Are there solutions of $a^n+n+b^n=c^n$ for $n>2$?

Question

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As is well-known, the Diophantine equation $a^n+b^n=c^n$ has many solutions when $n=2$ (Pythagorean triples) but none when $n>2$ (the Fermat-Wiles Theorem). If one includes in the equation an extra term $\pm k$, yielding the equation:

$$a^n \pm {k} +b^n=c^n\qquad(1)$$

then obviously solutions of (1) can easily be found. What is perhaps surprising however is that there exist solutions in small values of $a,b,c,k$, for example:

$$5^2-1+5^2=7^2$$

$$6^3+1+8^3=9^3$$

$$5^3+2+6^3=7^3$$

$$9^3-1+10^3=12^3$$

$$13^5-12+16^5=17^5$$

A special case of (1) is when we add a requirement that $\pm k = n$, so that the equation becomes:

$$a^n+n+b^n=c^n\qquad(2)$$

It is easy to find solutions when $n=2$, for example:

$$3^2+2+5^2=6^2$$

$$5^2+2+13^2=14^2$$

In fact, given any Pythagorean triple of the form:

$$(2m+1)^2 + (2m^2+2m)^2=(2m^2+2m+1)^2$$

we have:

$$(2m+1)^2+2+(2m^2+2m+1)^2=(2m^2+2m+2)^2$$

Question Are there any solutions of (2) with $n>2$ and if so for which values of $n$ is this possible?

Answer 1

This is what I would try first.

Focus on $n=3$, the next case and (possibly therefore) next simplest. Rewrite the equation as:

$$3=c^3-a^3-b^3$$

The point of this is that it makes a link with the problem, that has been much studied, of expressing integers as a sum of the cubes of three (positive or negative) integers. For $n=3$, this problem has two very well-known solutions:

$$3=1^3+1^3+1^3$$

$$3=4^3+4^3-5^3$$

Unfortunately, these do not help us: what we need is to express $3$ as a sum of the cubes of one positive and two negative integers. To look for any further known solutions, I googled "integer as sum of 3 cubes", and rather to my surprise I found this: https://news.mit.edu/2021/solution-3-sum-cubes-puzzle-0311

with the solution:

$$3=569936821221962380720^3 + (−569936821113563493509)^3 + (−472715493453327032)^3$$

which can be rearranged as:

$$569936821113563493509^3 + 3 + 472715493453327032^3 = 569936821221962380720^3$$

Answer 2

According to Fermat's little theorem (https://en.wikipedia.org/wiki/Fermat%27s_little_theorem), if $2n+1$ is prime then $x^{2n}\equiv 0,1 \pmod{2n+1}$, and therefore $x^n\equiv 0,\pm1 \pmod{2n+1}$. Hence $a^n+n+b^n=c^n$ has no positive integer solution where $2n+1$ is a prime number with $n>4$.

For instance, $a^n+n+b^n=c^n$ for $n=5,6,8,9$ has no positive integer solution.