Fermat's last theorem $\pm1$ I'm planning a challenge over on Code Golf.SE about integers $a, b, c \ge 0$ such that
$$a^n + b^n = c^n \pm 1$$
for a given integer $n > 2$. However, I'm interested in whether any non-trivial solutions to this exist for a given $n$. Here, I'm defining "non-trivial" solutions as triples $a, b, c$ such all three are unique and non-zero (i.e. to avoid $(a, 1, a)$ and $(a, 0, a)$, and related triples).
I've found this question which asks a related (and broader) question about the existence of such triples, and the accepted answer states

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.

However, this doesn't fully state whether there are a non-zero number of distinct, non-zero solutions.
This is a program which attempts to find such triples, with $0 \le a, b, c \le 100$, given an input $n$, but so far it hasn't found any for either $n = 4$ or $n = 5$, and it times out if you increase the upper limit by any significant amount.
Therefore, my question is:

*

*Can it be shown that, for all integers $n > 2$, the equation $a^n + b^n = c^n \pm 1$ has at least 1 non-trivial solution, for $a, b, c \ge 0$?

*If not, does expanding the range for $a, b, c$ to $\mathbb{Z}$ affect or change this?

 A: In a message "A conjecture related to Fermat's Last Theorem" sent to Number Theory List on Sep. 26, 2015, I wrote the following:
In 1936 K. Mahler discovered that
$$(9t^3+1)^3 + (9t^4)^3 - (9t^4+3t)^3 = 1.$$
Clearly,
$$|1^n+1^n-2^n| = 2^n-2\ \mbox{for every}\ n = 4,5,6,\ldots$$
and
$$13^5+16^5-17^5 = 371293+1048576-1419857 = 12 < 2^5-2.$$
Here I report my following conjecture which can be viewed as a
further refinement of Fermat's Last Theorem.
CONJECTURE (Sept. 24-25, 2015).
(i) For any integers $n > 3$ and $x,y,z > 0$ with $\{x,y\}\not= 
\{1,z\}$, we have
$$|x^n+y^n-z^n|\ge2^n-2,$$
unless $n = 5$, $\{x,y\} = \{13,16\}$ and $z = 17$.
(ii) For any integers $n > 3$ and $x,y,z > 0$ with $z\not\in\{x,y\}$,
there is a prime $p$ with
$$x^n+y^n < p < z^n\ \ \mbox{or}\ \  z^n < p < x^n+y^n, $$
unless $n = 5$, $\{x,y\} = \{13,16\}$ and $z = 17$.
(iii) For any integers $n > 3$, $x > y \ge0$ and $z > 0$ with $x\not=z$, there always exists a prime $p$ with
$$x^n-y^n < p < z^n\ \ \mbox{or}\ \ z^n < p < x^n-y^n.    $$
I have checked this new conjecture via Mathematica. For example, I
have verified part (i) of the conjecture for $n = 4,\ldots,10$ and $x,y,z=1,\ldots,1700$.
A: [EDITED] It is likely that there are no solutions at all for $n \ge 4$.  For $n \ge 5$ a solution would be a counterexample to the Lander, Parkin, and Selfridge conjecture. The best FLT "near miss" that I know of is $13^5 + 16^5 = 17^5 + 12$.
