22$, which is excluded mod $9$. – Noam D. Elkies Feb 26 '14 at 1:57
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    Sorry, you answered as I was typing, and anyway you wrote $21$, not $22$. For the record, trying for $|a^3+b^3-c^3| = 21$ with $0 \leq a,b \leq 100$ finds $$ 21 = 16^3 - 14^3 - 11^3 = 28^3 + 85^3 - 86^3 = 49^3 + 97^3 - 101^3. $$ – Noam D. Elkies Feb 26 '14 at 4:30
  • The generalization to $m=x_1^n+\cdots+x_k^n$ has been studied, especially the question of, given $n$, how big $k$ must be to guarantee a solution for every $m$. The keyphrase is "Waring's problem". – Gerry Myerson Jul 19 at 0:00

This is not an answer but a series of comments.

Here's an obstacle stopping you from straightforwardly generalizing the classical based on the Gaussian integers to, say, sums of two cubes. The problem is that $x^3 + y^3$ is not the norm of an element of a number field. If it were, the number field would be $\mathbb{Q}(\zeta_6)$, where $\zeta_6$ is a primitive sixth root of unity, but this is a quadratic number field and the norm of, say, $x + y \zeta_6$ is actually the quadratic factor $x^2 - xy + y^2$ of $x^3 + y^3$ rather than the whole thing.

The ring where norms give $x^3 + y^3$ is instead the ring $\mathbb{Q}[\zeta]/(\zeta^3 + 1)$, which unfortunately is not a field, so it's unclear in what sense the "algebraic integers" $\mathbb{Z}[\zeta]/(\zeta^3 + 1)$ inside it can have a reasonable theory of prime factorization. Moreover, the general element of this ring has the form $x + y \zeta + z \zeta^2$, with three parameters, and the corresponding norm is

$$x^3 + y^3 + z^3 - 3xyz$$

so even if one has understood this norm one still has to take into account the additional constraint that $z = 0$.

It's also worth pointing out that $m = x^3 + y^3$ describes a curve of genus $1$ for almost all values of $m$ and IIRC it's already unknown whether there exists an algorithm taking as input such a curve and determining whether it has an integer point or not. Bjorn Poonen has written about related topics; see, for example, this expository article.

Some other relevant stuff:

  • So, even the case $x^3 + y^3$ seems an open problem... I am also interested in what would be the state-of-the-art of such characterizations (maybe for some specific $n$ or the "sharpest" necessary/sufficient conditions?) – Campello Feb 26 '14 at 19:34
  • As far as I know, the primes of the form $x^3 + y^3$ (rationally!) have been characterised by Zagier as a consequence of Gross-Zagier. – TKe Jul 18 at 17:19

Let $K/\mathbb{Q}$ be an abelian extension of degree $n$, and let $N_{K/\mathbb{Q}}(x)$ denote the norm of an element $x \in \mathcal{O}_K$. Then, with respect to a basis of $\mathcal{O}_K$ over $\mathbb{Z}$ as a $\mathbb{Z}$-lattice, one can interpret $N_{K/\mathbb{Q}}(x)$ as a homogenenous polynomial in $n$-variables, say $N(x) = N(x_1, \cdots, x_n)$. The form $N$ splits completely over $\overline{\mathbb{Q}}$ (in fact it already splits over $K$, since $K$ is abelian, thus Galois over $\mathbb{Q}$) and is irreducible over $\mathbb{Q}$. In this case it is possible to determine which integers $m$ may be written in the form $m = N(a_1, \cdots, a_n)$ for integers $a_1, \cdots, a_n$ in a similar fashion as in the case with binary quadratic forms (which are complete norm forms of degree 2). This is a consequence of class field theory. Of course, I cheated here, since like in the case of quadratic forms one has to take into account the situation when the class number exceeds one ($x^2 + y^2$ corresponds to $\mathbb{Q}(i)$, which has class number one). This can be generalized by replacing class field theory with say Chebotarev's density theorem.

In general, there does not seem to be an easy way to describe the numbers which can be written as $m = x^n + y^n$ for $n \geq 3$. It is conjectured (see for example: that for $n \geq 5$, for each integer $m$ such that $x^n + y^n = m$ has a solution in integers $x,y$, it has exactly two solutions, namely $(x,y)$ and $(y,x)$. In the language of my paper with C.L. Stewart (, this asserts that each $m$ that can be so represented is essentially represented.

In that joint paper with Stewart I mentioned above, we showed that for any binary form $F$ with integer coefficients, non-zero discriminant, and degree $d \geq 3$ there exists a positive number $C_F$ such that the number of integers $m$ with $|m| \leq X$ and for which the Thue equation $F(x,y) = m$ has a solution is asymptotic to $C_F X^{2/d}$. For $d \geq 3$ this set is too thin to have a description like in the quadratic case (any such set has density like $X (\log X)^{-a}$ for some explicit rational number $a > 0$). There are a lot of subtleties with this set. For example, even for the binary cubic form $xy(x+y)$, the existence of an infinite sequence $\{m_s\}$ of cube-free integers such that the number of solutions to the equation $xy(x+y) = m_s$ tends to infinity as $s \rightarrow \infty$ would imply that the Mordell-Weil rank of elliptic curves is unbounded, which seems to be an ever more fantastical situation.

I should add that the projective curve

$$\displaystyle x^n + y^n = mz^n$$

has complex multiplication, and hence one can in principle determine the set of rational points on the curve via Minhyong Kim's non-abelian Chabauty method, as shown by Coates and Kim in this paper:

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