Let $\mathbb{N}=\{1,2,3,\ldots\}$ be the set of positive integers. For $n,k\in\mathbb{N}$ we define $$\text{Sol}(n,k) = \{(a,b,c)\in \mathbb{N}^3: |a^n + b^n - c^n| \leq k\}.$$ (The set $\text{Sol}(n,k)$ denotes the solutions of the inequality $|a^n + b^n - c^n| \leq k$ for fixed $n,k$.)

Moreover, for $j\in\mathbb{N}$ set $$\text{Inf}(j) =\{n\in \mathbb{N}: \text{Sol}(n,j) \text{ is infinite}\}.$$

Clearly, we have $\{1,2\}\subseteq \text{Inf}(j)$ for all $j\in\mathbb{N}$.


  1. Is there $j\in\mathbb{N}$ such that $\text{Inf}(j)\neq \{1,2\}$?
  2. Is there $j\in\mathbb{N}$ such that $\text{Inf}(j)$ is infinite, and what is the smallest such $j$?
  • 7
    $\begingroup$ To Q1: yes, trivially, fix $n$ and $a$ and set $j:=a^n$, then any $b=c$ gives you a solution. To Q2: yes, trivially, by same idea, $j=1$ (and $a=1$). To be honest, your question is not really appropriate for MO. $\endgroup$ – M.G. Aug 10 '15 at 8:31
  • 7
    $\begingroup$ @July That seems a bit harsh. Maybe a better comment is that the question only becomes interesting if one insists that there be infinitely many non-trivial solutions. Of course, then one needs to figure out what are the trivial cases. You've mentioned two of them. So let's insist that solutions have $a,b,c\ge2$ and $b\ne c$. $\endgroup$ – Joe Silverman Aug 10 '15 at 12:38
  • 2
    $\begingroup$ Apologies, it was not my intention to sound harsh. It was simply a quickly typed answer to the questions. I agree, a revised version of the question might indeed contain interesting, or even highly non-trivial math. Once some "obvious" classes of solutions are excluded, who knows, it might even turn out as hard as FLT itself. If a revised question is posted, I would also like to kindly ask the author to choose a different notation than Inf. $\endgroup$ – M.G. Aug 10 '15 at 13:03
  • 4
    $\begingroup$ Noam Elkies' work on Fermat near-misses: math.harvard.edu/~elkies/ferm.html $\endgroup$ – Gerry Myerson Aug 10 '15 at 13:34
  • 2
    $\begingroup$ An infinite family of integer solutions to $a^{3} + b^{3} - c^{3} = \pm 1$ was known to Ramanujan. $\endgroup$ – Jeremy Rouse Aug 10 '15 at 15:40

A 4-variable version of the infamous ABC Conjecture says the following: Let $a,b,c,d\in\mathbb{Z}$ be non-zero, satisfy $a+b+c+d=0$ and $\gcd(a,b,c,d)=1$, and no subsum of two or three of $a,b,c,d$ equal to $0$. Then for every $\epsilon>0$ there is a constant $K_\epsilon$ such that $$ \max\{|a|,|b|,|c|,|d|\} \le K_\epsilon \prod_{p\mid abcd} p^{1+\epsilon}. $$ Applying this to an expression of the form $a^n+b^n-c^n-k$ gives a very strong bound. Assuming that I haven't made an error (which is quite possible), I think that if $n\ge5$ (and assuming the ABCD conjecture), then for any $k$, the equation $$ a^n + b^n - c^n = k $$ has only finitely many solutions $a,b,c\in\mathbb{Z}$ with $|a|,|b|,|c|$ distinct and non-zero.

Actually, I guess the same (more or less) should be true for $n=4$. The point is that the surface $$ x^n+y^n-z^n=k $$ is of general type for $n\ge5$, so the Bombieri-Lang conjecture says that the solutions in rational numbers $(x,y,z)\in\mathbb{Q}^3$ are not Zariski dense (lie on a finite set of curves). This also follows from Vojta's conjecture. And for $n=4$, the equation defines an affine piece of a K3 surface, so Vojta's conjecture implies that the set of integer solutions likewise lies on a finite set of curves.

So your problem fits into a general framework, and for example, these statements are known if you replace $\mathbb Z$ by the ring of polynomials $\mathbb C[t]$. And as July suggests, you might want to read about how such problems are normally written, since your notation is not at all standard (and somewhat hard to parse).

  • $\begingroup$ I think you mean the n-conjecture (dropping pairwise coprimality). As stated, there is counterexample to the 4-variable ABC as shown here: mathoverflow.net/questions/185857/… $\endgroup$ – joro Aug 10 '15 at 15:17
  • $\begingroup$ For $a,b,c,d=[1, 3*x^2 - 3*x, x^3 - 3*x^2 + 3*x - 1, -x^3]$ the radical is $3x(x-1)$ and setting $x$ large power of small number makes it at most $3(x-1)$, while $|d| = x^3$. $\endgroup$ – joro Aug 10 '15 at 15:22
  • $\begingroup$ @joro Right, I guess I do want $a,b,c,d$ pairwise co-prime. Otherwise one has to apply Vojta's conjecture on a blow-up of $\mathbb P^3$ and figure out how the canonical bundle on the blow-up interacts with your ample divisor. $\endgroup$ – Joe Silverman Aug 10 '15 at 15:27
  • $\begingroup$ Silverman, the link in the first comment was confirmed to disprove pairwise coprime abcd by experts. Are you ready to bet that pairwise abcd still stands? ;-) $\endgroup$ – joro Aug 10 '15 at 15:35
  • 1
    $\begingroup$ @joro Ouch, you caught me. :) I tend to treat MO like a conversation about mathematics, so as per my comment to July, I generally assume that the statement "except for the usual and/or natural exceptions and counterexamples" is understood to be appended to statements. When writing articles or books, of course, this is not allowed, but it's a common convention when mathematicians talk. And anytime one is talking about solutions to Diophantine equations in dimension greater than 1, these days it's natural to omit a Zariski closed subset (then work down inductively by dimension). $\endgroup$ – Joe Silverman Aug 10 '15 at 17:44

Not counting the trivial solutions suggested by July, it is known that an integer cube or twice a cube is sum of three integer cubes in infinitely many ways via polynomial identities.

To get to the naturals, adjust the sign.

For $n=3$, one of the simplest is:

$$ (6kx^2)^3+(k(6x^3-1))^3-(k(6x^3+1))^3 = -2k^3$$


Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.