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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?

Adam Bailey
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Rajesh Bhowmick
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    Why do you think there is a simple answer to this? Without the $+n,$ it was a notoriously hard problem for $n>2,$ and I don't see why adding $n$ would make it any easier - we might find scattered solutions, by trial and error, I suppose, but a systemic approach to it seems difficult. – Thomas Andrews Mar 19 '24 at 16:19
  • For $n=2$, there are solutions, in fact, infinitely many solutions which I have shown. But For $n>2$ I'm unable to find any. So I put it here with some hope, that may be I will get some help or any solution with $n>2$. – Rajesh Bhowmick Mar 19 '24 at 16:32
  • For example, without $+n,$ we quickly can reduce the question to the question of odd prime $n$ and $n=2.$ And even then, the final result seems to require, as far as we know, a huge amount of highly abstract number theory. – Thomas Andrews Mar 19 '24 at 16:32
  • Right, but people will rarely volunteer to help solve a problem that might be harder than Fermat. Finding any solutions might just require a computer program doing trial by error, I suppose. But that isn't what you asked. "How did I solve" does night mean "find some solutions." – Thomas Andrews Mar 19 '24 at 16:34
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    In particular, you haven't even told us what you tried when looking for additional solutions. So we don't know where to start. – Thomas Andrews Mar 19 '24 at 16:36
  • I have given solutions for the case when $n=2$ for this particular equation, $a^n+n+b^n=c^n$, & that's what I've until now in my pocket to show. – Rajesh Bhowmick Mar 19 '24 at 16:45
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    Since $x^4\equiv 0,1 \pmod{16} ,a^4+4+b^4=c^4$ has no integer solution. – Tomita Mar 20 '24 at 02:05

2 Answers2

3

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:

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

Adam Bailey
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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.

Adam Bailey
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Tomita
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  • I've taken the liberty to editing your answer to try to clarify the reasoning. Please feel free to reverse my edit if it doesn't reflect your intentions. – Adam Bailey Mar 21 '24 at 11:28
  • @Adam, No problem, thanks. I'm not very good at English. – Tomita Mar 22 '24 at 00:13
  • There doesn't seem to be a simple way to show there cannot be a solution with $n=7$. In this case $2n+1=15$ is not prime, so it's worth trying to prove impossibility using modulus $4n+1=29$ which is prime, the 7th power residues mod $29$ being $0,1,12,17,28$. But this does not lead to a proof, since it allows a solution with $a^7 \equiv b^7 \equiv 17$ and $c^7 \equiv 12$ (note that $17+7+17 \equiv 12 \pmod{29}$. – Adam Bailey Mar 22 '24 at 14:40
  • @Adam, Only applicable when $2n+1$ is a prime number. – Tomita Mar 23 '24 at 01:40