# 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?
• What kind of weird messed-up language is your example written in? We're not on CodeGolf here, readability isn't considered a defect. Nov 26, 2020 at 17:11
• For $n = 4$ there are no solutions with $a, b \le 10^4$. I strongly suspect there are no solutions at all. (The folks at CodeGolf might be a bit unhappy if you challenge them to find an object which doesn't exist.) Nov 26, 2020 at 18:13
• Did you follow the links at that older question, card? In particular, the link to Noam Elkies' computations? Nov 26, 2020 at 22:24
• I think it's worth digging in, to get some idea of how unlikely one is to find any non-trivial examples of $a^n+b^n=c^n\pm1$ for $n\ge4$. Nov 26, 2020 at 22:42
• @JohnD.Cook The first line of the linked answer: “ A 4-variable version of the infamous ABC Conjecture says the following:” Nov 29, 2020 at 13:39

## 2 Answers

[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 $$13^5 + 16^5 = 17^5 + 12$$.

• "It is likely that there are no solutions at all." For $n \geq 4$, that is. For $10^3+9^3-12^3=1000+729-1728=1$.
– Joël
Nov 29, 2020 at 2:39
• Yes, I meant for $n \ge 4$. As Zhi-Wei Sun noted, there are infinitely many solutions for $n=3$. Editing. Nov 30, 2020 at 6:17

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 $$x,y,z=1,\ldots,1700$$.