How many "near-Fermat triples" are there?

Asked Aug 14 '15 at 09:17

Active Aug 14 '15 at 09:17

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Is the following statement true?

Claim

For every $n\in \mathbb N$, there is a constant $d$ such that there are infinitely many triples $a,b,c \in \mathbb N$ with

$$ | a^n + b^n - c^n | \le d $$

For $n=2$, we can take $d=0$ and use the Pythagorean triples. For $n>2$, there are no triples with $d=0$ (Fermat's last theorem).

asked Aug 14 '15 at 09:17

My intuition says it should be true for small $n$ because there are enough degrees of freedom. But probabilistically it looks quite bleak for large $n$. – user21820 Aug 14 '15 at 09:26
Related question recently asked on MO, http://mathoverflow.net/questions/214422/fermats-last-theorem-pm-k and in particular note math.harvard.edu/~elkies/ferm.html – Gerry Myerson Aug 14 '15 at 09:49
possible duplicate of Fermat's Last Theorem near misses? – Stefan4024 Aug 16 '15 at 16:21
@Stefan4024 That post doesn't seem to answer this specific question with $|\cdot |\le d$ – Aug 16 '15 at 21:07

How many "near-Fermat triples" are there?

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