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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 $j\in\mathbb{N}$.

Questions:

  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 $j$?
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    $\begingroup$ 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.
    Commented Aug 10, 2015 at 8:31
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    $\begingroup$ @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
    Commented Aug 10, 2015 at 12:38
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    $\begingroup$ 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.
    Commented Aug 10, 2015 at 13:03
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    $\begingroup$ Noam Elkies' work on Fermat near-misses: math.harvard.edu/~elkies/ferm.html $\endgroup$
    – Gerry Myerson
    Commented Aug 10, 2015 at 13:34
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    $\begingroup$ An infinite family of integer solutions to $a^{3} + b^{3} - c^{3} = \pm 1$ was known to Ramanujan. $\endgroup$
    – Jeremy Rouse
    Commented Aug 10, 2015 at 15:40

2 Answers 2

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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).

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  • $\begingroup$ 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
    Commented Aug 10, 2015 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
    Commented Aug 10, 2015 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
    Commented Aug 10, 2015 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
    Commented Aug 10, 2015 at 15:35
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    $\begingroup$ @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
    Commented Aug 10, 2015 at 17:44
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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:

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

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